721

CHAPTER 34 REFLECTION, REFRACTION, AND OPTICS

Select odd-numbered solutions, marked with a dagger (†), appear in the Student SolutionsManual, available for purchase. Answers to all solutions below are underscored.

†34-1. Ideally, each plane mirror will reflect the same energy flux of 0.10 W/cm2. To start the canvas onfire requires forty times this flux, 4.0 W/cm2. Therefore, 40 mirrors are required.

34-2. Light is reflected from the cat in all directions. Some of those rays will reflect from the mirrorsin such a way that they may enter a viewers eye and be seen. Note that the reflected image to theright is left-right inverted such that each point of the cat is equally distant from the mirror aseach corresponding point on the image. The image on the top is similarly left-right inverted. Theviewer sees light coming directly from the cat to his eye and also sees the virtual light rayscoming from the two images.

†34-3. Consider the drawing showing the incident ray oflight approaching a point on the surface of themirror at an angle θ with respect to a line drawnperpendicular to the surface at that point. From thelaw of reflection, the reflected ray leaves the pointin a direction that is an angle θ with respect to theperpendicular line. Since the angle between theplane of the surface and the perpendicular line is90°, the angles β are equal to 90° − θ.

θ θββ

CHAPTER 34

722

34-4. Since the image distance is equal to the object distance, the camera should be focused at adistance of 2(1.2 m) = 2.4 m.

†34-5. The drawing at the right shows Figure 34.73 withsome added features. The line normal to the surfaceis rotated by 10° as the mirror is rotated by 10° andshown as a dashed line. The net result is that thereflected ray makes an angle φ with respect to theoriginal normal line. After the mirror is rotated, theincident ray makes an angle of θ + 10° with respectto the rotated normal line. Since the angle ofreflection is equal to the angle of incidence, φ = θ +10° + 10° = θ + 20°. Therefore, with respect to theoriginal reflected ray, we have rotated it by 20°.

34-6. For two mirrors placed at an angle θ with respect to each other. The number of images formed bythe mirrors is

360 3601 1 560

Nθ

° °= − = − =°

34-7. For two mirrors placed at an angle θ with respect to each other. The number of images formed bythe mirrors is

360 3601 1 745

Nθ

° °= − = − =°

34-8. All light rays coming from the sun are assumed to beparallel to one another. The angle of incidence for therays is 50°; and they are reflected from the mirror at anangle of 50°. Consider a ray that strikes the mirror atthe very top. It is reflected to the left-most point of thepatch of sunlight on the floor. The width of the patchwill equal the width of the mirror. Therefore, the areaof the patch will be

( ) ( ) 20.50 m 0.50 m0.21 m

tan 50° tan 50°HA hW W⎛ ⎞= = = =⎜ ⎟

⎝ ⎠

10°

φ

50°

50°

H

h

CHAPTER 34

723

†34-9. Let’s define the following lengths that will aid us indetermining the necessary height of the mirror H.h : your height (1.80 m)h′ : the height of your eyes above the floorh′′ : the length between your eyes and the top of yourhead = h − h′The drawing shows the minimum height of the mirrorsuch that a ray of light from the top of your head willstrike the mirror in such a way that its reflection willenter your eye. Note that the top of the mirror is locatedone-half the distance between your eyes and the top ofyour head. Similarly, a ray coming from the bottom ofyour foot will reflect from the bottom-most point on themirror and enter your eye. The bottom of the mirror islocated exactly halfway between your eyes and thefloor. Thus, the height of the mirror is

( ) ( )1 1 1 1 1 1.80 m 0.90 m2 2 2 2 2

H h h h h h′ ′′ ′ ′′= + = + = = =

Similarly, for the width of the mirror, we can draw a view from above as shown.Assume that you are standing directly in front of the mirror, so thatyour midline is directly across from the midline of the mirror. Asbefore, a ray from each shoulder reflects from the correspondingleft or right edge of the mirror and into your eyes as shown. Fromthe diagram, we see that the minimum width of the mirror is

( )

1 1 12 2 2

1 12 22 2

1 0.50 m 0.06 m = 0.31 m2

W w w w

W w w w w w

⎛ ⎞′ ′= − +⎜ ⎟⎝ ⎠

⎛ ⎞′ ′ ′= − + = +⎜ ⎟⎝ ⎠

= +

34-10. The speed of light in a material is given by Equation 34.4cvn

=

where the indices of refraction n are given in Table 34.1. Therefore, for crown glass, ncg = 1.52and vcg is

88

cgcg

3.00 × 10 m/s 1.97 × 10 m/s1.52

cvn

= = =

Similarly, for light flint glass and heavy flint glass, the speeds are8

8lf

lf8

8hf

hf

3.00 × 10 m/s 1.90 × 10 m/s1.58

3.00 × 10 m/s 1.81 × 10 m/s1.66

cvncv

n

= = =

= = =

†34-11. The relationship between speed v, wavelength λ, and frequency f is v = λf, so the frequency of thelight is

W

ww′

θθ

h

h′′

h′

H

CHAPTER 34

724

814

9

3.00 × 10 m/s 4.74 × 10 Hz633 × 10 m

cfλ −= = =

When the light is passing through the crown glass (n = 1.52), the frequency is not changed, but itsspeed and wavelength do change. The speed of light in the crown glass is

