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Chapter 32 Optical Images Conceptual Problems 1 • Can a virtual image be photographed? If so, give an example. If not, explain why. Determine the Concept Yes. Note that a virtual image is ″seen″ because the eye focuses the diverging rays to form a real image on the retina. For example, you can photograph the virtual image of yourself in a flat mirror and get a perfectly good picture. 2 • Suppose the x, y, and z axes of a 3-D coordinate system are painted a different color. One photograph is taken of the coordinate system and another is taken of its image in a plane mirror. Is it possible to tell that one of the photographs is of a mirror image? Or could both photographs be of the real coordinate system from different angles? Determine the Concept Yes; the coordinate system and its mirror image will have opposite handedness. That is, if the coordinate system is a right-handed coordinate system, then the mirror image will be a left-handed coordinate system, and vice versa. 3 • [SSM] True or False (a) The virtual image formed by a concave mirror is always smaller than the

object. (b) A concave mirror always forms a virtual image. (c) A convex mirror never forms a real image of a real object. (d) A concave mirror never forms an enlarged real image of an object. (a) False. The size of the virtual image formed by a concave mirror when the object is between the focal point and the vertex of the mirror depends on the distance of the object from the vertex and is always larger than the object. (b) False. When the object is outside the focal point, the image is real. (c) True. (d) False. When the object is between the center of curvature and the focal point, the image is enlarged and real.

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4 • An ant is crawling along the axis of a convex mirror that has a radius of curvature R. At what object distances, if any, will the mirror produce (a) an upright image, (b) a virtual image, (c) an image smaller than the object, and (d) an image larger than the object? Determine the Concept Let s be the object distance and f the focal length of the mirror. (a) If Rs 2

1< , the image is virtual, upright, and larger than the object. (b) If Rs 2

1< , the image is virtual, upright, and larger than the object. (c) If Rs > , the image is real, inverted, and smaller than the object. (d) If RsR <<2

1 , the image is real, inverted, and larger than the object. 5 • [SSM] An ant is crawling along the axis of a concave mirror that has a radius of curvature R. At what object distances, if any, will the mirror produce (a) an upright image, (b) a virtual image, (c) an image smaller than the object, and (d) an image larger than the object? Determine the Concept (a) The mirror will produce an upright image for all object distances. (b) The mirror will produce a virtual image for all object distances. (c) The mirror will produce an image that is that is smaller than the object for all object distances. (d) The mirror will never produce an enlarged image. 6 •• Convex mirrors are often used for rearview mirrors on cars and trucks to give a wide-angle view. ″Warning, objects are closer than they appear.″ is written below the mirrors. Yet, according to a ray diagram, the image distance for distant objects is much shorter than the object distance. Why then do they appear more distant? Determine the Concept They appear more distant because in a convex mirror the angular sizes of the images are smaller than they would be in a flat mirror. The

angular size of the image is Ls

y+'' , where y′ is the height of the image, s′ is the

distance between the image and the mirror, and L is the distance between the observer and the mirror. For a flat mirror, y′ = y and s′ = −s.

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We wish to prove that, for a convex mirror: Ls

yLs

y+

<+'' (1)

If equation (1) is valid, then:

yLs

yLs +>

+

''

or

yL

ys

yL

ys

+>+''

' (2)

Because the lateral magnification is

ss

yym ''

−== :

sy

sy=

'' (3)

Combining this with equation (2) yields:

yL

yL>

'⇒ yy <' (4)

It follows that, if y′ < y, then:

ss <'

To show that ss <' we begin with

the mirror equation:

fss11

'1

=+

Solving for s′ yields: fs

fss−

='

Because f is negative and s is positive: fs

fss

+=' ⇒ ss <'

Thus, we have proven that, if equation (1) is valid, then equation (4) is valid. To prove that equation 1 is valid, we merely follow this proof in reverse order. 7 • As an ant on the axis of a concave mirror crawls from a great distance to the focal point of a concave mirror, the image of the ant moves from (a) a great distance toward the focal point and is always real, (b) the focal point to a great distance from the mirror and is always real, (c) the focal point to the center of curvature of the mirror and is always real, (d) the focal point to a great distance from the mirror and changes from a real image to a virtual image.

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Picture the Problem The ray diagram shows three object positions 1, 2, and 3 as the moves from a great distance to the focal point F of a concave mirror. The real images corresponding to each of these object positions are labeled with the same numeral. )(b is correct.

8 • A kingfisher bird that is perched on a branch a few feet above the water is viewed by a scuba diver submerged beneath the surface of the water directly below the bird. Does the bird appear to the diver to be closer to or farther from the surface than the actual bird? Explain your answer using a ray diagram. Picture the Problem The diagram shows two rays (from the bundle of rays) of light refracted at the air-water interface. Because the index of refraction of water is greater than that of air, the rays are bent toward the normal. The diver will, therefore, think that the rays are diverging from a point above the bird and so the bird appears to be farther from the surface than it actually is.

AirWater

.

Imageof thebird

.

The bird

9 • An object is placed on the axis of a diverging lens whose focal length has a magnitude of 10 cm. The distance from the object to the lens is 40 cm. The image is (a) real, inverted, and diminished, (b) real, inverted, and enlarged, (c) virtual, inverted, and diminished, (d) virtual, upright, and diminished, (e) virtual, upright, and enlarged. Picture the Problem We can use a ray diagram to determine the general features of the image. In the diagram shown, the parallel ray and central ray have been used to locate the image.

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cm 40=s

cm 10=f

>

>

From the diagram, we see that the image is virtual (only one of the rays from the head of the object actually pass through the head of the image), upright, and diminished. )(d is correct.

10 •• If an object is placed between the focal point of a converging lens and the optical center of the lens, the image is (a) real, inverted, and enlarged, (b) virtual, upright, and diminished, (c) virtual, upright, and enlarged, (d) real, inverted, and diminished. Picture the Problem We can use a ray diagram to determine the general features of the image. In the diagram shown, the ray parallel to the principle axis and the central ray have been used to locate the image. Note that the object has been placed farther to the right than suggested in the problem statement. This was done so that the image would not be so large and so far to the left as to make the diagram excessively large.

F F'

>

>

From the diagram, we see that the image is virtual (neither ray from the head of the object passes through the head of the image), upright, and enlarged. )(c is

correct. 11 • A converging lens is made of glass that has an index of refraction of 1.6. When the lens is in air, its focal length is 30 cm. When the lens is immersed in water, its focal length (a) is greater than 30 cm, (b) is between zero and 30 cm, (c) is equal to 30 cm, (d) has a negative value.

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Picture the Problem We can apply the lens maker’s equation to the air-glass lens and to the water-glass lens to find the ratio of their focal lengths. Apply the lens maker’s equation to the air-glass interface:

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

21air

11)16.1(1rrf

Apply the lens maker’s equation to the water-glass interface:

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

21water

11)33.16.1(1rrf

Divide the first of these equations by the second to obtain: 2.2

11)33.16.1(

11)16.1(

21

21

air

water =

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=

rr

rrf

f

or airwater 2.2 ff = and )(a is correct. 12 • True or false: (a) A virtual image cannot be displayed on a screen. (b) A negative image distance implies that the image is virtual. (c) All rays parallel to the axis of a spherical mirror are reflected through a

single point. (d) A diverging lens cannot form a real image from a real object. (e) The image distance for a converging lens is always positive. (a) True. (b) True. (c) False. Where the rays intersect the axis of a spherical mirror depends on how far from the axis they are reflected from the mirror. (d) True. (e) False. The image distance for a virtual image is negative. 13 • Both the human eye and the digital camera work by forming real images on light-sensitive surfaces. The eye forms a real image on the retina and the camera forms a real image on a CCD array. Explain the difference between the ways in which these two systems accommodate. That is, the difference between how an eye adjusts and how a camera adjusts (or can be adjusted) to form a focused image for objects at both large and short distances from the camera.

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Determine the Concept The muscles in the eye change the thickness of the lens and thereby change the focal length of the lens to accommodate objects at different distances. A camera lens, on the other hand, has a fixed focal length so that focusing is accomplished by varying the distance between the lens and the light-sensitive surface. 14 • If an object is 25 cm in front of the naked eye of a farsighted person, an image (a) would be formed behind the retina if it were not for the fact that that the light is blocked (by the back of the eyeball) and the corrective contact lens should be convex, (b) would be formed behind the retina if it were not for the fact that that the light is blocked (by the back of the eyeball) and the corrective contact lens should be concave, (c) is formed in front of the retina and the corrective contact lens should be convex, (d) is formed in front of the retina and the corrective contact lens should be concave. Determine the Concept The eye muscles of a farsighted person lack the ability to shorten the focal length of the lens in the eye sufficiently to form an image on the retina of the eye. A convex lens (a lens that is thicker in the middle than at the circumference) will bring the image forward onto the retina. )(a is correct.

15 •• Explain the following statement: A microscope is an object magnifier, but a telescope is an angle magnifier. Hint: Take a look at the ray diagram for each magnifier and use it to explain the difference in adjectives. Determine the Concept The objective lens of a microscope ordinarily produces an image that is larger than the object being viewed, and that image is angularly magnified by the eyepiece. The objective lens of a telescope, on the other hand, ordinarily produces an image that is smaller than the object being viewed (see Figure 32-52), and that image is angularly magnified by the eyepiece. The telescope never produces a real image that is larger than the object. Estimation and Approximation 16 • Estimate the location and size of the image of your face when you hold a shiny new tablespoon a foot in front of your face and with the convex side toward you. Picture the Problem We can model the spoon as a convex spherical mirror and use the mirror equation and the lateral magnification equation to estimate the location and size of the image of your face. The mirror equation is: fs's

111=+

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Because 2rf = , where r is the radius of curvature of the mirror: rs's

211=+ ⇒

srrss'

2−=

Let the distance from your face to the surface of the spoon be 15 cm. If the radius of curvature of the spoon is 2 cm, then:

cm 1.1cm) 15(2cm .02

cm) (15 cm) 0.2(−≈

−=s'

or spoon thebehind cm 1.1about

The lateral magnification equation is:

ss'

yy'm −== ⇒ y

ssy ⎟⎠⎞

⎜⎝⎛−=

''

Assume your face is 25 cm long. Substitute numerical values and evaluate y′:

( )

cm 2

cm 25cm 15cm 1.1'

≈

⎟⎠⎞

⎜⎝⎛ −−=y

17 • Estimate the focal length of the ″mirror″ produced by the surface of the water in the reflection pools in front of the Lincoln Memorial on a still night. Determine the Concept The surface of the ″mirror″ produced by the surface of the water in the Lincoln Memorial reflecting pool is approximately spherical with a radius of curvature approximately equal to the radius of Earth. The focal length of a spherical mirror is half its radius of curvature. Hence Earth2

1 Rf = 18 •• Estimate the maximum value that could be obtained for the magnifying power of a simple magnifier, using Equation 32-20. Hint: Think about the smallest focal length lens that could be made from glass and still be used as a magnifier. Picture the Problem Because the focal length of a spherical lens depends on its radii of curvature and the magnification depends on the focal length, there is a practical upper limit to the magnification. Use Equation 32-20 to relate the magnification M of a simple magnifier to its focal length f:

fx

M np=

Use the lens-maker’s equation to relate the focal length of a lens to its radii of curvature and the index of refraction of the material from which it is constructed:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

21

1111rr

nf

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For a plano-convex lens, r2 = ∞. Hence: 1

11r

nf

−= ⇒

11

−=

nrf

Substitute in the expression for M and simplify to obtain:

( )1

np1r

xnM

−=

Note that the smallest reasonable value for r1 will maximize M.

A reasonable smallest value for the radius of a magnifier is 1.0 cm. Use this value and n = 1.5 to estimate Mmax:

( )( ) 13cm0.1

cm2515.1max =

−=M

Plane Mirrors 19 • [SSM] The image of the object point P in Figure 32-57 is viewed by an eye, as shown. Draw rays from the object point that reflect from the mirror and enter the eye. If the object point and the mirror are fixed in their locations, indicate the range of locations where the eye can be positioned and still see the image of the object point. Determine the Concept Rays from the source that are reflected by the mirror are shown in the following diagram. The reflected rays appear to diverge from the image. The eye can see the image if it is in the region between rays 1 and 2.

20 • You are 1.62 m tall and want to be able to see your full image in a vertical plane mirror. (a) What is the minimum height of the mirror that will meet your needs? (b) How far above the floor should the bottom of mirror in (a) be placed, assuming that the top of your head is 14 cm above your eye level? Use a ray diagram to explain your answer.

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Determine the Concept A ray diagram showing rays from your feet and the top of your head reaching your eyes is shown to the right. (a) The mirror must be half your height, i.e., cm 81

(b) The top of the minimum-height mirror must be 7 cm below the top of your head, or 155 cm above the floor. The bottom of the mirror should be placed cm 47 cm 81 cm 155 =− above

the floor.

21 •• [SSM] (a) Two plane mirrors make an angle of 90º. The light from a point object that is arbitrarily positioned in front of the mirrors produces images at three locations. For each image location, draw two rays from the object that, after one or two reflections, appear to come from the image location. (b) Two plane mirrors make an angle of 60º with each other. Draw a sketch to show the location of all the images formed of an object on the bisector of the angle between the mirrors. (c) Repeat Part (b) for an angle of 120º. Determine the Concept (a) Draw rays of light from the object (P) that satisfy the law of reflection at the two mirror surfaces. Three virtual images are formed, as shown in the following figure. The eye should be to the right and above the mirrors in order to see these images.

Virtual image Virtual image

PVirtual image

(b) The following diagram shows selected rays emanating from a point object (P) located on the bisector of the 60° angle between the mirrors. These rays form the two virtual images below the horizontal mirror. The construction details for the

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two virtual images behind the mirror that is at an angle of 60° with the horizontal mirror have been omitted due to the confusing detail their inclusion would add to the diagram.

Virtual image

Virtual image

Virtual image

Virtual image

P

(c) The following diagram shows selected rays emanating from a point object (P) on the bisector of the 120° angle between the mirrors. These rays form the two virtual images at the intersection of the dashed lines (extensions of the reflected rays):

Virtual image

Virtual image

P

22 •• Show that the mirror equation (Equation 32-4 where f = r/2) yields the correct image distance and magnification for a plane mirror. Picture the Problem We can use Equation 32-4 and the definition of the lateral magnification of an image to show that the mirror equation yields the correct image distance and magnification for a plane mirror.