88

cgcg

3.00 × 10 m/s 1.97 × 10 m/s1.52

cvn

= = =

The wavelength of the light in crown glass is8

714

3.00 × 10 m/s 4.16 × 10 m or 416 nm(1.52) (4.74 × 10 Hz)

v cf nf

λ −= = = =

34-12.8

8

3.00 × 10 m/s 2.421.24 × 10 m/s

cnv

= = =

†34-13. The index of refraction for castor oil is 1.48, but it is only 1.36 in ethyl alcohol. To find the factorof the increase in speed, take the ratio of the speeds.

e e co

co e

co

1.48 1.091.36

cv n n

cv nn

= = = =

34-14. 1 1 2 2

1 112 1

2

sin sin

1sin sin sin sin 40 291.33

n n

nn

θ θ

θ θ− −

=

⎛ ⎞ ⎛ ⎞= = ° = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

†34-15. Since the index of refraction is inversely proportional to the speed of the waves in the media, wecan write a version of Snell’s Law for these sound waves as

1 1 2 2

1 21 2

1 122 1

1

sin sin 1 1 sin sin

1500 m/ssin sin sin sin 10 50340 m/s

n n

v v

vv

θ θ

θ θ

θ θ− −

=

=

⎛ ⎞ ⎛ ⎞= = ° = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

34-16. The maximum angle is 49°.

0 15 30 45 60 75 900

10

20

30

40

50

Ref

ract

ed A

ngle

Incident Angle

CHAPTER 34

725

†34-17. The Sun’s light passes through the space (n1 = 1.0000) between it and the Earth’s upperatmosphere making an angle of incidence θ1 on the atmosphere (n2 = 1.0003). The angle ofrefraction is θ2 = 39.000°. Snell’s Law gives the angle of incidence

1 121 2

1

1.0003 sin sin sin sin 39.000 39.0141.0000

nn

θ θ− −⎛ ⎞ ⎛ ⎞= = ° = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

34-18. 2.0 m 1.5 m1.33

ddn

′ = = =

34-19. Let’s use an index of refraction that falls in the middle of the two indices given.

( )12 1

2

sin sin 30 1.00 1.46, fused quartzsin sin 20.1

n n θθ

°= = =°

34-20. 5.0 cm 3.2 cm1.58

ddn

′ = = =

†34-21. The light rays are incident on the water-glass interface at an angle θ1 =30.0° where it refracts at an angle θ2 as shown. Given the three indices ofrefraction, the angle θ3 can be found by applying Snell’s Law at the twointerfaces.

( )

12 1

2

2 2 13 2 1

3 3 2

1 113 1

3

sin sin

sin sin sin

1.33sin sin sin sin 30.0 41.7 1.00

nn

n n nn n n

nn

θ θ

θ θ θ

θ θ− −

=

⎛ ⎞= = ⎜ ⎟

⎝ ⎠⎛ ⎞ ⎛ ⎞= = ° = °⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠34-22. The wavelength λ′ in the unknown clear material is

2nλλ′ =

where the index of refraction is1

2 12

sin = sin

n n θθ

Therefore,

( )121

2

550 nm 365 nmsin sin 451.00 sin 28sin

n n

λ λλ θθ

′ = = = =°°

34-23. 1 121 2

1

1.33 sin sin sin sin 30.0 421.00

nn

θ θ− −⎛ ⎞ ⎛ ⎞= = ° = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

34-24. The speed of light in the crown glass is c dvn t

= = where d is the thickness of the glass, n is the

index of refraction, and t is the transit time for the light in the glass. The difference in the transittimes for red and blue light is

( )

( )

blue red blue red

314

8

1.00 × 10 m 1.530 1.510 6.67 × 10 s3.00 × 10 m

dt t t n nc

−−

∆ = − = −

⎛ ⎞= − =⎜ ⎟⎝ ⎠

θ1

θ2

θ3

n1

n2

n3

CHAPTER 34

726

34-25. 1 1C

1 1sin sin 24.42.42n

θ − −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

34-26. 1 12 1

2

sin sin nn

θ θ− ⎛ ⎞= ⎜ ⎟

⎝ ⎠

( )23 2

3

12 11

3 2

1

sin sin 90

sin 90 sin sin

1.58 1.00sin 90 sin sin 70 1.27 1.00 1.58

nn

n nn n

θ θ

θ−

−

= ° −

⎛ ⎞⎛ ⎞= ° −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= ° − ° =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

This indicates that θ3 does not exist and total internal reflection occurs. The ray does not exit thesecond side of the prism.

34-27. 1 1 11 1 12 1

2 2 2

sin sin sin sin 30 sin 2 n n nn n n

θ θ− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = ° =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠For red light, n1 = 1.650, θ2 = 55.6°For green light, n1 = 1.660, θ2 = 56.1°For red light, n1 = 1.690, θ2 = 57.7°

34-28. 1 1C

1 1sin sin 41.81.50n

θ − −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

†34-29. (a) Consider light that is incident on an interface where n1 > n2. Depending on the angle ofincidence, some of the light will be reflected at the interface and some will be transmitted at arefracted angle θ2. Snell’s Law gives

1 1 2 2

21 2

1

sin sin

sin sin

n nnn

θ θ

θ θ

=

=

As the incident angle is varied with respect to the normal to the interface, a critical angle θc isreached where total internal reflection occurs and there is no transmitted light. The refractedangle is then 90°.

2 2c

1 1

sin sin 90n nn n

θ = ° =

(b) 2c

1

1 12c

1

sin

1.20sin sin 64.51.33

nn

nn

θ

θ − −

=

⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

34-30. C

C

1sin

1 1 1.2sin sin 56

n

n

θ

θ

=

= = =°

θ1θ2

90° − θ2

θ3

CHAPTER 34

727

34-31. 1 1C

1 1sin sin 41.11.52n

θ − −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

If the glass is immersed in water, then1 12

C1

1.33sin sin 61.01.52

nn

θ − −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

34-32. 1 12C

1

1.58sin sin1.52

nn

θ − −⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

The critical angle is undefined for this combination of materials. The critical angle for the flintglass-air interface is

1 1C

1 1sin sin 39.31.58n

θ − −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

The incidence angle within the crown glass is found via Snell’s Law.

1

(1.52) sin (1.58) sin (39.3 )1.58sin sin(39.3°) 41.11.52

θ

θ −

= °

⎛ ⎞= = °⎜ ⎟⎝ ⎠

So, for light that leaks from the crown glass at angles > 41.1°, the light will be totally internallyreflected.

34-33.

( )

2C

1

2 1 C

sin

sin 1.000 26 sin 89.4° 1.000 21

nn

n n

θ

θ

=

= = =34-34. Brewster’s Law (Equation 34.17) gives the relationship between the angle at which total

polarization occurs and the index of refraction.ptan tan (60.5 ) 1.77n θ= = ° =

†34-35. Brewster’s Law (Equation 34.17) gives the relationship between the angle at which totalpolarization occurs and the index of refraction.

( )p

1 1p

horizon p

tan

tan tan 1.33 53.1

90 90 53.1 36.9

n

n

θ

θθ θ

− −

=

= = = °

= ° − = ° − ° = °

34-36. The index of refraction of cubic zirconia may be found from the critical angle for total internalreflection.

C

C

1sin

1sin

n

n

θ

θ

=

=

Then, Brewster’s law gives the angle for total polarization,1 1 1

pC

1 1tan tan tan 65.6sin sin 27.0

nθθ

− − −⎛ ⎞ ⎛ ⎞= = = = °⎜ ⎟ ⎜ ⎟°⎝ ⎠⎝ ⎠

CHAPTER 34

728

†34-37. The incident light is refracted at the air-glass interface toan angle θ2 given by

1 112 1

2

1.0 sin sin sin sin 50 30.7 1.5 nn

θ θ− −⎛ ⎞ ⎛ ⎞= = ° = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

The hypotenuse of the triangle that has the side of length dshown in the drawing is

( ) ( )3 3(2.00 × 10 m) tan 50 tan 30.7 1.196 × 10 mL − −= ⎡ ° − ° ⎤ =⎣ ⎦To find d, the angle θ3 is found by applying Snell’s Law again.

1 123 2

3

1.5 sin sin sin sin 30.7 50 1.0 nn

θ θ− −⎛ ⎞ ⎛ ⎞= = ° = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

As expected, the final refracted ray is parallel to the initial incident ray. The distance d is then,

( )3

3 43

sin (90 )

sin (90 ) 1.196 × 10 m sin 40 7.7 × 10 m or 0.77 mm

dL

d L

θ

θ − −

° − =

= ° − = ° =

34-38. The first image occurs a distance l belowthe top surface of the glass (just as it doesfor a plane mirror). Let a represent thedistance between the reflected ray and thesecond parallel ray leaving the top surfaceas shown. The distance below the firstimage to the second image is representedby x. Here is a magnified image of therays within the glass:The incident ray approaches the surface atangle θ and is refracted at angle θ′. Thedistancea/2 = d tan θ′Then, a = 2d tan θ′The distance x is then given by

2 tan tan tan

a dx θθ θ

′= =

From Snell’s Law, there is a relationship between θ and θ′ and these angle are quite small, so thesmall angle approximation applies, i.e. tan θ ≈ θ and tan θ′ ≈ θ′. Therefore,

2 2 2d d dxn n

θ θθ θ

′ ′= = =

′

The distance from the top surface to the second image is 2dl x ln

+ = +

Similarly, the distance from the top surface to the third image is 2 42 2 d dl x l ln n

⎛ ⎞+ = + = +⎜ ⎟⎝ ⎠

t

d

θ1

θ3

θ2 L

x

a

θ′ θ′

a/2 a/2

d

CHAPTER 34

729

†34-39. Since the incident ray is polarized in the plane of incidence, no reflected ray will occur at theBrewster angle.

1 12B

1

1.33tan tan 53.11.00

nn

θ − −= = = °

34-40. (a) The lateral displacement of the ray that originatesfrom behind the glass is

1 22

sin ( )cos

td θ θθ

= −

and

1 2

1 2 1

sin ( )sin cos sin

d tl θ θθ θ θ

⎛ ⎞ −∆ = = ⎜ ⎟⎝ ⎠

When the angles are very small,cos 1 and sin θ θ θ≈ ≈Then,

1 2 2

1 1 1

1dl t tθ θ θθ θ θ

⎛ ⎞ ⎛ ⎞−∆ = = = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Applying Snell’s Law for small angles, θ1 = nθ2,11l tn