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The mirror equation is: fs's

111=+

Because 2rf = : rs's

211=+

For a plane mirror, ∞=r . Hence:

011=+

s's⇒ ss' −=

The lateral magnification of the image is given by:

1+=−

−=−=ss

ss'M

23 •• When two plane mirrors are parallel, such as on opposite walls in a barber shop, multiple images arise because each image in one mirror serves as an object for the other mirror. An object is placed between parallel mirrors separated by 30 cm. The object is 10 cm in front of the left mirror and 20 cm in front of the right mirror. (a) Find the distance from the left mirror to the first four images in that mirror. (b) Find the distance from the right mirror to the first four images in that mirror. (c) Explain why each more distant image becomes fainter and fainter. Determine the Concept (a) The first image in the mirror on the left is 10 cm behind the mirror. The mirror on the right forms an image 20 cm behind that mirror or 50 cm from the left mirror. This image will result in a second image 50 cm behind the left mirror. The first image in the left mirror is 40 cm from the right mirror and forms an image 40 cm behind the right mirror or 70 cm from the left mirror. That image gives an image 70 cm behind the left mirror. The fourth image behind the left mirror is 110 cm behind that mirror. (a) For the mirror on the left, the images are located 10 cm, 50 cm, 70 cm, and 110 cm behind the mirror on the left (b) For the mirror on the right, the images are located 20 cm, 40 cm, 80 cm, and 100 cm behind the mirror on the right. (c) The successive images are dimmer because the light travels farther to form them. The intensity falls of inversely with the square of the distance the light travels. In addition, at each reflection a small percentage of the light intensity is lost. Real mirrors are not 100% reflecting. Spherical Mirrors 24 • A concave mirror has a radius of curvature equal to 24 cm. Use ray diagrams to locate the image, if it exists, for an object near the axis at distances of

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(a) 55 cm, (b) 24 cm, (c) 12 cm, and (d) 8.0 cm from the mirror. For each case, state whether the image is real or virtual; upright or inverted; and enlarged, reduced, or the same size as the object. Picture the Problem The easiest rays to use in locating the image are (1) the ray parallel to the principal axis and passes through the focal point of the mirror, (2) the ray that passes through the center of curvature of the spherical mirror and is reflected back on itself, and (3) the ray that passes through the focal point of the spherical mirror and is reflected parallel to the principal axis. We can use any two of these rays emanating from the top of the object to locate the image of the object. (a) The ray diagram follows. The image is real, inverted, and reduced.

CF

(b) The ray diagram is shown to the right. The image is real, inverted, and the same size as the object.

C F

Object

Image

(c) The ray diagram is shown to the right. The object is at the focal plane of the mirror. The emerging rays are parallel and do not form an image.

C F

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980

(d) The ray diagram is shown to the right. The image is virtual, erect, and enlarged.

C F

25 • [SSM] (a) Use the mirror equation (Equation 32-4 where f = r/2) to calculate the image distances for the object distances and mirror of Problem 24. (b) Calculate the magnification for each given object distance. Picture the Problem In describing the images, we must indicate where they are located, how large they are in relationship to the object, whether they are real or virtual, and whether they are upright or inverted. The object distance s, the image

distance s′, and the focal length of a mirror are related according to ,111fs's

=+

where rf 21= and r is the radius of curvature of the mirror. In this problem,

f = 12 cm because r is positive for a concave mirror. (a) Solve the mirror equation for s′:

fsfss−

=' where 2rf =

When s = 55 cm: ( )( )

cm15

cm 35.15cm12cm55cm55cm12

=

=−

=s'

When s = 24 cm: ( )( ) cm24

cm12cm24cm24cm12

=−

=s'

When s = 12 cm: ( )( ) undefined is

cm12cm12cm12cm12'

−=s

When s = 8.0 cm: ( )( )

m2.0

cm0.24cm12cm0.8cm0.8cm12'

−=

−=−

=s

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(b) The lateral magnification of the image is: s

s'm −=

When s = 55 cm, the lateral magnification of the image is:

28.0cm55

cm35.15−=−=−=

ss'm

When s = 24 cm, the lateral magnification of the image is:

0.1cm42cm24

−=−=−=ss'm

When s = 12 cm: undefined. is m

When s = 8.0 cm, the lateral magnification of the image is:

0.3cm8.0cm24'

=−

−=−=ssm

Remarks: These results are in excellent agreement with those obtained graphically in Problem 24. 26 • A convex mirror has a radius of curvature that has a magnitude equal to 24 cm. Use ray diagrams to locate the image, if it exists, for an object near the axis at distances of (a) 55 cm, (b) 24 cm, (c) 12 cm, (d) 8.0 cm, (e) 1.0 cm from the mirror. For each case, state whether the image is real or virtual; upright or inverted; and enlarged, reduced, or the same size as the object. Picture the Problem The easiest rays to use in locating the image are (1) the ray parallel to the principal axis and passes through the focal point of the mirror, (2) the ray that passes through the center of curvature of the spherical mirror and is reflected back on itself, and (3) the ray that passes through the focal point of the spherical mirror and is reflected parallel to the principal axis. We can use any two of these rays emanating from the top of the object to locate the image of the object. Note that, due to space limitations, the following ray diagrams are not to the same scale as those in Problem 24. (a) The ray diagram follows. The image is virtual, upright, and reduced.

CF

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(b) The ray diagram follows. The image is virtual, upright, and reduced.

CF

(c) The ray diagram follows. The image is virtual, upright and reduced.

CF

(d) The ray diagram follows. The image is virtual, upright, and reduced.

CF

(e) The ray diagram follows. The image is virtual, upright, and reduced.

CF

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27 • (a) Use the mirror equation (Equation 32-4 where f = r/2) to calculate the image distances for the object distances and mirror of Problem 26. (b) Calculate the magnification for each given object distance. Picture the Problem In describing the images, we must indicate where they are located, how large they are in relationship to the object, whether they are real or virtual, and whether they are upright or inverted. The object distance s, the image

distance s′, and the focal length of a mirror are related according to ,111fs's

=+

where rf 21= and r is the radius of curvature of the mirror. In this problem,

f = −12 cm because r is negative for a convex mirror. (a) Solve the mirror equation for s′: fs

fss'−

=

When s = 55 cm: ( )( )

( )cm9.9

cm85.9cm12cm55cm55cm12

−=

−=−−

−=s'

When s = 24 cm: ( )( )

( ) cm0.8cm12cm24cm24cm12

−=−−

−=s'

When s = 12 cm: ( )( )

( ) cm0.6cm12cm12cm12cm12

−=−−

−=s'

When s = 8.0 cm: ( )( )

( ) cm8.4cm12cm0.8cm0.8cm12

−=−−

−=s'

(b) The lateral magnification is given by:

ss'm −=

When s = 55 cm, the lateral magnification of the image is:

18.0cm55

cm85.9=

−−=m

When s = 24 cm, the lateral magnification of the image is:

33.0cm42cm0.8

=−

−=m

When s = 12 cm, the lateral magnification of the image is:

50.0cm21

cm0.6=

−−=m

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When s = 8.0 cm, the lateral magnification of the image is:

60.0cm8.0

cm80.4=

−−=m

Remarks: These results are in excellent agreement with those obtained graphically in Problem 26. 28 •• Use the mirror equation (Equation 32-4 where f = r/2) to prove that a convex mirror cannot form a real image of a real object, no matter where the object is placed. Picture the Problem We can solve the mirror equation for 1/s′ and then examine the implications of f < 0 and s > 0. Solve the mirror equation for 1/s′:

sffs

sfs'−

=−=111

For a convex mirror:

0<f

For s > 0, the numerator is positive and the denominator negative. Consequently:

0 01<⇒< s'

s'

29 • [SSM] A dentist wants a small mirror that will produce an upright image that has a magnification of 5.5 when the mirror is located 2.1 cm from a tooth. (a) Should the mirror be concave or convex? (b) What should the radius of curvature of the mirror be? Picture the Problem We can use the mirror equation and the definition of the lateral magnification to find the radius of curvature of the dentist’s mirror. (a) The mirror must be concave. A convex mirror always produces a diminished virtual image. (b) Express the mirror equation: rfs's

2111==+ ⇒

ssss'r+

='

2 (1)

The lateral magnification of the mirror is given by: s

s'm −= ⇒ mss' −=

Substitute for s′ in equation (1) to obtain:

mmsr

−−

=1

2

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Substitute numerical values and evaluate r:

( )( ) cm1.55.51

cm1.25.52=

−−

=r

30 •• Convex mirrors are used in many stores to provide a wide angle of surveillance for a reasonable mirror size. Your summer job is at a local convenience store that uses the mirror shown in Figure 32-58. This setup allows you (or the clerk) to survey the entire store when you are 5.0 m from the mirror. The mirror has a radius of curvature equal to 1.2 m. Assume all rays are paraxial. (a) If a customer is 10 m from the mirror, how far from the mirror is his image? (b) Is the image in front of or behind the mirror? (c) If the customer is 2.0 m tall, how tall is his image? Picture the Problem We can use the mirror equation and the relationship between the focal length of a mirror and its radius of curvature to find the location of the image. We can then use the definition of the lateral magnification of the mirror to find the height of the image formed in the mirror. (a) Solve the mirror equation for s′:

fsfss'−

= where rf 21=

Substitute for f and simplify to obtain: rs

rsrs

rss'−

=−

=22

121

Substitute numerical values and evaluate s′:

( )( )( ) ( ) cm6.56

m2.1m102m10m2.1

−=−−

−=s'

and mirror. thefrom cm57 is image the

(b) Because the image distance is negative:

mirror. thebehind is image the

(c) The lateral magnification of the mirror is given by Equation 32-5:

ss'

yy'm −== ⇒ y

ss'y' −=

Substitute numerical values and evaluate y′:

( ) cm11m0.2m10

m566.0=

−−=y'

31 •• A certain telescope uses a concave spherical mirror that has a radius equal to 8.0 m. Find the location and diameter of the image of the moon formed by this mirror. The moon has a diameter of 3.5 × 106 m and is 3.8 × 108 m from Earth.

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Picture the Problem We can use the mirror equation to locate the image formed in this mirror and the expression for the lateral magnification of the mirror to find the diameter of the image. Solve the mirror equation for the location of the image of the moon:

fsfss'−

=

Because rf 21= :

rsrs

rsrs

s'−

=−

=22

121

Substitute numerical values and evaluate s′:

( )( )( ) m0.4

m0.8m108.32m108.3m0.8

8

8

=−×

×=s'

Express the lateral magnification of the mirror: s

s'yy'm −== ⇒ y

ss'y' −=

Substitute numerical values and evaluate y′:

( )cm7.3

m105.3m108.3

m0.4 68

−=

××

−=y'

The 3.7-cm-diameter image is 4.0 m in front of the mirror.

32 •• A piece of a thin spherical shell that has a radius of curvature of 100 cm is silvered on both sides. The concave side of the piece forms a real image 75 cm from the piece. The piece is then turned around so that its convex side faces the object. The piece is moved so that the image is now 35 cm from the piece on the concave side. (a) How far was the piece moved? (b) Was it moved toward the object or away from the object? Picture the Problem We can use the mirror equation to find the focal length of the piece of the thin spherical shell and then apply it a second time to find the object position after the piece has been moved. (a) Solve the mirror equation for f to obtain:

ss'ss'f+

=

Substitute numerical values and evaluate f:

( )( ) cm86.42cm100cm75cm75cm100

=+

=f

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987

Solving the mirror equation for s yields: fs'

fs's−

=

With the piece turned around, f = −42.86 cm and s′ = − 35 cm. Substitute numerical values and evaluate s:

( )( )( ) cm191

cm86.42cm35cm35cm86.42

≈−−−

−−=s

The distance d the mirror moved is:

cm91cm100 cm191 =−=d

(b) The piece was moved away from the object. 33 •• Two light rays parallel to the optic axis of a concave mirror strike that mirror as shown in Figure 32-59. This mirror has a radius of curvature equal to 5.0 m. They then strike a small spherical mirror that is 2.0 m from the large mirror. The light rays finally meet at the vertex of the large mirror. Note: The small mirror is shown as planar, so as not to give away the answer, but it is not actually planar. (a) What is the radius of curvature of the small mirror? (b) Is that mirror convex or concave? Explain your answer. Picture the Problem Denote the larger mirror on the right using the numeral 1, the smaller mirror using the numeral 2, and the distance between the mirrors by d. Applying the mirror equation to the two mirrors will lead us to an expression for the radius of curvature of the small mirror. (a) Apply the mirror equation to mirror 1: 111

211r'ss

=+

Because ∞=1s :

11

21r's

= ⇒ 121

1 r's =

The image formed by mirror #1 serves as a virtual object for mirror 2. Apply the mirror equation to mirror 2 to obtain:

222

211r'ss

=+

or, because 121

12 rds'ds −=−= ,

22121

211r'srd

=+−

Because the image distance for mirror 2 is d:

21

212

2rdrd

=+−

Chapter 32

988

Substituting numerical values yields: ( ) 2

2m 0.2

1m 0.5m 0.22

2r

=+−

Finally, solve for r2 to obtain:

m3.12 −=r

(b) Because 022

12 <= rf , the small mirror is convex .

Images Formed by Refraction 34 •• A very long 1.75-cm-diameter glass rod has one end ground and polished to a convex spherical surface that has a 7.20-cm radius. The glass material has an index of refraction of 1.68. (a) A point object in air is on the axis of the rod and 30.0 cm from the spherical surface. Find the location of the image and state whether the image is real or virtual. (b) Repeat Part (a) for a point object in air, on the axis, and 5.00 cm from the spherical surface. Draw a ray diagram for each case. Picture the Problem We can use the equation for refraction at a single surface to find the images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions. Use the equation for refraction at a single surface to relate the image and object distances:

rnn

s'n

sn 1221 −

=+ (1)

Solving for s′ yields:

sn

rnnns'

112

2

−−

=

(a) Substitute numerical values (s = 30.0 cm, n1 = 1.00, n2 = 1.68 and r = 7.20 cm) and evaluate s′:

cm 27

0.3000.1

20.700.168.168.1

=−

−=s'

where the positive distance tells us that the image is 27 cm in back of the surface and is real.

Optical Images

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In the following ray diagram, the object is at P and the image at P′.

Air GlassCP P'

n =1.68

(b) Substitute numerical values (s = 5.00 cm, n1 = 1.00, n2 = 1.68 and r = 7.20 cm) and evaluate s′:

cm 16

00.500.1

20.700.168.168.1

−=−

−=s'

where the minus sign tells us that the image is 16 cm in front of the surface and is virtual.

In the following ray diagram, the object is at P and the image at P′.

Air GlassCPP'

n =1.68

35 • [SSM] A fish is 10 cm from the front surface of a spherical fish bowl of radius 20 cm. (a) How far behind the surface of the bowl does the fish appear to someone viewing the fish from in front of the bowl? (b) By what distance does the fish’s apparent location change (relative to the front surface of the bowl) when it swims away to 30 cm from the front surface? Picture the Problem The diagram shows two rays (from the bundle of rays) of light refracted at the water-air interface. Because the index of refraction of air is less than that of water, the rays are bent away from the normal. The fish will, therefore, appear to be closer than it actually is. We can use the equation for refraction at a single surface to find the distance s′. We’ll assume that the glass bowl is thin enough that we can ignore the refraction of the light passing through it.

Fish

WaterAir

Image ofFish

r

Chapter 32

990

(a) Use the equation for refraction at a single surface to relate the image and object distances:

rnn

s'n

sn 1221 −

=+

Solving for s′ yields:

sn

rnnns'

112

2

−−

=

Substitute numerical values (s = −10 cm, n1 = 1.33, n2 = 1.00 and r = 20 cm) and evaluate s′:

cm .68

cm 1033.1

cm 2033.100.1

00.1=

−−

−=s'

and the image is 8.6 cm from the front surface of the bowl.

(b) For s = 30 cm: cm 9.35

cm 3033.1

cm 2033.100.1

00.1=

−−

−=s'

The change in the fish’s apparent location is:

cm 27cm 6.8cm 5.93 ≈−

36 •• A very long 1.75-cm-diameter glass rod has one end ground and polished to a concave spherical surface that has a 7.20-cm radius. The glass material has an index of refraction of 1.68. A point object in air is on the axis of the rod and 15.0 cm from the spherical surface. Find the location of the image and state whether the image is real or virtual. Draw a ray diagram. Picture the Problem We can use the equation for refraction at a single surface to find the images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions. Use the equation for refraction at a single surface to relate the image and object distances:

rnn

s'n

sn 1221 −

=+ (1)

Solving for s′ yields:

sn

rnnns'

112

2

−−

=

Optical Images

991

Substitute numerical values (s = 15.0 cm, n1 = 1.00, n2 = 1.68, and r = −7.20 cm) and evaluate s′:

10

cm 0.1500.1

cm 20.700.168.1

68.1−=

−−

−=s'

where the minus sign tells us that the image is 10 cm in front of the surface of the rod and is virtual.

In the following ray diagram, the object is at P and the virtual image is at P′.

CAir Glass

P P'

n =1.68

37 •• [SSM] Repeat Problem 34 for when the glass rod and the object are immersed in water and (a) the object is 6.00 cm from the spherical surface, and (b) the object is 12.0 cm from the spherical surface. Picture the Problem We can use the equation for refraction at a single surface to find the images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions. Use the equation for refraction at a single surface to relate the image and object distances:

rnn

s'n

sn 1221 −

=+ (1)

Solving for s′ yields:

sn

rnnns'

112

2

−−

=

(a) Substitute numerical values (s = 6.00 cm, n1 = 1.33, n2 = 1.68, and r = 7.20 cm) and evaluate s′:

cm 7.9

cm 00.633.1

cm 20.733.168.1

68.1−=

−−

=s'

where the negative distance tells us that the image is 9.7 cm in front of the surface and is virtual.

Chapter 32

992

In the following ray diagram, the object is at P and the virtual image is at P′.

GlassCPP'

Water33.11 =n 68.12 =n

(b) Substitute numerical values (s = 12.0 cm, n1 = 1.33, n2 = 1.68, and r = 7.20 cm) and evaluate s′:

cm 27

cm 0.1233.1

cm 20.733.168.1

68.1−=

−−

=s'

where the negative distance tells us that the image is 27 cm in front of the surface and is virtual.

In the following ray diagram, the object is at P and the virtual image is at P′.

GlassCPP' Water

33.11 =n 68.12 =n

38 •• Repeat Problem 36 for when the glass rod and the object are immersed in water and the object is 20 cm from the spherical surface. Picture the Problem We can use the equation for refraction at a single surface to find the images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions. Use the equation for refraction at a single surface to relate the image and object distances:

rnn

s'n

sn 1221 −

=+ (1)

Solving for s′ yields:

sn

rnnns'

112

2

−−

=

Optical Images

993

Substitute numerical values (s = 20 cm, n1 = 1.33, n2 = 1.68, and r = −7.20 cm) and evaluate s′:

15

cm 0233.1

cm 20.733.168.168.1

−=−

−−

=s'

where the minus sign tells us that the image is 15 cm in front of the surface of the rod and is virtual

In the following ray diagram the object is at P and the virtual image is at P′.

CGlass

P P'Water

33.11 =n 68.12 =n

39 •• A rod that is 96.0-cm long is made of glass that has an index of refraction equal to 1.60. The rod has its ends ground to convex spherical surfaces that have radii equal to 8.00 cm and 16.0 cm. An object is in air on the long axis of the rod 20.0 cm from the end that has the 8.00-cm radius. (a) Find the image distance due to refraction at the 8.00-cm-radius surface. (b) Find the position of the final image due to refraction at both surfaces. (c) Is the final image real or virtual? Picture the Problem (a) We can use the equation for refraction at a single surface to find the images due to refraction at the ends of the glass rod. (b) The image formed by the refraction at the first surface will serve as the object for the second surface. (c) The sign of the final image distance will tell us whether the image is real or virtual. (a) Use the equation for refraction at a single surface to relate the image and object distances at the surface whose radius is 8.00 cm:

rnn

s'n

sn 1221 −

=+

Solving for s′ yields:

sn

rnnns'

112

2

−−

=

Chapter 32

994

Substitute numerical values (s = 20.0 cm, n1 = 1.00, n2 = 1.60, and r = 8.00 cm) and evaluate s′:

cm 64

cm 0.0200.1

cm 00.800.160.1

60.1=

−−

=s'

(b) The object for the second surface is 96 cm − 64 cm = 32 cm from the surface whose radius is −16.0 cm. Substitute numerical values (note that now n1 = 1.60 and n2 = 1.00) and evaluate s′:

cm 80

cm 2360.1

cm 0.1660.100.100.1

−=−

−−

=s'

(c) The final image is inside the rod and 96 cm − 80 cm = 16 cm from the surface whose radius of curvature is 8.00 cm and is virtual. 40 •• Repeat Problem 39 for an object in air on the axis of the glass rod 20.0 cm from the end that has the 16.0-cm radius. Picture the Problem (a) We can use the equation for refraction at a single surface to find the images due to refraction at the ends of the glass rod. (b) The image formed by the refraction at the first surface will serve as the object for the second surface. (c) The sign of the final image distance will tell us whether the image is real or virtual. (a) Use the equation for refraction at a single surface to relate the image and object distances at the surface that has a 16.0 cm radius:

rnn

s'n

sn 1221 −

=+

Solving for s′ yields:

sn

rnnns'

112

2

−−

=

Substitute numerical values (s = 20.0 cm, n1 = 1.00, n2 = 1.60 and r = 16.0 cm) evaluate s′:

cm 103.1cm 128cm 0.02

00.1cm 6.01

00.160.160.1

2×−=−=

−−

=s'

where the minus sign tells us that the image is 1.3 × 102 cm to the left of the surface whose radius of curvature is 16.0 cm.

Optical Images

995

(b) The object for the second surface is 96 cm + 128 cm = 224 cm from the surface whose radius of curvature is −8.00 cm. Substitute numerical values (note that now n1 = 1.60 and n2 = 1.00) and evaluate s′:

cm15

cm 7.14

cm 24260.1

cm .00860.100.1

00.1

=

=−

−−

=s'

(c) The final image is outside the rod, 15 cm from the end whose radius of curvature is 8.00 cm, and is real. Thin Lenses 41 • [SSM] A double concave lens that has an index of refraction equal to 1.45 has radii whose magnitudes are equal to 30.0 cm and 25.0 cm. An object is located 80.0 cm to the left of the lens. Find (a) the focal length of the lens, (b) the location of the image, and (c) the magnification of the image. (d) Is the image real or virtual? Is the image upright or inverted? Picture the Problem We can use the lens-maker’s equation to find the focal length of the lens and the thin-lens equation to locate the image. We can use

ss'm −= to find the lateral magnification of the image.

(a) The lens-maker’s equation is: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

21air

1111rrn

nf

where the numerals 1 and 2 denote the first and second surfaces, respectively.

Substitute numerical values to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−⎟⎠⎞

⎜⎝⎛ −=

cm0.251

cm0.3011

00.145.11

f

Solving for f yields: cm30cm3.30 −=−=f

(b) Use the thin-lens equation to relate the image and object distances:

fs's111

=+ ⇒fs

fss'−

=

Chapter 32

996

Substitute numerical values and evaluate s′:

( )( )( )

cm22

cm 98.21cm3.30cm0.80cm0.80cm3.30

−=

−=−−

−=s'

The image is 22 cm from the lens and on the same side of the lens as the object.

(c) The lateral magnification of the image is given by: s

s'm −=

Substitute numerical values and evaluate m:

0.27cm0.80

cm98.21=

−−=m

(d) upright. and virtual is image the0, and 0 Because >< ms' 42 • The following thin lenses are made of glass that has an index of refraction equal to 1.60. Make a sketch of each lens and find each focal length in air: (a) r1 = 20.0 cm and r2 = 10.0 cm, (b) r1 = 10.0 cm and r2 = 20.0 cm, and (c) r1 = –10.0 cm and r2 = –20.0 cm. Picture the Problem We can use the lens-maker’s equation to find the focal length of each of the lenses described in the problem statement. The lens-maker’s equation is: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

21air

1111rrn

nf

where the numerals 1 and 2 denote the first and second surfaces, respectively.

(a) For r1 = 20.0 cm and r2 = 10.0 cm: ⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛ −=

cm0.101

cm0.2011

00.160.11

f

and cm33−=f

A sketch of the lens is shown to the right:

Optical Images

997

(b) For r1 = 10.0 cm and r2 = 20.0 cm: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛ −=

cm0.201

cm0.1011

00.160.11

f

and cm33=f

A sketch of the lens is shown to the right:

(c) For r1 = −10.0 cm and r2 = −20.0 cm:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−

⎟⎠⎞

⎜⎝⎛ −=

cm0.201

cm0.1011

00.160.11

f

and cm33−=f

A sketch of the lens is shown to the right:

Remarks: Note that the lenses that are thicker on their axis than on their circumferences are positive (converging) lenses and those that are thinner on their axis are negative (diverging) lenses. 43 • The following four thin lenses are made of glass that has an index of refraction of 1.5. The radii given are magnitudes. Make a sketch of each lens and find each focal length in air: (a) double-convex that has radii of curvature equal to 15 cm and 26 cm, (b) plano-convex that has a radius of curvature equal to 15 cm, (c) double concave that has radii of curvature equal to 15 cm, and (d) plano-concave that has a radius of curvature equal to 26 cm. Picture the Problem We can use the lens-maker’s equation to find the focal length of each of the lenses. The lens-maker’s equation is: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

21air

1111rrn

nf

where the numerals 1 and 2 denote the first and second surfaces, respectively.

Chapter 32

998

(a) Because the lens is double convex, the radius of curvature of the second surface is negative. Hence r1 = 15 cm and r2 = −26 cm:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎠⎞

⎜⎝⎛ −=

cm261

cm1511

00.150.11

f

and cm19=f

A double convex lens is shown to the right:

(b) Because the lens is plano-convex, the radius of curvature of the second surface is negative. Hence r1 = ∞ and r2 = −15 cm:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−∞

⎟⎠⎞

⎜⎝⎛ −=

cm15111

00.150.11

f

and cm30=f

A plano-convex lens is shown to the right:

(c) Because the lens is double concave, the radius of curvature of its first surface is negative. Hence r1 = −15 cm and r2 = +15 cm:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎠⎞

⎜⎝⎛ −=

cm151

cm1511

00.150.11

f

and cm15−=f

A double concave lens is shown to the right:

(d) Because the lens is plano-concave, the radius of curvature of its second surface is positive. Hence r1 = ∞ and r2 = +26 cm:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

∞⎟⎠⎞

⎜⎝⎛ −=

cm26111

00.150.11

f

and cm52−=f

A plano-concave lens is shown to the right:

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999

44 • Find the focal length of a glass lens that has an index of refraction equal to 1.62, a concave surface that has a radius of curvature of magnitude 100 cm, and a convex surface that has a radius of curvature of magnitude 40.0 cm. Picture the Problem We can use the lens-maker’s equation to find the focal length of the lens. The lens-maker’s equation is: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

21air

1111rrn

nf

where the numerals 1 and 2 denote the first and second surfaces, respectively.

Substitute numerical values to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−−

⎟⎠⎞

⎜⎝⎛ −=

cm0.401

cm10011

00.162.11

f

Solving for f yields: m 1.1=f

45 •• [SSM] (a) An object that is 3.00 cm high is placed 25.0 cm in front of a thin lens that has a power equal to 10.0 D. Draw a ray diagram to find the position and the size of the image and check your results using the thin-lens equation. (b) Repeat Part (a) if the object is placed 20.0 cm in front of the lens. (c) Repeat Part (a) for an object placed 20.0 cm in front of a thin lens that has a power equal to –10.0 D. Picture the Problem We can find the focal length of each lens from the definition of the power P, in diopters, of a lens (P = 1/f ). The thin-lens equation can be applied to find the image distance for each lens and the size of the image

can be found from the magnification equation ss'

yy'm −== .

(a) The parallel and central rays were used to locate the image in the diagram shown below.

F

'F

>

>

The image is real, inverted, and diminished.

Chapter 32

1000

Solving the thin-lens equation for s′ yields:

fs's111

=+ ⇒ (1) fs

fss'−

=

Use the definition of the power of the lens to find its focal length:

cm 10.0 m100.0m0.10

111 ==== −P

f

Substitute numerical values and evaluate s′:

( )( ) cm7.16cm0.10cm0.25cm0.25cm0.10

=−

=s'

Use the lateral magnification equation to relate the height of the image y′ to the height y of the object and the image and object distances:

(2) yss'y'

ss'

yy'm −=⇒−==

Substitute numerical values and evaluate y′:

( ) cm00.2cm00.3cm0.25cm7.16

−=−=y'

cm7.16=s' , cm00.2−=y' . Because s′ > 0, the image is real, and because

y′/y = −0.67 cm, the image is inverted and diminished. These results confirm those obtained graphically. (b) The parallel and central rays were used to locate the image in the following diagram.

F

'F

>

>

The image is real and inverted and appears to be the same size as the object.

Use the definition of the power of the lens to find its focal length:

cm 10.0 m100.0m0.10

11 === −f

Substitute numerical values in equation (1) and evaluate s′:

( )( ) cm0.20cm0.10cm0.20cm0.20cm0.10

=−

=s'

Substitute numerical values in equation (2) and evaluate y′:

( ) cm00.3cm00.3cm0.20cm0.20

−=−=y'

Optical Images

1001

Because s′ > 0 and y′ = −3.00 cm, the image is real, inverted, and the same size as the object. These results confirm those obtained from the ray diagram.

(c) The parallel and central rays were used to locate the image in the following diagram.

F

'F

>

>

The image is virtual, upright, and diminished.

Use the definition of the power of the lens to find its focal length:

cm 0.01 m100.0m0.10

11 −=−=

−= −f

Substitute numerical values in equation (1) and evaluate s′:

( )( )( ) cm67.6

cm0.10cm0.20cm0.20cm0.10

−=−−

−=s'

Substitute numerical values in equation (2) and evaluate y′:

( ) cm00.1cm00.3cm0.20cm67.6

=−

−=y'

Because s′ < 0 and y′ = 1.00 cm, the image is virtual, erect, and about one-third the size of the object. These results are consistent with those obtained graphically. 46 •• The lens-maker’s equation has three design parameters. They consist of the index of refraction of the lens and the radii of curvature for its two surfaces. Thus, there are many ways to design a lens that has a particular focal length in air. Use the lens-maker’s equation to design three different thin converging lenses, each having a focal length of 27.0 cm and each made from glass that has an index of refraction of 1.60. Sketch each of your designs. Picture the Problem We can use the lens-maker’s equation to obtain a relationship between the two radii of curvature of the lenses we are to design.

Chapter 32

1002

For a thin lens of focal length 27.0 cm and index of refraction of 1.60:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

21

11160.1cm0.27

1rr

or

cm2.16111

21

=−rr

One solution is a plano-convex lens (one with a flat surface and a convex surface). Let r2 = ∞. Then

cm2.161 =r and ∞=2r .

Another design is a double convex lens (one with both surfaces convex and radii of curvature that are equal in magnitude) obtained by letting r2 = −r1. Then cm4.321 =r and

cm4.322 −=r .