⎛ ⎞∆ = −⎜ ⎟⎝ ⎠

(b) 1 11 1 2.0 mm 0.67 mm1.5

l tn

⎛ ⎞ ⎛ ⎞∆ = − = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(c) 1 11 1 8.0 mm 2.7 mm1.5

l tn

⎛ ⎞ ⎛ ⎞∆ = − = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

34-41. The maximum reflection at the back of the sphere occurswhen a light ray enters at the top of the sphere as shown toa refracted angle θ2 = 45°, which is deduced from thegeometry shown. In this case, θ1 = 90° and

1

2

sin sin 90° 2 2sin sin 45° 2

n θθ

= = = =

The index of refraction must be at least 1, therefore1 2n≤ ≤

34-42. We can apply Snell’s Law to sound waves as wedo for light, with the index of refraction inverselyproportional to the speed of the wave in the givenmedium. Therefore,

2 1C

1 2

sin n vn v

θ = =

Consider the drawing of this situation. Thedistance d from the ship to the point at which thesound wave will undergo total reflection is

t

d

θ1

θ3

θ2

L

∆l

θ2

h

d

θC

CHAPTER 34

730

1 11C

2

340 m/s tan tan sin (200 m) tan sin 46.5 m = 47 m1500 m/s

vd h hv

θ − −⎡ ⎤⎛ ⎞ ⎡ ⎤⎛ ⎞= = = =⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎝ ⎠⎣ ⎦†34-43. Consider the drawing of the situation. A ray of light

originating in water (n = 1.33) is incident on the water-oilinterface. It is refracted at an angle θ′. If the ray is to betotally reflected at the oil-air interface, then θ′ ≤ θC. Themaximum value of θ occurs when θ′ = θC. Applying Snell’sLaw gives,

max

1 1max

sin sin 1 sin sin 1

1 1sin sin 48.81.33

C

n n

n n nn

n

θ θ

θ θ

θ − −

′ ′=

⎛ ⎞′ ′= = =⎜ ⎟′⎝ ⎠⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

The maximum angle has no dependence on n′.

34-44. The shark can be seen when the ray of light from theshark approaches the water-air interface at an angle lessthan the critical angle. This maximum angle was foundin the previous problem to be θC = 48.8°. Therefore,using the right triangle shown, the horizontal distance is

(5.0 m) tan (5.0 m) tan 48.8 5.7 mCx θ= = ° =†34-45. (a) Since the critical angle is given by

1C sin n

nθ − ′⎛ ⎞= ⎜ ⎟

⎝ ⎠the index of refraction n′ of the solution is

( ) ( )C sin 1.6640 sin 57.295 1.4002n n θ′ = = ° =(b) From the table, we see that the index increases by 0.0002 for each 0.10% increase inconcentration. Therefore, this is a linear relationship between index and concentrations (at leastfor the range of values given). The value found in part (a) is halfway between the index valuesgiven for concentrations 40.20 and 40.30%, therefore, after interpolating, the unknownconcentration is 40.25%.

34-46. Consider the angles shown in the drawing. For total internalreflection at the glass-air interface, the angle β = 90.0°. Fromthe drawing, the following relationships are present:

45.0 90.0 18045.0 (i)

= 90.0 45.0 (ii)

α θα θθ α θ

′+ ° + ° − = °′= ° +

′′ ′° − = ° −The critical value of θ′′ is given by

1 1C

1 sin sin 41.811.50

nn

θ − −′⎛ ⎞ ⎛ ⎞′′ = = = °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Applying equation (ii) gives C= 45.0 45.0 41.81 3.19 .θ θ′ ′′° − = ° − ° = °At the air-glass interface,

θ

θ′ θC

nn′

θθ′

θ′′αβ

45° 45°

90°

5.0 m

x

θC

CHAPTER 34

731

( )

1 2

1 12

1

sin sin

1.50sin sin sin sin 3.19 4.791.00

n n

nn

θ θ

θ θ− −

′=

⎛ ⎞ ⎛ ⎞′= = ° = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

†34-47. (a) At the first air-water interface, the ray is refracted at theangle θ′. Note that all normals to the surface of a sphereintersect at the center of the sphere, D. Because of this, as thedrawing shows, there are two equal, isosceles triangles. Thereflected ray at B has equal angles θ′ about the normal. Becausethe incident angle at the final water-air interface is θ′, Snell’sLaw gives the final refracted angle θ, which is equal to theinitial incident angle.(b) The drawing shows the angle of deflection at each of thethree points. The angles are evident from the drawing.(c) At the critical angle θC,

2 4 0

12

d dd ddd

θθ θθθ

′∆ = − =

′=

However, from Snell’s Law at the air-water interface,1 1 2 2 sin sin

sin sin n n

nθ θ

θ θ=

′=Taking the derivative of this, givescos cos

cos 1 cos 2

d n ddd n

θ θ θ θθ θθ θ

′ ′=′

= =′

Also, 2 2C C C2

1cos 1 sin 1 sinn

θ θ θ⎛ ⎞′ ′= − = − ⎜ ⎟⎝ ⎠

Therefore,

2C2

cos cos 1 cos 21 1 sin

n

C C

C

dd n

n

θ θθθ θ

θ

′= = =

′ ⎛ ⎞− ⎜ ⎟⎝ ⎠

2C2

1 1cos 1 sin2 nC nθ θ⎛ ⎞= − ⎜ ⎟

⎝ ⎠

( )2 2 2 2 2C C2

1 1 1cos 1 sin sin4 n 4C n nθ θ θ⎛ ⎞⎛ ⎞= − = −⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