A third possibility is a double convex lens with unequal curvature, e.g., let r2 = −12.0 cm. Then

cm89.61 −=r and

cm0.122 −=r .

47 •• Repeat Problem 46, but for a diverging lens that has a focal length in air of the same magnitude. Picture the Problem We can use the lens-maker’s equation to obtain a relationship between the two radii of curvature of the lenses we are to design. For a thin lens of focal length −27.0 cm and index of refraction of 1.60:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

− 21

11160.1cm0.27

1rr

or

cm2.16111

21 −=−

rr

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1003

One solution is a plano-concave lens (one with a flat surface and a concave surface), Let r2 = ∞. Then

cm2.161 −=r and ∞=2r .

Another design is a biconcave lens (one with both surfaces concave) by letting r2 = −r1. Then

cm4.321 −=r and

cm4.322 =r .

A third possibility is a lens with r2 = −8.10 cm. Then cm40.51 −=r

and cm10.82 −=r .

48 •• (a) What is meant by a negative object distance? Describe a specific situation in which a negative object distance can occur. (b) Find the image distance and the magnification for a thin lens in air when the object distance is –20 cm and the lens is a converging lens that has a focal length of 20 cm. Describe the image—is it virtual or real, upright or inverted? (c) Repeat Part (b) if the object distance is, instead, –10 cm, and the lens is diverging and has a focal length (magnitude) of 30 cm. Picture the Problem The parallel and central rays were used to locate the image in the diagram shown below. The power P of the lens, in diopters, can be found

from P = 1/f and the lateral magnification of the image from ss'm −= .

(a) A negative object distance means that the object is virtual; i.e., that extensions of light rays, and not the light rays themselves, converge on the object rather than diverge from it. A virtual object can occur in a two-lens system when converging rays from the first lens are incident on the second lens before they converge to form an image. (b) The thin-lens equation is: fs's

111=+ ⇒

fsfss'−

=

Chapter 32

1004

Substitute numerical values and evaluate s′:

( )( )( ) cm10

cm20cm20cm20cm20

=−−−

=s'

The lateral magnification is given by:

50.0cm20

cm 10'=

−−=−=

ssm

The parallel and central rays were used to locate the image in the following ray diagram:

F 'F>

>

s′ = 10 cm, m = 0.50. Because s′ > 0, the image is real, and because m > 0, the image is erect. In addition, it is one-half the size of the virtual object.

(c) Proceed as in Part (b) with s = −10 cm and f = −30 cm:

( )( )( ) cm15

cm30cm10cm10cm30

=−−−−−

=s'

and

5.1cm01

cm15=

−−=−=

ss'm

The parallel and central rays were used to locate the image in the following ray diagram:

F 'F

>>

Object Image

s′ > 0 and m = 1.5. Because s′ > 0, the image is real, and because m > 0, the image is erect. In addition, it is one and one-half times the size of the virtual object.

Optical Images

1005

49 •• [SSM] Two converging lenses, each having a focal length equal to 10 cm, are separated by 35 cm. An object is 20 cm to the left of the first lens. (a) Find the position of the final image using both a ray diagram and the thin-lens equation. (b) Is the final image real or virtual? Is the final image upright or inverted? (c) What is the overall lateral magnification of the final image? Picture the Problem We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens. (a) The parallel, central, and focal rays were used to locate the image formed by the first lens and the parallel and central rays to locate the image formed by the second lens.

Apply the thin-lens equation to express the location of the image formed by the first lens:

11

111 fs

sf's−

= (1)

Substitute numerical values and evaluate 's1 :

( )( ) cm20cm10cm20cm20cm10

1 =−

='s

Find the lateral magnification of the first image:

0.1cm20cm201

1 −=−=−=s'sm

Because the lenses are separated by 35 cm, the object distance for the second lens is 35 cm − 20 cm = 15 cm. Equation (1) applied to the second lens is:

22

222 fs

sf's−

=

Substitute numerical values and evaluate 's2 :

( )( ) cm30cm10cm15cm15cm10

2 =−

='s

and the final image is cm85 to the

right of the object.

Chapter 32

1006

Find the lateral magnification of the second image:

0.2cm51cm302

2 −=−=−=s'sm

(b) Because 0 2 >s' , the image is real and because m = m1m2 = 2.0, the image is erect and twice the size of the object.

(c) The overall lateral magnification of the image is the product of the magnifications of each image:

( )( ) 0.20.20.121 =−−== mmm

50 •• Repeat Problem 49 but with the second lens replaced by a diverging lens that has a focal length equal to −15 cm. Picture the Problem We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens. (a) The parallel, central, and focal rays were used to locate the image formed by the first lens and the parallel and central rays to locate the image formed by the second lens.

Apply the thin-lens equation to express the location of the image formed by the first lens:

11

111 fs

sf's−

= (1)

Substitute numerical values and evaluate 's1 :

( )( ) cm20cm10cm20cm20cm10

1 =−

='s

Find the lateral magnification of the first image:

0.1cm20cm201

1 −=−=−=s

s'm

Because the lenses are separated by 35 cm, the object distance for the second lens is 35 cm − 20 cm = 15 cm. Equation (1) applied to the second lens is:

22

222 fs

sf's−

=

Optical Images

1007

Substitute numerical values and evaluate 's2 :

( )( )( ) cm5.7

cm15cm15cm15cm15

2 −=−−

−='s

and the final image is cm48 to the

right of the object.

Find the lateral magnification of the second image:

50.0cm51

cm5.722 =

−−=−=

s'sm

(b) Because 0 2 <s' and m = m1m2 = −0.50, the image is virtual, inverted, and half the height of the object.

(c) The overall lateral magnification of the image is the product of the magnifications of each image:

( )( ) 50.050.00.121 −=−== mmm

51 •• (a) Show that to obtain a magnification of magnitude |m| using a converging thin lens of focal length f, the object distance must be equal to ( )fm 11 −+ . (b) You want to use a digital camera which has a lens whose focal length is 50.0 mm to take a picture of a person 1.75 m tall. How far from the camera lens should you have that person stand so that the image size on the light receiving sensors of your camera is 24.0 mm? Picture the Problem We can use the thin-lens equation and the definition of the lateral magnification to derive the result. (a) Start with the thin-lens equation:

fs's111

=+

Express the magnitude of the lateral magnification of the image and solve for s′: (Note that this ensures that s′ is positive for this situation, as it must be because the image is real.)

ss'm −= ⇒ sms' =

Substitute for s′ to obtain: fsms111

=+ ⇒ ( )fms 11 −+=

(b) The magnitude of the magnification m is: 0137.0

m75.1mm24.0

==−=yy'm

Chapter 32

1008

Substitute numerical values and evaluate s:

( )( ) m70.30137.0

mm0.5010137.0=

+=s

52 •• A converging lens has a focal length of 12.0 cm. (a) Using a spreadsheet program or graphing calculator, plot the image distance as a function of the object distance, for object distances ranging from 1.10f to 10.0f where f is the focal length. (b) On the same graph used in Part (a), but using a different y axis, plot the magnification of the lens as a function of the object distance. (c) What type of image is produced for this range of object distances—real or virtual, upright or inverted? (d) Discuss the significance of any asymptotic limits your graphs have. Picture the Problem We can plot the first graph by solving the thin-lens equation for the image distance s′ and the second graph by using the definition of the magnification of the image. (a) and (b) Solve the thin-lens equation for s′ to obtain: fs

fss'−

=

The magnification of the image is given by: s

s'm −=

A spreadsheet program to calculate s′as a function of s is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Content/Formula Algebraic Form B1 12 f A4 13.2 s A5 A4 + 1 s + Δs B4 $B$1*A4/(A4 − $B$1)

fsfs−

C5 −B4/A4 ss'

−

A B C 1 f= 12 cm 2 3 s s' m 4 13.2 132.00 −10.00 5 14.2 77.45 −5.45 6 15.2 57.00 −3.75 7 16.2 46.29 −2.86 8 17.2 39.69 −2.31 9 18.2 35.23 −1.94

Optical Images

1009

108 117.2 13.37 −0.11 109 118.2 13.36 −0.11 110 119.2 13.34 −0.11 111 120.2 13.33 −0.11

A graph of s′ and m as functions of s follows.

0

20

40

60

80

100

120

140

0 20 40 60 80 100 120

s (cm)

s ' (c

m)

-12.0

-10.0

-8.0

-6.0

-4.0

-2.0

0.0

m

s' m

(c) The images are real and inverted for this range of object distances. (d) The equations for the horizontal and vertical asymptotes of the graph of s′ versus s are s′ = f and s = f, respectively. These indicate that as the object moves away from the lens image distance the image moves toward the first focal point, and that as the object moves toward the second focal point the image distance becomes large without limit. The equations for the horizontal and vertical asymptotes of the graph of m versus s are m = 0 and s = f, respectively. These indicate that as the object moves away from the lens, the image size of the image approaches zero, and that as the object moves toward the second focal point the image becomes large without limit. 53 •• A converging lens has a focal length of 12.0 cm. (a) Using a spreadsheet program or graphing calculator, plot the image distance as a function of the object distance, for object distances ranging from 0.010f to s = 0.90f, where f is the focal length. (b) On the same graph used in Part (a), but using a different y axis, plot the magnification of the lens as a function of the object distance. (c) What type of image is produced for this range of object distances—real or virtual, upright or inverted? (d) Discuss the significance of any asymptotic limits your graphs have.

Chapter 32

1010

Picture the Problem We can plot the first graph by solving the thin-lens equation for the image distance s′ and the second graph by using the definition of the magnification of the image. (a) and (b) Solve the thin-lens equation for s′ to obtain: fs

fss'−

=

The magnification of the image is given by: s

s'm −=

A spreadsheet program to calculate s′as a function of s is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Content/Formula Algebraic Form B1 12 f A4 0.12 s A5 A4 + 0.1 s + Δs B4 $B$1*A4/(A4 − $B$1)

fsfs−

C5 −B4/A4 ss'

−

A B C 1 f= 12 cm 2 3 s s' m 4 0.12 −0.12 1.01 5 0.22 −0.22 1.02 6 0.32 −0.33 1.03 7 0.42 −0.44 1.04 8 0.52 −0.54 1.05 9 0.62 −0.65 1.05

108 10.52 −85.30 8.11 109 10.62 −92.35 8.70 110 10.72 −100.50 9.37 111 10.82 −110.03 10.17

Optical Images

1011

A graph of s′ and m as functions of s follows.

-110

-90

-70

-50

-30

-10

10

0 2 4 6 8 10 12

s (cm)

s' (c

m)

0.0

2.0

4.0

6.0

8.0

10.0

12.0

m

s'm

(c) The images are virtual and erect for this range of object distances. (d) The equation for the vertical asymptote of the graph of s′ versus s is f = s. This indicates that, as the object moves toward the second focal point, the magnitude of the image distance becomes large without limit. The equation for the vertical asymptote of the graph of m versus s is s = f. This indicates that, as the object moves toward the second focal point, the image becomes large without limit. In addition, as s approaches zero, s′ approaches negative infinity and m approaches 1, so as the object moves toward the lens the image becomes the same size as the object and the magnitude of the image distance increases without limit 54 •• An object is 15.0 cm in front of a converging lens that has a focal length equal to 15.0 cm. A second converging lens that also has a focal length equal to15.0 cm is located 20.0 cm in back of the first. (a) Find the location of the final image and describe its properties (for example, real and inverted) and (b) draw a ray diagram to corroborate your answers to Part (a). Picture the Problem We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens. (a) Apply the thin-lens equation to express the location of the image formed by the first lens:

11

111 fs

sf's−

= (1)

Chapter 32

1012

Substitute numerical values and evaluate 's1 :

( )( )∞=

−=

cm15.0cm15.0cm0.15cm0.15

1's

i.e., no image is formed by the first lens.

With 's1 = ∞, the thin-lens equation applied to the second lens becomes:

22

11f's

= ⇒ cm0.1522 == f's

The overall magnification of the two-lens system is the magnification of the second lens:

00.1cm 15.0cm 0.15

1

22 −=−=−==

s'smm

The final image is 50 cm from the object, real, inverted, and the same size as the object. (b) A corroborating ray diagram is shown below:

Lens 1 Lens 2

'1F 1F 2F '2F

> >> >

55 •• [SSM] An object is 15.0 cm in front of a converging lens that has a focal length equal to 15.0 cm. A diverging lens that has a focal length whose magnitude is equal to 15.0 cm is located 20.0 cm in back of the first. (a) Find the location of the final image and describe its properties (for example, real and inverted) and (b) draw a ray diagram to corroborate your answers to Part (a). Picture the Problem We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens. Apply the thin-lens equation to express the location of the image formed by the first lens:

11

111 fs

sf's−

= (1)

Substitute numerical values and evaluate 's1 :

( )( )∞=

−=

cm15.0cm15.0cm0.15cm0.15

1's

Optical Images

1013

With 1s' = ∞, the thin-lens equation applied to the second lens becomes:

22

11f's

= ⇒ cm0.1522 −== f's

The overall magnification of the two-lens system is the magnification of the second lens:

00.1cm 15.0cm 0.15

1

22 =

−−=−==

s'smm

The final image is 20 cm from the object, virtual, erect, and the same size as the object. A corroborating ray diagram follows:

Lens 1 Lens 2

'1F 1F 2F '2F

>>

>>

56 ••• In a convenient form of the thin-lens equation used by Newton, the object and image distances x and x′ are measured from the focal points F and F’, and not from the center of the lens. (a) Indicate x and x′ on a sketch of a lens and show that if x = s – f and x′ = s′ – f, the thin-lens equation (Equation 32-12) can be rewritten as 2fxx' = . (b) Show that the lateral magnification is given by m = –x′/f = –f/x. Picture the Problem We can substitute x = s − f and x′ = s′ − f in the thin-lens equation and the equation for the lateral magnification of an image to obtain Newton’s equations. (a) The variables x, f, s ,and s′ are shown in the following sketch:

'F F

>>

x x'ff

s s'

Object Image

Chapter 32

1014

Express the thin-lens equation: fs's111

=+

If x = s − f and x′ = s′ − f, then:

ffx'fx111

=+

++

Expand this expression to obtain: ( ) ( )( )

2

2fx'fxfxx'

fx'fxfxx'f+++=

++=++

or, simplifying, 2fxx' = (1)

(b) The lateral magnification is: s

s'm −=

or, because x = s − f and x′ = s′ − f,

fxfx'm

++

−=

Solve equation (1) for x:

x'fx

2

=

Substitute for x and simplify to obtain: ( ) f

x'

x'x'fffx'

fx'f

fx'm −=++

−=+

+−= 2

The lateral magnification is also given by:

fxfxm

++

−='

From equation (1) we have: x

fx2

'=

Substitute to obtain:

xf

xfx

xff

fx

fxf

m −=⎟⎠⎞

⎜⎝⎛ +

⎟⎠⎞

⎜⎝⎛ +

−=+

+−=

1

12

Optical Images

1015

57 ••• [SSM] In Bessel’s method for finding the focal length f of a lens, an object and a screen are separated by distance L, where L > 4f. It is then possible to place the lens at either of two locations, both between the object and the screen, so that there is an image of the object on the screen, in one case magnified and in the other case reduced. Show that if the distance between these two lens locations is D, then the focal length is given by ( ) LDLf 22

41 −= . Hint: Refer to Figure

32-60. Picture the Problem The ray diagram shows the two lens positions and the corresponding image and object distances (denoted by the numerals 1 and 2). We can use the thin-lens equation to relate the two sets of image and object distances to the focal length of the lens and then use the hint to express the relationships between these distances and the distances D and L to eliminate s1, s1′, s2, and s2′ and obtain an expression relating f, D, and L. Relate the image and object distances for the two lens positions to the focal length of the lens:

f'ss111

11

=+ and f'ss111

22

=+

Solve for f to obtain: 'ss

'ss'ss

'ssf22

22

11

11

+=

+= (1)

The distances D and L can be expressed in terms of the image and object distances:

'ss'ssL 2211 +=+= and

's'sssD 2112 −=−=

Substitute for the sums of the image and object distances in equation (1) to obtain:

L'ss

L'ss

f 2211 ==

From the hint:

'ss 21 = and 21 s's =

Hence L = s1 + s2 and: 12sDL =− and 22sDL =+

Take the product of L − D and L + D to obtain:

( )( )'ssss

DLDLDL

1121

22

44 ==−=+−

From the thin-lens equation:

fL'srsss 44 1121 ==

Substitute to obtain: 224 DLfL −= ⇒

LDLf

4

22 −=

Chapter 32

1016

58 •• You are working for an optician during the summer. The optician needs to measure an unknown focal length and you suggest using Bessel’s method (see Problem 56) which you used during a physics lab. You set the object-to-image distance at 1.70 m. The lens position is adjusted to get a sharp image on the screen. A second sharp image is found when the lens is moved a distance of 72 cm from its first location. (a) Sketch the ray diagram for the two locations. (b) Find the focal length of the lens using Bessel’s method. (c) What are the two locations of the lens with respect to the object? (d) What are the magnifications of the images when the lens is in the two locations? Picture the Problem We can use results obtained in Problem 57 to find the focal length of the lens and the two locations of the lens with respect to the object. Let 1 refer to the first lens position and 2 to the second lens position. (a) The distances defined in Problem 57 have been identified in the following ray diagram. Note that, for clarity, the two images have been slightly offset from the plane of the screen.