( )

2 2 2C

2 2 2 2C

1 1cos sin4 4

3 1 1cos sin cos 4 4 4

C

C C

n

n

θ θ

θ θ θ

+ =

+ + =

θ′θ′

θ′

θ

θ

θ′BD

A

C

θ′θ′

θ′

θ

θ

θ′BD

A

C

θ − θ′

π − 2θ′

θ′

CHAPTER 34

732

( )

2 2

2 2

3 1 1cos 4 4 4

1cos 13

C

C

n

n

θ

θ

= −

= −

(d) Using the result from part (c)

( ) ( )1 2 1 21 1cos 1 cos (1.330) 13 3

59.585 59 35

C nθ − −= − = −

′= ° = °Then, C2( ) (180 2 )θ θ θ′ ′∆ = − + ° −where θ′ is found from Snell’s Law at the air–waterinterface,

C

1C

sin sin 1sin sin

n

n

θ θ

θ θ−

′=

⎛ ⎞′ = ⎜ ⎟⎝ ⎠

Therefore,

( )

1 1C C C

1C C

1

1 12 sin sin 180 2 sin sin

12 180 4 sin sin

12 59.585 180 4 sin sin 59.585 137.483 137 291.33

n n

n

θ θ θ

θ θ

− −

−

−

⎡ ⎤⎛ ⎞ ⎛ ⎞∆ = − + ° −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⎛ ⎞= + ° − ⎜ ⎟⎝ ⎠

⎛ ⎞ ′= ° + ° − ° = ° = °⎜ ⎟⎝ ⎠

(e) As shown in the drawing, violet light exits the drop and follows a trajectory that never entersthe eye, if the red light does reach the eye. The violet ray that the observer sees must come from adrop that is lower, so red appears higher in the sky than violet.The critical angle for violet light is

( ) ( )1 2 1 21 1cos 1 cos (1.342) 13 3

59.888 59 53

C nθ − −= − = −

′= ° = °Then,

( )

1 1C C C

1C C

1

1 12 sin sin 180 2 sin sin

12 180 4 sin sin

12 59.888 180 4 sin sin 59.888 139.235 139 141.342

n n

n

θ θ θ

θ θ

− −

−

−

⎡ ⎤⎛ ⎞ ⎛ ⎞∆ = − + ° −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⎛ ⎞= + ° − ⎜ ⎟⎝ ⎠

⎛ ⎞ ′= ° + ° − ° = ° = °⎜ ⎟⎝ ⎠

34-48. The radius of curvature is equal to twice the focal length, which may be determined usingEquation 34.21.1 1 1s s f

+ =′

CHAPTER 34

733

1 1 14.0 cm

1 4.0 cm28.0 cm

f

f R

R

+ =∞

= =

=†34-49. Three rays are drawn from the top of the

object, represented by the tip of an arrow.Ray 1 is directed parallel to the principalaxis and reflected along a line that ifextended behind the mirror, passesthrough the focal point f of the convexmirror. Ray 2 is directed toward the focalpoint. That ray reflects and is directedparallel to the principal axis. Ray 3 isdirected toward the center of curvature ata distance 2f behind the mirror. That rayreflects back on itself.The three reflected rays do not intersect on the object side of the mirror. If those reflected rays areextended behind the mirror, they intersect and form a virtual, upright, and reduced image at apoint between the mirror and the focal point. Applying the mirror equation gives

1

1 1 1

1 1 115.0 cm 40.0 cm

1 1 10.9 cm40.0 cm 15.0 cm

s s f

s

s−

+ =′

+ =′ −

⎛ ⎞′ = − = −⎜ ⎟−⎝ ⎠The virtual image is formed 10.9 cm behind the mirror. Note that when applying the mirrorequation to a convex mirror, the focal length is negative.

34-50. 1 1 1

1 1 1

2

s s f

s s fs f R

+ =′

+ =

= =The object should be placed at the radius of curvature for the concave mirror.

†34-51. Since f = R/2, the mirror equation may be written,1 1 2s s R

+ =′

If a real image is to be formed, then s′ must be positive.1 2 1 0

12

s R s

s R

= − >′

>

CHAPTER 34

734

For a virtual image, s′ is negative.1 2 1 0

12

s R s

s R

= − <′

<

34-52. The object (the candle) is placed withinthe focal length of the concave mirrorand forms an enlarged, upright, virtualimage.1 1 1

1 1 120.0 cm 15.0 cm60.0 cm

s f s

ss

= −′

= −′′ = −

†34-53. Since the radius of the ornament (a sphere) is 6.0 cm, the focal length is f = −3.0 cm. The negativesign is used because the focal length is on the opposite side of the reflecting surface than theobject (the candy cane). The mirror equation is used to find the image distance.1 1 1

1 1 13.0 cm 15.0 cm2.5 cm

s f s

ss

= −′

= −′ −′ = −

34-54. The moon is so far away that its object distance is essentially infinite. We’ll apply the mirrorequation for the two instances, recognizing that the focal length of the mirror does not change.Therefore,