1st lens position 2nd lens position

L

f

Screen

D

> >

>>

(b) From Problem 57 we have: L

DLf4

22 −=

Substitute numerical values and evaluate f:

( ) ( )( ) cm35

cm1704cm72cm170 22

=−

=f

(c) Solve the thin-lens equation for the image distance to obtain:

sffss−

=' (1)

Optical Images

1017

In Problem 57 it was established that:

12sDL =− and 22sDL =+

Solve for s1 and s2: 21

DLs −= and

22DLs +

=

Substitute numerical values and evaluate s1 and s2:

cm492

cm72cm1701 =

−=s

and

cm1212

cm72cm1702 =

+=s

(d) The magnification when the lens was in its first position is given by: 1

1

1

11 s

sDss'

m−

−=−=

Substituting numerical values yields:

5.2cm49

cm 49cm 1701 −=

−−=m

The magnification when the lens was in its second position is given by:

2

2

2

22 s

sLss'

m−

−=−=

Substitute numerical values and evaluate m2:

41.0cm211

cm 121cm 1702 −=

−−=m

59 ••• An object is 17.5 cm to the left of a lens that has a focal length of +8.50 cm. A second lens, which has a focal length of −30.0 cm, is 5.00 cm to the right of the first lens. (a) Find the distance between the object and the final image formed by the second lens. (b) What is the overall magnification? (c) Is the final image real or virtual? Is the final image upright or inverted? Picture the Problem The ray diagram shows four rays from the head of the object that locate images I1 and I2. We can use the thin-lens equation to find the location of the image formed in the positive lens and then, knowing the separation of the two lenses, determine the object distance for the second lens and apply the thin lens a second time to find the location of the final image.

>

>

>

>

1F

'1F

'2F 2F

Lens 1 Lens 2

Image 1 Image 2

Chapter 32

1018

(a) Express the object-to-image distance d:

'ssd 21 cm5 ++= (1)

Apply the thin-lens equation to the positive lens: 111

111f'ss

=+ ⇒11

111 fs

sf's−

=

Substitute numerical values and evaluate 's1 :

( )( ) cm53.16cm 50.8cm5.17cm5.17cm50.8

1 =−

='s

Find the object distance for the negative lens:

cm11.53cm16.53cm00.5cm00.5 12

−=−=−= 'ss

The image distance s2′ is given by: 22

222 fs

sf's−

=

Substitute numerical values and evaluate s2′:

( )( )( ) cm73.18

cm0.30cm53.11cm53.11cm0.30

2 =−−−−−

='s

Substitute numerical values in equation (1) and evaluate d: cm41

cm73.18cm00.5cm5.17

=

++=d

and the final image is 18.7 cm to the right of the second lens.

(b) The overall lateral magnification is given by:

21mmm =

Express m1 and m2: 1

11 s

'sm −= and 2

22 s

'sm −=

Substitute for m1 and m2 to obtain:

21

21

2

2

1

1

ss''ss

s's

s'sm =⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

Substitute numerical values and evaluate m:

( )( )( )( ) 53.1

cm53.11cm5.17cm73.18cm53.16

−=−

=m

(c) Because m < 0, the image is inverted. Because 2's > 0, the image is real.

Optical Images

1019

Aberrations 60 • Chromatic aberration is a common defect of (a) concave and convex lenses, (b) concave lenses only, (c) concave and convex mirrors, (d) all lenses and mirrors. Determine the Concept Chromatic aberrations are a consequence of the differential refraction of light of differing wavelengths by lenses. )(a is correct.

61 • Discuss some of the reasons why most telescopes that are used by astronomers are reflecting rather than refracting telescopes. Determine the Concept Reasons for the preference for reflectors include: 1) no chromatic aberrations, 2) cheaper to shape one side of a piece of glass than to shape both sides, 3) reflectors can be more easily supported from rear instead of edges, preventing sagging and focal length changes, and 4) support from rear makes larger sizes easier to handle. 62 • A symmetric double-convex lens has radii of curvature equal to 10.0 cm. It is made from glass that has an index of refraction equal to 1.530 for blue light and to 1.470 for red light. Find the focal length of this lens for (a) red light and (b) blue light. Picture the Problem We can use the lens-maker’s equation to find the focal length the this lens for the two colors of light. The lens-maker’s equation relates the radii of curvature and the index of refraction to the focal length of the lens:

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

21air

1111rrn

nf

(a) For red light: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−⎟⎠⎞

⎜⎝⎛ −=

cm0.101

cm0.1011

00.1470.11

redf

and cm6.10red =f

(b) For blue light: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−⎟⎠⎞

⎜⎝⎛ −=

cm0.101

cm0.1011

00.1530.11

bluef

and cm43.9blue =f

Chapter 32

1020

The Eye 63 • Find the change in the focal length of the eye when an object originally at 3.0 m is brought to 30 cm from the eye. Picture the Problem We can apply the thin-lens equation for the two values of s to find Δf. Express the change Δf in the focal length:

m0.3m30.0 == −= ss fffΔ

Solve the thin-lens equation for s: ss'

ss'f+

=

Substitute for f to obtain:

m0.3m0.3

m0.3m0.3

m30.0m30.0

m30.0m30.0

ss's's

ss's's

f+

−+

=Δ

Because m 30.0m 3.0 s's' = :

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−

+=

m0.3m0.3

m0.3

m30.0m30.0

m30.0m0.3 ss'

sss'

ss'fΔ

Substitute numerical values and evaluate Δf:

( ) mm1.7cm172.0cm300cm5.2

cm300cm30cm5.2

cm30cm5.2 −=−=⎥⎦

⎤⎢⎣

⎡+

−+

=fΔ

64 • A farsighted person requires lenses that have powers equal to 1.75 D to read comfortably from a book that is 25.0 cm from his eyes. What is that person’s near point without the lenses? Picture the Problem We can use the thin-lens equation and the definition of the power of a lens to express the near point distance as a function of P and s. From the thin-lens equation we have:

Pfs's==+

111⇒

1−=

Psss'

Substitute numerical values and evaluate s′: ( )( ) cm4.44

1m250.0m75.1cm0.25

1 −=−

= −s'

Optical Images

1021

The person’s near point without lenses is cm 44.4 .

65 • [SSM] If two point objects close together are to be seen as two distinct objects, the images must fall on the retina on two different cones that are not adjacent. That is, there must be an unactivated cone between them. The separation of the cones is about 1.00 μm. Model the eye as a uniform 2.50-cm-diameter sphere that has a refractive index of 1.34. (a) What is the smallest angle the two points can subtend? (See Figure 32-61.) (b) How close together can two points be if they are 20.0 m from the eye? Picture the Problem We can use the relationship between a distance measured along the arc of a circle and the angle subtended at its center to approximate the smallest angle the two points can subtend and the separation of the two points 20.0 m from the eye. (a) Relate θmin to the diameter of the eye and the distance between the activated cones:

m00.2mineye μθ ≈d ⇒eye

minm00.2

dμθ =

Substitute numerical values and evaluate θmin:

rad0.80cm50.2

m00.2min μμθ ==

(b) Let D represent the separation of the points R = 20.0 m from the eye to obtain:

( )( )mm60.1

rad0.80m0.20min

=

== μθRD

66 • Suppose the eye were designed like a camera that has a lens of fixed focal length equal to 2.50 cm that could move toward or away from the retina. Approximately how far would the lens have to move to focus the image of an object 25.0 cm from the eye onto the retina? Hint: Find the distance from the retina to the image behind it for an object at 25.0 cm. Picture the Problem We can use the thin-lens equation to find the distance from the lens to the image and then take their difference to find the distance the lens would have to be moved. Express the distance d that the lens would have to move:

fs'd −=

Solve the thin-lens equation for s′: fs

fss'−

=

Chapter 32

1022

Substitute for s′ to obtain: ffs

fsd −−

=

Substitute numerical values and evaluate d:

( )( )

cm28.0

cm50.2cm50.2cm0.25cm0.25cm50.2

=

−−

=d

That is, the lens would have to move cm28.0 toward the object.

67 •• [SSM] The Model Eye I: A simple model for the eye is a lens that has a variable power P located a fixed distance d in front of a screen, with the space between the lens and the screen filled by air. This ″eye″ can focus for all values of object distance s such that xnp ≤ s ≤ xfp where the subscripts on the variables refer to ″near point″ and ″far point″ respectively. This ″eye″ is said to be normal if it can focus on very distant objects. (a) Show that for a normal ″eye,″ of this type, the required minimum value of P is given by dP 1min = . (b) Show that the maximum value of P is given by dxP 11 npmax += . (c) The difference between the maximum and minimum powers, symbolized by A, is defined as minmax PPA −= and is called the accommodation. Find the minimum power and accommodation for this model eye that has a screen distance of 2.50 cm, a far point distance of infinity and a near point distance of 25.0 cm. Picture the Problem The thin-lens equation relates the image and object distances to the power of a lens. (a) Use the thin-lens equation to relate the image and object distances to the power of the lens:

Pfs's==+

111 (1)

Because s′ = d and, for a distance object, s = ∞: ds'

P 11min ==

(b) If npx is the closest distance an

object could be and still remain in clear focus on the screen, equation (1) becomes:

dxP 11

npmax +=

(c) Use our result in (a) to obtain: D0.40

cm50.21

min ==P

Optical Images

1023

Use the results of (a) and (b) to express the accommodation of the model eye:

npnpminmax

1111xddx

PPA =−+=−=

Substitute numerical values and evaluate A: D00.4

cm0.251

==A

68 •• The Model Eye II: In an eye that exhibits nearsightedness, the eye cannot focus on distant objects. (a) Show that for a nearsighted model eye capable of focusing out to a maximum distance xfp, the minimum value of the power P is greater than that of a normal eye (that has a far point at infinity) and is given by dxP 11 fpmin += . (b) To correct for nearsightedness, a contact lens may be placed directly in front of the model-eye’s lens. What power contact lens would be needed to correct the vision of a nearsighted model eye that has a far point distance of 50.0 cm? Picture the Problem The thin-lens equation relates the image and object distances to the power of a lens. (a) Use the thin-lens equation to relate the image and object distances to the power of the lens:

Pfs's==+

111

Because s′ = d and s = xfp: dx

P 11

fpmin +=

(b) To correct for the nearsightedness of this eye, we need a lens that will form an image 25.0 cm in front of the eye of an object at the eye’s far point:

D00.2

cm0.251

cm0.501

min

−=

−+=P

69 •• The Model Eye III: In an eye that exhibits farsightedness, the eye may be able to focus on distant objects but cannot focus on close objects. (a) Show that for a farsighted model eye capable of focusing only as close as a distance

′xnp , the maximum value of the power P is given by dxP 11 '

npmax += . (b) Show that, compared to a model eye capable of focusing as close as a distance xnp (where xnp <

′xnp ), the maximum power of the farsighted lens is too small by

'npnp 11 xx − . (c) What power contact lens would be needed to correct the vision

of a farsighted model eye, with ′xnp = 150 cm, so that the eye may focus on objects as close as 15 cm?