1 1 2 2

1 2 2

2

2

1 1 1

1 1 1 1

1 1 1 1

1 1 106.0 cm 8.0 cm24 cm

s s f

s s s s

s s s

ss

+ =′

+ = +′ ′

+ − =′ ′∞

+ − =′

′ =

34-55. 1 1 1

1 1 125.0 cm 10.0 cm

16.7 cm

s f s

ss

= −′

= −− −

=34-56. The only case of the image forming on the same side of the spoon as the object is when the

woman is facing the concave side of the spoon and her eye is placed within the focal distance ofthe mirror. The relevant mirror equation is

f2f s′s

CHAPTER 34

735

1 1 2s s R

+ =′

For the two cases, we have1 1 2Concave:

7.0 cm1 1 2 Convex:

2.0 cm

s R

s R

+ =

+ = −−

For these two equations with two unknowns, we find1 1 1 1

2.0 cm 7.0 cm2 1 1

7.0 cm 2.0 cm3.11 cm = 3.1 cm

s s

ss

+ = − −−

= − +

=and

1 1 2 3.11 2.0 cm

11.2 cmR

R

+ = −−

=†34-57. The magnification M of the mirror gives

1.5

1.5 1.5(20.0 cm) = 30.0 cm

sMs

s s

′= − =

′ = − = − −Therefore, by knowing both the image and object distances the mirror equation can be used tofind the radius of curvature.1 1 2

1 1 220.0 cm 30.0 cm

120.0 cm

s s R

RR

+ =′

+ =−

=The mirror is concave.

34-58. 1 1 2

1 2 1 2 14.5 cm 15 cm

1.956 cm = 1.96 cm

s s R

s R ss

+ =′

= − = −′ −′ = − −

1.956 cm 0.1315 cm

M −= − = −

The image size is 1/M = 7.6 times smaller than your face.34-59. (a) Consider the first mirror.

1 1 1s s f

+ =′

CHAPTER 34

736

1 1 1 1 115.0 cm 20.0 cm

60.0 cms f ss

= − = −′′ =

(b) The image formed by the first mirror becomes the object for the second mirror. The objectdistance is s = 80.0 cm − 60.0 cm = 20.0 cm. The final image is formed at1 1 1 1 1

12.0 cm 20.0 cm30.0 cm

s f ss

= − = −′′ =

34-60. For the concave mirror, the image distance is found from the mirror equation.1 1 1

1 1 1 1 130.0 cm 40.0 cm

120.0 cm

s s f

s f ss

+ =′

= − = −′′ =

This image then becomes the object for the convex mirror.1 1 1 1 1

30.0 cm 70.0 cm21.0 cm

s f ss

= − = −′ −′ = −

†34-61. Applying the thin lens formula Equation 34.27, we find

( ) ( )1 2

1 1 1 1 11 1.45 110 mm 6.0 mm

8.3 mm

nf R Rf

⎛ ⎞ ⎛ ⎞= − + = − +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

=

34-62. (a) ( ) ( )1 2

1 1 1 1 11 1.58 115 cm

25.9 cm = 26 cm

nf R Rf

⎛ ⎞ ⎛ ⎞= − + = − +⎜ ⎟ ⎜ ⎟∞⎝ ⎠⎝ ⎠=

(b) 1 1 1

1 1 1 1 125.9 cm 40.0 cm

= 73 cm

s s f

s f ss

+ =′

= − = −′′

34-63. ( ) ( ) ( )1 2

2 11 1 1 21 1n

n nf R R R R

−⎛ ⎞ ⎛ ⎞= − + = − =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

( ) ( )2 1 2 1.52 1 (20 cm) 20.8 cm 21 cmR n f= − = − = =

34-64. ( )

( ) ( )

1 2

2 1

2

1 1 11

1 1 1 1 11 16 cm 1.66 1 27 cm

17 cm

nf R R

R f n RR

⎛ ⎞= − +⎜ ⎟

⎝ ⎠

= − = −− −

=

CHAPTER 34

737

34-65. ( ) ( )1 2

1 1 1 1 11 1.58 116 cm 12 cm

12 cm

nf R Rf

⎛ ⎞ ⎛ ⎞= − + = − +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

=

34-66. ( ) ( )1 2

1 1 1 1 11 1.43 12.5 cm 3.0 cm

35 cm

nf R Rf

⎛ ⎞ ⎛ ⎞= − + = − +⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠=

†34-67. The ray diagram for this situation is similar to that shown in Figure 34.51. Apply Equation 34.28using the sign conventions illustrated in Figure 34.48. Assuming the object is placed 15 cm to theleft of the concave lens with f = −20 cm,1 1 1

1 1 1 1 120 cm 15 cm

8.57 cm = 8.6 cm

s s f

s f ss

+ =′

= − = −′ −′ = − −

The image forms 8.6 cm to the left of the lens. The image is virtual since it is formed by theintersection of virtual rays on the left side of the lens. The magnification is given by Equation34.30,

8.57 cm 0.5715 cm

sMs′ −= − = − =

The image is smaller and upright.