Chapter 32

1024

Picture the Problem The thin-lens equation relates the image and object distances to the power of a lens. (a) Use the thin-lens equation to relate the image and object distances to the power of the lens:

Pfs's==+

111

Because s′ = d and s = x′np: dx

P 1'1'np

max += (1)

(b) For a normal eye: dx

P 11

npmax += (2)

The amount by which the power of the lens is too small is the difference between equations (2) and (1):

npnp

npnpmaxmax

11

1111

x'x

dx'dxP'P

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−+=−

(c) For xnp = 15 cm and x′np = 150 cm:

D0.6

cm1501

cm151

maxmax

=

−=− P'P

70 •• A person who has a near point of 80 cm needs to read from a computer screen that is only 45 cm from her eye. (a) Find the focal length of the lenses in reading glasses that will produce an image of the screen at a distance of 80 cm from her eye. (b) What is the power of the lenses? Picture the Problem We can use the thin-lens equation to find f and the definition of the power of a lens to find P. (a) Solve the thin-lens equation for f: ss'

ss'f+

=

Noting that s′ < 0, substitute numerical values and evaluate f:

( )( )

m0.1

cm103cm45cm80cm80cm45

=

=+−−

=f

(b) Use the definition of the power of a lens to obtain:

D 97.0m03.1

11===

fP

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71 •• A nearsighted person cannot focus clearly on objects that are more distant than 2.25 m from her eye. What power lenses are required for her to see distant objects clearly? Picture the Problem We can use the thin-lens equation to find f and the definition of the power of a lens to find P. Express the required power of the lens:

fP 1=

The thin-lens equation is: fs's111

=+

For s = ∞: fs'

11= ⇒ s'f =

Substitute for f to obtain:

s'P 1=

Substitute for s′ and evaluate P: D444.0

m25.21

==P

72 •• Because the index of refraction of the lens of the eye is not very different from that of the surrounding material, most of the refraction takes place at the cornea, where the index changes abruptly from 1.00 (air) to approximately 1.38. (a) Modeling the cornea, aqueous humor, lens and vitreous humor is a transparent homogeneous solid sphere that has an index of refraction of 1.38, calculate the sphere’s radius if it focuses parallel light on the retina a distance 2.50 cm away. (b) Do you expect your result to be larger or smaller than the actual radius of the cornea? Explain your answer. Picture the Problem We can use the equation for refraction at a single surface (Equation 32-6) with s = ∞ to find the radius of the cornea modeled as a homogeneous sphere with an index of refraction of 1.38. (a) Use Equation 32-6 to relate the radius of the cornea to its index of refraction and that of air:

rnn

s'n

sn 1221 −

=+

Because n2 = n, n1 = 1, and s = ∞: r

ns'n 1−= ⇒

( ) s'nn

ns'r ⎟⎠⎞

⎜⎝⎛ −=

−=

111

Chapter 32

1026

Substitute numerical values and evaluate r:

( ) cm688.0cm50.238.111 =⎟

⎠⎞

⎜⎝⎛ −=r

(b) The calculated value for the distance to the retina is too large. The eye is not a homogeneous sphere, and in a real eye additional refraction occurs at the lens. By modeling the eye as a homogeneous solid sphere we are ignoring the refraction that takes place at the lens. 73 •• The near point of a certain person’s eyes is 80 cm. Reading glasses are prescribed so that he can read a book at 25 cm from his eye. The glasses are 2.0 cm from the eye. What diopter lenses should be used in the glasses? Picture the Problem We can use the definition of the power of a lens and the thin-lens equation to find the power of the lens that should be used in the glasses. Express the power of the lens that should be used in the glasses:

glasseseyelenseye

11ff

PPP +=+= (1)

Because the glasses are 2.0 cm from the eye:

cm78cm2.0cm80 −=+−=s' and

cm23cm2.0cm25 =−=s

Apply the thin-lens equation to the eye with s′ = ∞:

eye

11fs

= ⇒ sf =eye

Apply the thin-lens equation to the glasses with s = ∞:

glasses

11fs'

= ⇒ s'f =glasses

Substitute for eyef and glassesf in

equation (1) to obtain:

s'sP 11

+=

Substitute numerical values and evaluate P:

D1.3m78.0

1m23.0

1=

−+=P

74 ••• At age 45, a person is fitted for reading glasses that have a power equal to 2.10 D in order to read at 25 cm. By the time she reaches the age of 55, she discovers herself holding her newspaper at a distance of 40 cm in order to see it clearly with her glasses on. (a) Where was her near point at age 45? (b) Where is her near point at age 55? (c) What power is now required for the lenses of her reading glasses so that she can again read at 25 cm? Assume the glasses are placed 2.2 cm from her eyes.

Optical Images

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Picture the Problem We can use the thin-lens equation and the distance from her eyes to her glasses to derive an expression for the location of her near point. (a) Express her near point, xnp, at age 45 in terms of the location of her glasses:

cm2.2np += s'x (1)

Because the glasses are 2.2 cm from her eye:

cm 22.8cm 2.2cm 25 =−=s

Apply the thin-lens equation to the glasses:

Pfs's

==+glasses

111

Solving for s′ yields:

sPPs

ss' 11

1 −=

−=

Substitute in equation (1) to obtain:

cm2.211

np +−

=

sP

x (2)

Substitute numerical values and evaluate xnp:

cm46

cm2.2

m228.01m10.2

11

np

=

+−

=−

x

(b) At age 55: cm .873cm 2.2cm 40 =−=s

Substitute numerical values in equation (2) and evaluate s′:

m9.1

cm2.2

m378.01m10.2

11

np

=

+−

=−

x

(c) Solve the thin-lens equation for f: ss'

ss'f+

=

Chapter 32

1028

From the definition of P: s's

ss'f

P +==

1

For s = 22.8 cm and s′ = 183.3 cm:

( )( ) D9.4cm8.22cm3.183cm8.22cm3.183

=+

=P

The Simple Magnifier 75 • [SSM] What is the magnifying power of a lens that has a focal length equal to 7.0 cm when the image is viewed at infinity by a person whose near point is at 35 cm? Picture the Problem We can use the definition of the magnifying power of a lens to find the magnifying power of this lens. The magnifying power of the lens is given by: f

xM np=

Substitute numerical values and evaluate M:

0.5cm0.7cm35

==M

76 •• A lens that has a focal length equal to 6.0 cm is used as a simple magnifier by one person whose near point is 25 cm and by another person whose near point is 40 cm. (a) What is the effective magnifying power of the lens for each person? (b) Compare the sizes of the images on the retinas when each person looks at the same object with the magnifier. Picture the Problem Let the numerals 1 and 2 denote the 1st and 2nd persons, respectively. We can use the definition of magnifying power to find the effective magnifying power of the lens for each person. The relative sizes of the images on the retinas of the two persons is given by the ratio of the effective magnifying powers. (a) The magnifying power of the lens is given by:

fx

M np=

Substitute numerical values and evaluate M1 and M2:

2.417.4cm0.6cm25

1 ===M

and

7.667.6cm0.6cm40

2 ===M

Optical Images

1029

If xnp = 25 cm, then M = 4.2, and if xnp = 40 cm, then M = 6.7. (b) From the definition of magnifying power we have:

2

1

2

1

2

1

yy

fyfy

MM

==

Substitute for M1 and M2 and evaluate the ratio of y1 to y2:

63.067.617.4

2

1 ==yy

For a person with xnp = 40 cm, the image on the retina is 1.6 times larger than it is for a person with xnp = 25 cm. 77 •• [SSM] In your botany class, you examine a leaf using a convex 12-D lens as a simple magnifier. What is the angular magnification of the leaf if image formed by the lens (a) is at infinity and (b) is at 25 cm? Picture the Problem We can use the definition of angular magnification to find the expected angular magnification if the final image is at infinity and the thin-lens equation and the expression for the magnification of a thin lens to find the angular magnification when the final image is at 25 cm. (a) Express the angular magnification when the final image is at infinity:

Pxf

xM np

np ==

where P is the power of the lens.

Substitute numerical values and evaluate M:

( )( ) 0.3m12cm25 1 == −M

(b) Express the magnification of the lens when the final image is at 25 cm:

ss'm −=

Solve the thin-lens equation for s: fs'

fs's−

=

Substitute for s and simplify to obtain:

s'P

fs'

ffs'

fs'fs's'm

−=

+−=−

−=

−

−=

1

1

Chapter 32

1030

Substitute numerical values and evaluate m:

( )( ) 0.4m12m25.01 1 =−−= −m

The Microscope 78 •• Your laboratory microscope objective has a focal length of 17.0 mm. It forms an image of a tiny specimen at 16.0 cm from its second focal point. (a) How far from the objective is the specimen located? (b) What is the magnifying power for you if your near point distance is 25.0 cm and the focal length of the eyepiece is 51.0 mm? Picture the Problem We can use the thin-lens equation to find the location of the object and the expression for the magnifying power of a microscope to find the magnifying power of the given microscope for a person whose near point is at 25.0 cm. (a) Using the thin-lens equation, relate the object distance s to the focal length of the objective lens f0:

0

111fs's

=+ ⇒0

0

fs's'fs−

=

The image distance for the image formed by the objective lens is:

cm17.7cm16.0cm70.10 =+=+= Lfs'

Substitute numerical values and evaluate s:

( )( ) mm 8.18cm70.1cm7.17cm7.17cm70.1

=−

=s

(b) Express the magnifying power of a microscope:

e

np

0 fx

fLM −=

Substitute numerical values and evaluate M: 1.46

cm10.5cm0.25

cm1.70cm0.16

−=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−=M

79 •• [SSM] Your laboratory microscope objective has a focal length of 17.0 mm. It forms an image of a tiny specimen at 16.0 cm from its second focal point. (a) How far from the objective is the specimen located? (b) What is the magnifying power for you if your near point distance is 25.0 cm and the focal length of the eyepiece is 51.0 mm? Picture the Problem The lateral magnification of the objective is

oo fLm −= and the magnifying power of the microscope is .eoMmM =

Optical Images

1031

(a) The lateral magnification of the objective is given by:

oo f

Lm −=

Substitute numerical values and evaluate mo:

198.18mm5.8cm16

o −=−=−=m

(b) The magnifying power of the microscope is given by:

eoMmM = where Me is the angular magnification of the lens.

Substitute numerical values and evaluate M:

( )( ) 2109.1108.18 ×−=−=M

80 •• A crude, symmetric handheld microscope consists of two 20-D lenses fastened at the ends of a tube 30 cm long. (a) What is the tube length of this microscope? (b) What is the lateral magnification of the objective? (c) What is the magnifying power of the microscope? (d) How far from the objective should the object be placed? Picture the Problem We can find the tube length from the length of the tube to which the lenses are fastened and the focal lengths of the objective and eyepiece. We can use their definitions to find the lateral magnification of the objective and the magnifying power of the microscope. The distance of the object from the objective can be found using the thin-lens equation. (a) The tube length L is given by:

cm20m202m30.0 1

eo

=−=

−−=

−

ffDL

(b) The lateral magnification of the objective mo is given by:

0.4cm5.0cm20

oo −=−=−=

fLm

(c) The magnifying power of the microscope is given by:

e

npoeo f

xmMmM ==

Substitute numerical values and evaluate M:

( ) 20cm0.5cm254 −=−=M

(d) From the thin-lens equation we have:

oo

111f'ss o

=+

where Lf's += oo

Chapter 32

1032

Substitute for 'so to obtain:

ooo

111fLfs

=+

+ ⇒( )

LLff

s+

= ooo

Substitute numerical values and evaluate so:

( )( )

cm3.6

cm20cm20cm0.5cm0.5

o

=

+=s

81 •• A compound microscope has an objective lens that has a power of 45 D and an eyepiece that has a power of 80 D. The lenses are separated by 28 cm. Assuming that the final image is formed by the microscopy is 25 cm from the eye, what is the magnifying power? Picture the Problem The magnifying power of a compound microscope is the product of the magnifying powers of the objective and the eyepiece. Express the magnifying power of the microscope in terms of the magnifying powers of the objective and eyepiece:

eommM = (1)

The magnification of the eyepiece is given by:

11 npee

npe +=+= xP

fx

m

The magnification of the objective is given by:

oo f

Lm −=

where eo ffDL −−=

Substitute for L to obtain: o

eoo f

ffDm −−−=

Substitute for me and mo in equation (1) to obtain: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−−+=

o

eonpe 1

fffDxPM

Substitute numerical values and evaluate M:

( )( )[ ] 230cm22.2

cm25.1cm22.2cm281m25.0D80 −=⎟⎟⎠

⎞⎜⎜⎝

⎛ −−−+=M

Optical Images

1033

82 ••• A microscope has a magnifying power of 600. The eyepiece has an angular magnification of 15.0. The objective lens is 22.0 cm from the eyepiece. Calculate (a) the focal length of the eyepiece, (b) the location of the object so that it is in focus for a normal relaxed eye, and (c) the focal length of the objective lens. Picture the Problem We can find the focal length of the eyepiece from its angular magnification and the near point of a normal eye. The location of the object such that it is in focus for a normal relaxed eye can be found from the lateral magnification of the eyepiece and the magnifying power of the microscope. Finally, we can use the thin-lens equation to find the focal length of the objective lens. (a) The focal length of the eyepiece is related to its angular magnifying power:

e

npe f

xM = ⇒

e

npe M

xf =

Substitute numerical values and evaluate fe:

cm67.10.15cm0.25

e ==f

(b) Relate s to s′ through the lateral magnification of the objective:

ss'm −=o ⇒

oms's −=

Relate the magnifying power of the microscope M to the lateral magnification of its objective m0 and the angular magnification of its eyepiece Me:

eoMmM = ⇒e

o MMm =

Substitute for mo to obtain: M

s'Ms e−=

Evaluate s′:

cm20.33cm1.67cm22cm22 e

=−=−= fs'

Substitute numerical values and evaluate s:

( )( ) cm508.0600

0.15cm33.20=

−−=s

(c) Solving the thin-lens equation for fo yields: ss'

ss'f+

=o

Chapter 32

1034

Substitute numerical values and evaluate fo:

( )( )

cm496.0

cm508.0cm33.20cm33.20cm508.0

o

=

+=f

The Telescope 83 • [SSM] You have a simple telescope that has an objective which has a focal length of 100 cm and an eyepiece which has a focal length of 5.00 cm. You are using it to look at the moon, which subtends an angle of about 9.00 mrad. (a) What is the diameter of the image formed by the objective? (b) What angle is subtended by the image formed at infinity by the eyepiece? (c) What is the magnifying power of your telescope? Picture the Problem Because of the great distance to the moon, its image formed by the objective lens is at the focal point of the objective lens and we can use

θofD = to find the diameter D of the image of the moon. Because angle subtended by the final image at infinity is given by ,MM θθθ == oe we can solve (b) and (c) together by first using M = −fo/fe to find the magnifying power of the telescope. (a) Relate the diameter D of the image of the moon to the image distance and the angle subtended by the moon:

θ'sD o=

Because the image of the moon is at the focal point of the objective lens:

oo f's = and θofD =

Substitute numerical values and evaluate D:

( )( ) mm00.9mrad00.9cm100 ==D

(b) and (c) Relate the angle subtended by the final image at infinity to the magnification of the telescope and the angle subtended at the objective:

θθθ MM == oe

Express the magnifying power of the telescope:

e

o

ffM −=

Optical Images

1035

Substitute numerical values and evaluate M and θe:

0.20cm00.5cm100

−=−=M

and ( ) mrad 180mrad00.90.20e =−=θ

84 • The objective lens of the refracting telescope at the Yerkes Observatory has a focal length of 19.5 m. The moon subtends an angle of about 9.00 mrad. When the telescope is used to look at the moon, what is the diameter of the image of the moon formed by the objective? Picture the Problem Because of the great distance to the moon, its image formed by the objective lens is at the focal point of the objective lens and we can use

θofD = to find the diameter D of the image of the moon. Relate the diameter D of the image of the moon to the image distance and the angle subtended by the moon:

θ'sD o=

Because the image of the moon is at the focal point of the objective lens:

oo f's = and θofD =

Substitute numerical values and evaluate D:

( )( ) cm7.61mrad00.9m5.19 ==D

85 •• The 200-in (5.08-m) diameter mirror of the reflecting telescope at Mt. Palomar has a focal length of 16.8 m. (a) By what factor is the light-gathering power increased over the 40.0-in (1.02-m) diameter refracting lens of the Yerkes Observatory telescope? (b) If the focal length of the eyepiece is 1.25 cm, what is the magnifying power of the 200-in telescope? Picture the Problem Because the light-gathering power of a mirror is proportional to its area, we can compare the light-gathering powers of these mirrors by finding the ratio of their areas. We can use the ratio of the focal lengths of the objective and eyepiece lenses to find the magnifying power of the Palomar telescope.