34-68. If the object distance can be considered infinite; and the image has ,hf

θ ′ = then

( ) ( ) rad150 mm 0.52 1.36 mm180

h f πθ ⎛ ⎞′= = ° =⎜ ⎟°⎝ ⎠

34-69. 1 1 1

1 1 1 1 10.130 m 2.0 m

0.139 m = 14 cm

s s f

s f ss

+ =′

= − = −′

′ =2.0 m 14

0.139 msMs′

= − = − =

34-70. 2.0

2.0

sMs

s s

′= − =

′ = −

( )

1 1 12.0

2 1 2

1 1 20 cm 10 cm2 2

s s f

s s f

s f

+ =−

− =

= = =

CHAPTER 34

738

†34-71. The ray diagram drawn for this situationshows that the image will be inverted, real,and enlarged. Applying the lens formulagives,1 1 1

1 1 1 1 118 cm 30 cm

45 cm

s s f

s f ss

+ =′

= − = −′′ =

34-72. The Gaussian lens makers formula is derived in most optics texts and may also be found on theinternet. The problem asks for an argument for the modification of Equation 34.27 when the lensis surrounding by a medium other than air. When Snell’s Law applied to the interface between thesurrounding medium and the lens gives 0 0sin sin .n nθ θ= This gives,

00

sin sin nn

θ θ=

If the lens is in air, n0 = 1; and you have the situation that gives Equation 34.27. Since we havereplaced n by n/n0 in Snell’s Law and Snell’s Law is the basis for Equation 34.27, it makes senseto simply make the same substitution to give:

0 1 2

1 1 11nf n R R

⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠†34-73. Figure 34.81 shows two converging lenses in contact. Assume an object is placed a distance s to

the left of the first lens, the image will be formed at a distance s1 to the right of the first lens. Thedistances are related by the lens formula.

1 1

1 1 1f s s

= +

This image then becomes the object for the second lens. The final image is formed at a distances2, which may again be found using the lens formula with the proper sign convention.

2 1 2 2 1

1 1 1 1 1f s s s s

= + = −−

Since both equations contain s1, we can further relate both equations.

1 1

2 2 1 2 1

1 2 1 2

1 1 1

1 1 1 1 1 1

1 1 1 1 1

s f s

f s s s f s

f f s s f

= −

= − = − +

+ = + =

34-74. (a) 1 1 1 1 125 cm 80 cm

36.36 cm = 36 cms f ss

= − = −′′ =

(b) The image formed in part (a) becomes the object in part (b). The object distance for themirror will be 60 cm − 36.36 cm = 23.64 cm. Then, using the mirror equation, we can find wherethe second image is formed.

f f

CHAPTER 34

739

m

1 1 1 1 120 cm 23.64 cm

= 130 cms f ss

= − = −′′ ′′′

(c) The image formed from the mirror becomes an object for the lens again. The object distanceis negative since it is located 129.89 cm − 60 cm = 69.89 cm to the left of the lens, which in thiscase is a negative distance with regard to the lens formula.1 1 1 1 1

25 cm 69.89 cm18.41 cm = 18 cm

s f ss

= − = +′′′ ′′′′′ =

The final image forms 18 cm to the right of the lens.

34-75. (a) 1 1 1 1 130 cm 20 cm

60 cm s f ss

= − = −′′ = −

(b) 1 1 1 1 130 cm 70 cm

21 cms f ss

= − = − −′′ ′′′ = −

34-76. 1 1 1 13 32

3

s f ff

s f

= − =′

′ =This image distance is 2f to the right of the second lens.1 1 1 3

2 223

s f f f

s f

= + =′′

′′ =

The final image is formed at a distance 2f/3 to the left of the left lens.†34-77. Consider Figure 34.84. The usual image and object distances can be written: s = x + f and s′ =

x′ + f. Using these expressions in the lens formula gives,

( ) ( )

( )( ) ( )

2

2

2 2

2

1 1 1 1 1 2

1 2

2

x f x f x x ff s s x f x f x f x f xx xf x f f

x x ff xx f x x f

f x x f xx f x x f

xx f

′ ′+ + + + += + = + = =′ ′ ′ ′ ′+ + + + + + +

′+ +=′ ′+ + +

′ ′ ′+ + = + + +

′ =

34-78. From Problem 34-73,

1 2

2 1

2

1 1 1

1 1 1 1 122.0 cm 12.0 cm

26.4 cm

f f f

f f ff

+ =

= − = −

= −

CHAPTER 34

740

†34-79. First find the image for the mirror with the object placed at 15 cm in front of the mirror. Theimage is formed at1 1 1 2 1 2 1

10.0 cm 15.0 cm3.75 cm

s f s R ss

= − = − = −′ −′ = −

This indicates that the image is formed 3.75 cm behind the mirror, which is 23.75 cm from thelens. This image becomes the object for the lens. We then apply the lens formula to find thedistance to the final image.1 1 1 1 1

25.0 cm 23.75 cm475 cm

s f ss

= − = −′′ = −

The final image is 475 cm behind the lens, which is also behind the mirror.34-80. Assume the lens with a focal length f1 = 40 cm is to the left of the lens with a focal length

f2 = 60 cm. If parallel rays of light approach from the left, the first lens focuses the light at a point40 cm to its right, which is 30 cm to the right of the right lens. That image becomes the object forthe lens on the right, which focuses the light at a distance given by the lens formula.1 1 1 1 1

60 cm 30 cm20 cm

s f ss

= − = −′ −′ =

This is located 20 cm to the right of the right lens. In this case of parallel rays of lightapproaching from the left, the effective focal length measured from the center of the two lenssystem is located atfleft = 20 cm − 5 cm = 15 cm to the right of the system centerIf parallel rays of light approach from the right, the right lens focuses the light at a point 60 cm toits left, which is 50 cm to the left of the left lens. That image becomes the object for the lens onthe left, which focuses the light at a distance given by the lens formula.