Chapter 32

1036

(a) Express the ratio of the light-gathering powers of the Palomar and Yerkes mirrors:

2

mirror Yerkes

2

mirrorPalomar

2

mirror Yerkes

2

mirrorPalomar

mirror Yerkes

mirrorPalomar

Yerkes

Palomar

4

4

d

d

d

d

A

A

PP

=

==π

π

Substitute numerical values and evaluate PPalomar/PYerkes:

( )( )

0.25in0.40in200

2

2

Yerkes

Palomar ==PP

(b) Express the magnifying power of the Palomar telescope:

e

o

ffM −=

Substitute numerical values and evaluate M:

1340cm25.1m8.16

−=−=M

86 •• An astronomical telescope has a magnifying power of 7.0. The two lenses are 32 cm apart. Find the focal length of each lens. Picture the Problem We can use the expression for the magnifying power of a telescope and the fact that the length of a telescope is the sum of focal lengths of its objective and eyepiece lenses to obtain simultaneous equations in fo and fe. The magnifying power of the telescope is given by:

7e

o −=−=ffM

The length of the telescope is the sum of the focal lengths of the objective and eyepiece lenses:

cm32eo =+= ffL

Solve these equations simultaneously to obtain:

cm28o =f and cm4e =f

87 •• [SSM] A disadvantage of the astronomical telescope for terrestrial use (for example, at a football game) is that the image is inverted. A Galilean telescope uses a converging lens as its objective, but a diverging lens as its eyepiece. The image formed by the objective is at the second focal point of the eyepiece (the focal point on the refracted side of the eyepiece), so that the final image is virtual, upright, and at infinity. (a) Show that the magnifying power is given by M = –fo/fe, where fo is the focal length of the objective and fe is that of the

Optical Images

1037

eyepiece (which is negative). (b) Draw a ray diagram to show that the final image is indeed virtual, upright, and at infinity. Picture the Problem The magnification of a telescope is the ratio of the angle subtended at the eyepiece lens to the angle subtended at the objective lens. We can use the geometry of the ray diagram to express both θe and θo. (b) The ray diagram is shown below:

>

>

>

>

Lens 1 Lens 2

1F

2F '1F

o

e

θ

θh

(a) Express the magnifying power M of the telescope: o

e

θθ

=M

Because the image formed by the objective lens is at the focal point,

1F' :

oo f

h=θ

where we have assumed that θo << 1 so that tan θo ≈ θo.

Express the angle subtended by the eyepiece:

ee f

h=θ where fe is negative.

Substitute to obtain:

e

o

o

e

ff

fhfh

M ==

and e

o

ffM −= is positive.

Remarks: Because the object for the eyepiece is at its focal point, the image is at infinity. As is also evident from the ray diagram, the image is virtual and upright.

Chapter 32

1038

88 •• A Galilean telescope (see Problem 87) is designed so that the final image is at the near point, which is 25 cm (rather than at infinity). The focal length of the objective is 100 cm and the focal length of the eyepiece is –5.0 cm. (a) If the object distance is 30.0 m, where is the image of the objective? (b) What is the object distance for the eyepiece so that the final image is at the near point? (c) How far apart are the lenses? (d) If the object height is 1.5 m, what is the height of the final image? What is the angular magnification? Picture the Problem We can use the thin-lens equation to find the image distance for the objective lens and the object distance for the eyepiece lens. The separation of the lenses is the sum of these distances. We can use the definition of the angular magnification and the angles subtended at the objective and eyepiece lenses to find the height of the final image. (a) Solve the thin-lens equation for 'so :

oo

ooo fs

sf's−

=

Substitute numerical values and evaluate 'so :

( )( ) cm103m00.1m0.30m0.30m00.1

o =−

='s

(b) Solve the thin-lens equation for se: ee

eee f's

'sfs−

=

Noting that se′ = −25 cm, substitute numerical values and evaluate se:

( )( )( )

cm3.6

cm 25.6cm0.5cm25cm25cm0.5

ee

−

−=−−−−−

=s

where the minus sign tells us that the object of the eyepiece is virtual.

(c) Express the separation D of the lenses:

eo s'sD +=

Substitute numerical values and evaluate D:

cm97cm6cm103 ≈−=D

(d) Express the height h′ of the final image in terms of the magnification M of the telescope:

Mhh' =

Optical Images

1039

The magnification of the telescope is the product of the magnifications of the objective and eyepiece lenses:

e

e

o

oeo s

'ss

'smmM ==

Substitute for M to obtain: hs

'ss

'sh'e

e

o

o=

Substitute numerical values and evaluate h′: ( )

cm21

m5.1cm25.6

cm25cm3000cm45.103

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛=h'

Express the angular magnification of the telescope: o

e

θθ

=M

The angle subtended by the object is: o

o sh

=θ

The angle subtended by the image is:

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

e

1e tan

sh'θ

Substitute for θe and oθ and simplify to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛

= −

−

e

1o

o

e

1

tantan

sh'

hs

sh

sh'

M

Substitute numerical values and evaluate M: 26

cm25.6cm7.20tan

m5.1m30 1 =⎟⎟

⎠

⎞⎜⎜⎝

⎛= −M

89 ••• If you look into the wrong end of a telescope, that is, into the objective, you will see distant objects reduced in angular size. For a refracting telescope that has an objective with a focal length equal to 2.25 m and an eyepiece with a focal length equal to 1.50 cm, by what factor is the angular size of the object changed? Picture the Problem The roles of the objective and eyepiece lenses are reversed. Express the magnifying power of the ″wrong end″ telescope:

o

e

ffM −=

Chapter 32

1040

Substitute numerical values and evaluate M:

31067.6m25.2

cm50.1 −×−=−=M

General Problems 90 • To focus a camera, the distance between the lens and the image-sensing surface is varied. A wide-angle lens of a camera has a focal length of 28 mm. By how much must the lens move to change from focusing on an object at infinity to an object at a distance of 5.00 m in front of the camera? Picture the Problem We can express the distance Δs that the lens must move as the difference between the image distances when the object is at 30 m and when it is at infinity and then express these image distances using the thin-lens equation. Express the distance Δs that the lens must move to change from focusing on an object at infinity to one at a distance of 5 m:

∞−=Δ s's's 5

Solve the thin-lens equation for s′ to obtain:

fsfss'−

=

Substitute 5s' and ∞s' and simplify to obtain:

⎥⎦

⎤⎢⎣

⎡−

−=

−−

−=

−−

−=Δ

∞

∞

∞

1

1

5

5

5

5

5

5

fssf

sff

fsfs

fsfs

fsfss

Substitute numerical values and evaluate Δs: ( )

mm16.0

1m028.0m00.5

m0.5mm28

=

⎥⎦

⎤⎢⎣

⎡−

−=Δs

91 • A converging lens that has a focal length equal to 10 cm is used to obtain an image that is twice the height of the object. Find the object and image distances if (a) the image is to be upright and (b) the image is to be inverted. Draw a ray diagram for each case.

Optical Images

1041

Picture the Problem We can use the thin-lens and magnification equations to obtain simultaneous equations that we can solve to find the image and object distances for the two situations described in the problem statement. (a) Use the thin-lens equation to relate the image and object distances to the focal length of the lens:

fs's111

=+

Because the image is twice as large as the object:

ss'm −= ⇒ ss' 2−=

Substitute for s' to obtain: fss

1211

=−

+ ⇒ fs 21=

Substitute numerical values and evaluate s and s′:

( ) cm0.5cm1021 ==s

and ( ) cm10cm0.52 −=−=s'

(b) If the image is inverted, then: ss' 2= and

fss1

211=+ ⇒ fs 2

3=

Substitute numerical values and evaluate s and s′:

( )[ ] cm15cm10321 ==s

and ( ) cm30cm152 ==s'

The ray diagrams for (a) (left) and (b) (right) follow:

92 •• You are given two converging lenses that have focal lengths of 75 mm and 25 mm. (a) Show how the lenses should be arranged to form a telescope. State which lens to use as the objective, which lens to use as the eyepiece, how far apart to place the lenses and, what angular magnification you expect. (b) Draw a ray diagram to show how rays from a distant object are refracted by the two lenses.

Chapter 32

1042

Picture the Problem In an astronomical telescope the eyepiece (short focal length) and objective (long focal length) lenses are separated by the sum of their focal lengths. Given these two lenses, we’ll use the 25 mm lens as the eyepiece lens and the 75 mm lens as the objective lens and mount them 100 mm apart. The

angular magnification is then 3mm25mm75

e

0 −=−=−=ff

M .

(a) The lens with a focal length of 75 mm should be the objective. The two lenses should be separated by 100 mm. The angular magnification is –3. (b) A ray diagram showing how rays from a distant object are refracted by the two lenses follows. A real and inverted image of the distant object is formed by the objective lens near its second focal point. The eyepiece lens forms an enlarged and inverted image of the image formed by the objective lens.

Objective Eyepiece

Light from

distant object

f fo e

Image formed by

objective lens

Image formed by

eyepiece lens is

at infinity 93 •• [SSM] (a) Show how the same two lenses in Problem 92 should be arranged to form a compound microscope that has a tube length of 160 mm. State which lens to use as the objective, which lens to use as the eyepiece, how far apart to place the lenses, and what overall magnification you expect to get, assuming the user has a near point of 25 cm. (b) Draw a ray diagram to show how rays from a close object are refracted by the lenses. Picture the Problem Because the focal lengths appear in the magnification formula as a product, it would appear that it does not matter in which order we use them. The usual arrangement would be to use the shorter focal length lens as the objective but we get the same magnification in the reverse order. What difference does it make then? None in this problem. However, it is generally true that the smaller the focal length of a lens, the smaller its diameter. This condition makes it harder to use the shorter focal length lens, with its smaller diameter, as the eyepiece lens.

Optical Images

1043

In a compound microscope, the lenses are separated by:

0e ffL ++=δ

Substitute numerical values and evaluate δ:

cm26cm2.5cm7.5cm16 =++=δ

The overall magnification of a compound microscope is given by:

e

np

0e0 f

xfLMmM −==

Substitute numerical values and evaluate M:

21cm2.5cm25

cm7.5cm16

−=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−=M

The lens with a focal length of 25 mm should be the objective. The two lenses should be separated by 210 mm. The angular magnification is –21. (b) A ray diagram showing how rays from a near-by object are refracted by the lenses follows. A real and inverted image of the near-by object is formed by the objective lens at the first focal point of the eyepiece lens. The eyepiece lens forms an inverted and virtual image of this image at infinity.

Eyepiece

Objective

o ef fL

94 •• On a vacation, you are scuba-diving and using a diving mask that has a face plate and that bulges outward with a radius of curvature of 0.50 m. As a result, a convex spherical surface exists between the water and the air in the mask. A fish swims by you 2.5 m in front of your mask. (a) How far in front of the mask does the fish appear to be? (b) What is the lateral magnification of the image of the fish?

Chapter 32

1044

Picture the Problem We can use the equation for refraction at a single surface to locate the image of the fish and the expression for the magnification due to refraction at a spherical surface to find the magnification of the image. (a) Use the equation describing refraction at a single surface to relate the image and object distances:

rnn

s'n

sn 1221 −

=+

Solving for s′ yields: ( ) rnsnn

rsns'112

2

−−=

Substitute numerical values and evaluate s′:

( )( )( )( )( ) ( )( )

cm84m839.0

m50.033.1m5.233.11m5.2m50.01

−=−=

−−=s'

Note that the fish appears to be only 84 cm in front of your mask.

(b) Express the magnification due to refraction at a spherical surface:

sns'nm

2

1−=

Substitute numerical values and evaluate m:

( )( )( )( ) 45.0

m5.21m839.033.1

=−

−=m

Note that the fish appears to be smaller than it actually is.

95 •• [SSM] A 35-mm digital camera has a rectangular array of CCDs (light sensors) that is 24 mm by 36 mm. It is used to take a picture of a person 175-cm tall so that the image just fills the height (24 mm) of the CCD array. How far should the person stand from the camera if the focal length of the lens is 50 mm? Picture the Problem We can use the thin-lens equation and the definition of the magnification of an image to determine where the person should stand. Use the thin-lens equation to relate s and s′:

fs's111

=+

Optical Images

1045

The magnification of the image is given by:

21037.1cm175cm4.2 −×−=−=−=

ss'm

and mss' −=

Substitute for s' to obtain:

fmss111

=− ⇒ fm

s ⎟⎠⎞

⎜⎝⎛ −=

11

Substitute numerical values and evaluate s:

( ) m7.3mm501037.1

11 2 =⎟⎠⎞

⎜⎝⎛

×−−= −s

96 •• A 35-mm camera that has interchangeable lenses is used to take a picture of a hawk that has a wing span of 2.0 m. The hawk is 30 m away. What would be the ideal focal length of the lens used so that the image of the wings just fills the width of the light sensitive area of the camera, which is 36 mm? Picture the Problem We can use the thin-lens equation and the definition of the magnification of an image to determine the ideal focal length of the lens. Use the thin-lens equation to relate s and s′:

fs's111

=+

The magnification of the image is given by:

21080.1cm002cm6.3 −×−=−=−=

ss'm

and mss' −=

Substitute for s' to obtain:

fmss111

=− ⇒

m

sf 11−=

Substitute numerical values and evaluate f:

cm53

1080.111

m30

2

=

×−−

=

−

f

97 •• An object is placed 12.0 cm in front of a lens that has a focal length equal to 10.0 cm. A second lens that has a focal length equal to 12.5 cm is placed 20.0 cm in back of the first lens. (a) Find the position of the final image. (b) What is the magnification of the image? (c) Sketch a ray diagram showing the final image.

Chapter 32

1046

Picture the Problem Let the numeral 1 refer to the first lens and the numeral 2 to the second lens. We apply the thin-lens equation twice; once to locate the image formed by the first lens and a second time to find the image formed by the second lens. The magnification of the image is the product of the magnifications produced by the two lenses. (a) Solve the thin-lens equation for the location of the image formed by the first lens:

11

111 fs

sf's−

=

Substitute numerical values and evaluate 's1 :

( )( ) cm60cm10cm12cm12cm10

1 =−

='s

Because the second lens is 20 cm to the right of the first lens:

cm40cm60cm202 −=−=s

Solve the thin-lens equation for the location of the image formed by the second lens:

22

222 fs

sf's−

=

Substitute numerical values and evaluate 's2 :

( )( ) cm5.9cm5.12cm40cm40cm5.12

2 =−−−

='s

i.e., the final image is 9.5 cm in back of the second lens.

(b) Express the magnification of the final image:

21

21

2

2

1

121 ss

''sss

'ss'smmm =⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−==

Substitute numerical values and evaluate m:

( )( )( )( ) 2.1

cm40cm12cm52.9cm60

−=−

=m

i.e., the final image is about 20% larger than the object and is inverted.

(c) The enlarged, inverted image formed by the first lens serves as a virtual object for the second lens. The image formed from this virtual object is the real, inverted image shown in the diagram.

Optical Images

1047

FFFF

''

1

1

2 2

Lens 1 Lens 2

98 •• (a) Show that if fa is the focal length of a thin lens in air, its focal length in water is given by ( )( ) ( ) awaaww fnnnnnnf −−= where nw is the index of refraction of water, n is that of the lens material and na is that of air. (b) Calculate the focal length in air and in water of a double concave lens that has an index of refraction of 1.50 and radii of magnitude 30 cm and 35 cm. Picture the Problem We can apply the equation for refraction at a surface to both surfaces of the lens and add the resulting equations to obtain an equation relating the image and object distances to the indices of refraction. We can then use the lens maker’s equation to complete the derivation of the given relationship between f ′ and f. (a) Relate s and s′ at the water-lens interface: 1

w

1

w

rnn

'sn

sn −

=+

Relate s and s′ at the lens-water interface: 2

w

1 rnn

s'n

'sn −

=+−

Add these equations to obtain: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=⎟

⎠⎞

⎜⎝⎛ +

21ww

1111rr

nns's

n

Let s'sf111

w

+= to obtain:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

21w

w

w 11rr

nnfn

(1)

Chapter 32

1048

The lens-maker’s equation is: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

21aa

1111rrn

nf

and

( ) aa

a

aa

21 1

111fnn

n

fnnrr −

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=−

Substitute for 21

11rr

− in equation

(1) to obtain:

( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=aa

aw

w

w

fnnn

nnfn

Solving for f ′ yields: ( )( ) a

wa

aww f

nnnnnn

f−−

=

(b) Use the lens-maker’s equation to find the focal length of the lens in air:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎠⎞

⎜⎝⎛ −=

cm351

cm3011

00.150.11

af

and cm32cm3.32 −=−=f

Use the result derived in Part (a) to find fw:

( )( ) ( )

m3.1

cm3.3233.150.1

150.133.1

−=

−−

−=wf

99 •• While parked in your car, you see a jogger in your rear view mirror, which is convex and has a radius of curvature whose magnitude is equal to 2.00 m. The jogger is 5.00 m from the mirror and is approaching at 3.50 m/s. How fast is the image of the jogger moving relative to the mirror? Picture the Problem The speed of the jogger as seen in the mirror is

.dtds'v' = We can use the mirror equation to derive an expression for v′ in terms of f and ds/dt. Solve the mirror equation for s′:

111

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

sfs' (1)

Optical Images

1049

Differentiate s′ with respect to time to obtain:

dtds

ssf

sfdtd

dtds'v'

⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−==

−

−

2

2

1

111

11

Simplify this result to obtain:

vss'v'

2

⎟⎠⎞

⎜⎝⎛−= (2)

Rewrite equation (1) in terms of r:

112 −

⎟⎠⎞

⎜⎝⎛ −=

srs'

Find s′ when s = 5.00 m:

m8333.0m00.5

1m00.2

21

−=⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

−

s'

Use equation (2) to find v' when

v = 3.50 m/s: ( )

cm/s72.9

m/s50.3m5.00

m8333.02

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−=v'

100 •• A 2.00-cm-thick layer of water (n = 1.33) floats on top of a 4.00-cm-thick layer of carbon tetrachloride (n = 1.46) in a tank. How far below the top surface of the water does the bottom of the tank appear, according to an observer looking down from above at normal incidence? Picture the Problem Let the numeral 1 denote the CCl4-H2O interface and the numeral 2 the H2O-air interface. We can locate the final image by applying the equation for refraction at a single surface to both interfaces. The ray diagram shown below shows a spot at the bottom of the tank and the rays of light emanating from it that form the intermediate and final images.

Air

H2O

CCl4

1air =n

33.1OH2=n

46.14CCl =n

2.00 cm

cm 00.41 =s

1s'

2s'

2s

Chapter 32

1050

Use the equation for refraction at a single surface to relate s and s′ at the CCl4-H2O interface:

rnn

'sn

sn

4224 CClOH

1

OH

1

CCl −=+

or, because r = ∞,

01

OH

1

CCl 24 =+'s

ns

n⇒

4

2

CCl

1OH1 n

sn's −= (1)

Substitute numerical values and evaluate s1′:

( )( ) cm64.346.1

cm00.433.11 −=−='s

The depth of this image, as viewed from the H2O-air interface is:

( ) cm64.5cm3.64cm00.2cm00.2 12

=−−=−= 'ss

At the H2O-air interface equation (1) becomes: O2H

2air2 n

sn's −=

Substitute numerical values and evaluate s2′:

( )( ) cm24.433.1

cm64.512 −=−='s

cm.4.24 isdepth apparent The 101 ••• [SSM] An object is 15.0 cm in front of a thin converging lens that has a focal length equal to 10.0 cm. A concave mirror that has a radius equal to 10.0 cm is 25.0 cm in back of the lens. (a) Find the position of the final image formed by the mirror-lens combination. (b) Is the image real or virtual? Is the image upright or inverted? (c) On a diagram, show where your eye must be to see this image. Picture the Problem We can use the thin-lens equation to locate the first image formed by the converging lens, the mirror equation to locate the image formed in the mirror, and the thin-lens equation a second time to locate the final image formed by the converging lens as the rays pass back through it.

(a) Solve the thin-lens equation for s1′: fs

fs's−

=1

11

Substitute numerical values and evaluate s1′:

( )( ) cm30cm0.10cm0.15cm0.15cm0.10

1 =−

='s

Because the image formed by the converging lens is behind the mirror:

cm5cm30cm252 −=−=s

Optical Images

1051

Solve the mirror equation for s2′: fs

fs's−

=2

22

Substitute numerical values and evaluate s2′:

( )( ) cm50.2cm5cm5cm5cm5

2 =−−−

='s and the

image is 22.5 cm from the lens; i.e., s3 = 22.5 cm.

Solve the thin-lens equation for s3′: fs

fs's−

=3

33

Substitute numerical values and evaluate s3′:

( )( ) cm18cm0.10cm5.22cm5.22cm0.10

3 =−

='s

The final position of the image is 18 cm from the lens, on the same side as the original object. (b) The ray diagram is shown below. The numeral 1 represents the object. The parallel and central rays from 1 are shown; one passes through the center of the converging lens, the other is paraxial and then passes through the focal point F′. The two rays intersect behind the mirror, and the image formed there, identified by the numeral 2, serves as a virtual object for the mirror. Two rays are shown emanating from this virtual image, one through the center of the mirror, the other passing through its focal point (halfway between C and the mirror surface) and then continuing as a paraxial ray. These two rays intersect in front of the mirror, forming a real image, identified by the numeral 3. Finally, the image 3 serves as a real object for the lens; again we show two rays, a paraxial ray that then passes through the focal point F and a ray through the center of the lens. These two rays intersect to form the final upright, real, and diminished image, identified as 4.

23CF14

F'

Final

image

Object

Image formed

by the lens

Real object

for the final

image

(c) To see this image the eye must be to the left of the final image.

Chapter 32

1052

102 ••• When a bright light source is placed 30 cm in front of a lens, there is an upright image 7.5 cm from the lens. There is also a faint inverted image 6.0 cm from the lens on the incident-light side due to reflection from the front surface of the lens. When the lens is turned around, this weaker, inverted image is 10 cm in front of the lens. Find the index of refraction of the lens. Picture the Problem The mirror surfaces must be concave to create inverted images on reflection. Therefore, the lens is a diverging lens. Let the numeral 1 denote the lens in its initial orientation and the numeral 2 the lens in its second orientation. We can use the mirror equation to find the magnitudes of the radii of the lens’ surfaces, the thin-lens equation to find its focal length, and the lens maker’s equation to find its index of refraction. Solve the mirror equation for 1r :

11

111

2s's'ssr

+=

Substitute numerical values and evaluate 1r :

( )( ) cm0.10cm30cm0.6cm0.6cm302

1 =+

=r

Solve the mirror equation for 2r : 22

222

2s's'ssr

+=

Substitute numerical values and evaluate 2r :

( )( ) cm0.15cm30cm10cm10cm302

2 =+

=r

Solve the thin-lens equation for f: ss'

ss'f+

=

Substitute numerical values and evaluate f:

( )( ) cm0.10cm30cm5.7cm5.7cm30

−=+−−

=f

Solve the lens-maker’s equation for n to obtain: air

21

a

11n

rrf

nn +

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

Because the lens is a diverging lens, r1 = −10 cm and r2 = 15 cm. Substitute numerical values and evaluate n:

( )

6.1

00.1

cm151

cm101cm10

00.1

=

+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−

=n

Optical Images

1053

103 ••• A concave mirror that has a radius of curvature equal to 50.0 cm is oriented with its axis vertical. The mirror is filled with water that has an index of refraction equal to 1.33 and a maximum depth of 1.00 cm. At what height above the vertex of the mirror must an object be placed so that its image is at the same position as the object? Picture the Problem Assume that the object is very small compared to r so that all incident and reflected rays traverse 1 cm of water. The problem involves two refractions at the air−water interface and one reflection at the mirror. Let the numeral 1 refer to the first refraction at the air−water interface, the numeral 2 to the reflection in the mirror surface, and the numeral 3 to the second refraction at the water−air interface. Use the equation for refraction at a single surface to relate s1 and s1′: r

nn's

nsn 12

1

2

1

1 −=+

or, because r = ∞,

01

2

1

1 =+'s

nsn

⇒1

121 n

sn's −=

Let n2 = n. Because n1 = 1: 11 ns's −=

Find the object distance for the mirror:

112 1s1 ns's +=−= where 1 has units of cm.

Solve the mirror equation for s2′: 1

22

12−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

sr's

Substitute for s2: 1

12 1

12−

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=nsr

's

Find the object distance s3 for the water−air interface:

1

123 1

1211−

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−−=−=nsr

'ss

Use the equation for refraction at a single surface to relate s3 and s3′:

rnn

'sn

sn 12

3

2

3

1 −=+

or, because r = ∞,

03

2

3

1 =+'s

nsn

⇒1

323 n

sn's −=

Chapter 32

1054

Because n2 = 1 and n1 = n:

nnsr

ns's

1

133

1121

−

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−−−=−=

Equate s3′ and s1:

nnsr

s

1

11

1121

−

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−−−=

Simplifying yields:

01221

21 =

−+

−+

nrs

nrs

Substitute numerical values and simplify:

( )0

33.1cm0.50cm00.1

33.1cm0.50cm00.2

2

121

=−

+

−+ ss

or 070.2709.36 1

21 =−− ss

where s1 is in cm.

Solve for the positive value of s1 to obtain:

cm371 =s

104 ••• A concave side of a lens has a radius of curvature of magnitude equal to 17.0 cm, and the convex side of the lens that has a radius of curvature of magnitude equal to 8.00 cm. The focal length of the lens in air is 27.5 cm. When the lens is placed in a liquid that has an unknown index of refraction, the focal length increases to 109 cm. What is the index of refraction of the liquid? Picture the Problem We can use the lens maker’s equation, in conjunction with the result given in Problem 108, to find the index of refraction of the liquid. Solve the lens-maker’s equation for n: air

21

air

11n

rrf

nn +

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

Substitute numerical values and evaluate n:

( )55.100.1

cm00.81

cm0.171cm5.27

00.1=+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−

=n

Optical Images

1055

From Problem 98, the focal length of the lens in the liquid, fL, is related to the focal length of the lens in air, f, according to:

( )( ) f

nnnnnn

fLair

airLL −

−=

Solving for nL yields: ( ) Lairair

LairL fnfnn

nfnn

+−=

Substitute numerical values and evaluate nL:

( )( )( )( )( ) ( )( ) 4.1

cm10900.1cm5.2700.155.1cm10900.155.1

L =+−

=n

105 ••• A solid glass ball of radius 10.0 cm has an index of refraction equal to 1.500. The right half of the ball is silvered so that it acts as a concave mirror (Figure 32-63). Find the position of the final image formed for an object located at (a) 40.0 cm, and (b) 30.0 cm to the left of the center of the ball. Picture the Problem The problem involves two refractions and one reflection. We can use the refraction at spherical surface equation and the mirror equation to find the images formed in the two refractions and one reflection. Let the numeral 1 refer to the first refraction at the air-glass interface, the numeral 2 to the reflection from the silvered surface, and the numeral 3 refer to the refraction at the glass-air interface. (a) The image and object distances for the refraction at the surface of the ball are related according to:

rnn

'sn

sn 12

1

2

1

1 −=+

Solve for s1′ to obtain: ( ) rnsnn

rsn's1112

121 −−=

Substitute numerical values and evaluate s1′:

( )( )( )( )( ) ( )( )

cm0.90cm0.101cm0.301500.1

cm0.30cm0.10500.11

=−−

='s

The object for the mirror surface is behind the mirror and its distance from the surface of the mirror is:

cm70.0cm90.0cm0.202 −=−=s

Chapter 32

1056

Use the mirror equation to relate s2 and s2′: r'ss

211

22

=+ ⇒rs

rs's−

=2

22 2

Substitute numerical values and evaluate s2′:

( )( )( ) cm67.4

cm0.10cm0.702cm0.70cm0.10

2 =−−

−='s

The object for the second refraction at the glass-air interface is in front of the mirrored surface and its distance from the glass-air interface is:

cm15.3cm4.67cm0.203 =−=s

The image and object distances for the second refraction are related according to:

rnn

'sn

sn 12

3

2

3

1 −=+

Solve for s3′ to obtain: ( ) rnsnn

rsn's1312

323 −−=

Noting that r = −10.0 cm, substitute numerical values and evaluate s3′:

( )( )( )( )( ) ( )( ) cm76.6

cm0.10500.1cm3.151500.1cm3.15cm0.101

3 −=−−−

−='s

The final image is 3.2 cm to the left of the center of the ball.

(b) With s1 = 20.0 cm: ( )( )( )

( )( ) ( )( )undefined

cm0.101cm0.201500.1cm0.20cm0.10500.1

1

=−−

='s

Because 's1 is undefined, no image is formed when the object is 20.0 cm to the left of the glass ball. (Alternatively, an image is formed an infinite distance to the left of the ball.) 106 ••• (a) Show that a small change dn in the index of refraction of a lens material produces a small change in the focal length df given approximately by df/f = –dn/(n – nair). (b) Use this result to estimate the focal length of a thin lens for blue light, for which n = 1.530, if the focal length for red light, for which n = 1.470, is 20.0 cm. Picture the Problem We can solve the lens maker’s equation for f and then differentiate with respect to n and simplify to obtain df/f = −dn/(n − nair).

Optical Images

1057

(a) The lens maker’s equation is: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

21air

1111rrn

nf

(1)

Differentiating both sides gives: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−=−

21air2

11rrn

dnfdf (2)

Diving equation (2) by equation (1) and simplifying yields:

air

air

air

air

air

21air

21air2

1111

11

1n

nnndn

nnndn

rrnn

rrndn

f

fdf

−=

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=−

Simplifying both sides yields:

airnndn

fdf

−−=

(b) Express the focal length for blue light in terms of the focal length for red light:

fff Δ+= redblue (1)

Approximate df/f by Δf/f and dn by Δn to obtain:

airnnn

ff

−−≈

ΔΔ

Solving for Δf yields : airnnnff

−−=

ΔΔ

Substitute for Δf in equation (1) to obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

−−=

airredred

air

redredblue

1nn

nf

nnnf

ff

Δ

Δ

Substitute numerical values and evaluate bluef :

( ) cm1700.1470.1

470.1530.11cm0.20blue =⎟⎠⎞

⎜⎝⎛

−−

−=f

Chapter 32

1058

107 ••• [SSM] The lateral magnification of a spherical mirror or a thin lens is given by m = –s′/s. Show that for objects of small horizontal extent, the longitudinal magnification is approximately –m2. Hint: Show that ds′/ds = – s′2/s2. Picture the Problem Examine the amount by which the image distance s′ changes due to a change in s. Solve the thin-lens equation for s′: 1

11−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

sfs'

Differentiate s′ with respect to s:

22

2

22

11

11111 m

ss'

ssf

sfdsd

dsds'

−=−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−

.length a have willlength ofobject an of image The 2 sms Δ−Δ