1

1 1 1 1 140 cm 50 cm

22.22 cm = 22 cms f ss

= − = −′ −′ =

This is located 22 cm to the left of the left lens. In this case of parallel rays of light approachingfrom the right, the effective focal length measured from the center of the two lens system islocated atfright = 22 cm − 5 cm = 17 cm to the left of the system center

†34-81. From the drawing, the following relationships can be shown:θ1 = α + β (i)θ2 =β − α′ (ii)Snell’s Law applied at the interface is

1 1 2 2 sin sin n nθ θ=The angles are generally quite small; and the small angle approximation yields,

1 1 2 2.n nθ θ= Multiplying equation (i) by n1 gives n1θ1 = n1α + n1β; and equation (ii) multiplied n2

gives n2θ2 = n2β − n2α′. These two equations are equal to each other, so

CHAPTER 34

741

( )1 1 2 2

1 1 2 2

1 2 2 1

+ n n n nh h hn n n ns R s

n n n ns s R

α β β α′= −

+ − = −′

−+ =′

In the second line above, we have used the small angle approximation that tan α = α = h/s andsimilar relations for the other angles.

34-82. At the first interface, the result of Problem 34-81 gives,1 2 2 1

1 1 1

n n n ns s R

−+ =′

(i)

Similarly, at the second interface,2 1 1 2

2 2 2

n n n ns s R

−+ =′

(ii)

Assume the thickness of the lens is d, then the object distance at the second interface is properlyrelated to the image distance resulting from refraction at the first interface as s2 = d − s1′.Substitute this relationship into equation (ii) and add equations (i) and (ii) together.

1 2 2 1 2 1 1 2

1 1 1 2 1 2

n n n n n n n ns s d s s R R

− −+ + + = +′ ′ ′−

In the thin lens approximation d is approximately zero, so

( )

1 2 2 1 2 1 1 2

1 1 1 2 1 2

1 12 1

1 2 1 2

1 1 (iii)

n n n n n n n ns s s s R R

n n n ns s R R

− −+ − + = +′ ′ ′

⎛ ⎞+ = − −⎜ ⎟′ ⎝ ⎠

Since n1 is 1 for air and s1 = s, the original object distance, and s2′ = s′, the final image distance,we can simplify equation (iii) by letting n2 = n to give

( )1 2

1 1 1 11ns s R R

⎛ ⎞+ = − −⎜ ⎟′ ⎝ ⎠

The focal length is defined for an object placed at infinity or for an image at infinity, so

θ2

α′α

θ1

βh

CHAPTER 34

742

1 1 1 1 1f s s

= + = +′∞ ∞

Therefore,

( )1 2

1 1 11 (34.27)

and1 1 1 (34.28)

nf R R

f s s

⎛ ⎞= − −⎜ ⎟

⎝ ⎠

= +′

†34-83. Apply the equation from Problem 34-81,1 2 2 1n n n ns s R

−+ =′

at the first air-glass interface, with n1 = 1 (for air) and s = ∝ (since the Sun is very far away andthe light comes in as parallel rays).

2 2 1 1

1.60 1.60 1.00 17.5 cm

20.0 cm

n n n ns R s

ss

−= −′

−= −′ ∞

′ =The image distance then becomes the object distance for the second interface.

2 2 1 1

1.00 1.00 1.60 1.007.5 cm 20.0 cm 15.0 cm

3.57 cm

n n n ns R s

ss

−= −′

−= −′ −

′ = −The final image is observed at 7.5 cm − 3.57 cm = 3.9 cm to the right of the center of the ball.

34-84. The small segment is considered to be part of a lens with aradius of curvature R and focal length f. The lens makers’formula (Equation 34.27) is applied to this lens segment.

( ) ( )

( )1 2 1

1 1 1 1 1 11 1

1

nn nf R R R R

R f n

⎛ ⎞ ⎛ ⎞ −= − + = − + =⎜ ⎟ ⎜ ⎟∞⎝ ⎠ ⎝ ⎠= −

Consider the drawing.

1 1

sin

sin sin( 1)

rR

r rR f n

θ

θ − −

=

⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟ −⎝ ⎠ ⎝ ⎠†34-85. The camera is designed to focus the image at the plane of the film in the camera, so the image

distance is adjustable so that a range of object distances can be accommodated. When the lens is50 mm or 5.0 cm from the film, the object distance is1 1 1 1 1 0

5.0 cm 5.0 cms f ss

= − = − =′

= ∞

Rθ

θ

r

f

CHAPTER 34

743

When the lens is 62 mm or 6.2 cm from the film, the object distance is1 1 1 1 1

5.0 cm 6.2 cm25.8 cm

s f ss

= − = −′

=So, the camera can form a sharp image for object distances ranging from 25.8 cm to infinity.

34-86. When the object is 20 cm from the lens,1 1 1 1 1

1.5 cm 20 cm1.6 cm

s f ss

= − = −′′ =

When the object is at infinity,1 1 1 1 1

1.5 cm1.5 cm

s f ss

= − = −′ ∞′ =

†34-87. The f number is defined as the focal length of the lens divided by its diameter, f = f/d. The irischanges the effective diameter of the lens, but not its focal length. Since we want to relate the fnumber to an exposure time and the f number is related to the effective diameter, let’s define anew parameter called the exposure E, which is the product of the amount of light incident on thefilm (which is inversely proportional to the area) and the exposure time, E = At. For the first fnumber, 1.7, the exposure time is 1/250 s, therefore the exposure is

( )222 11

1 1 1 1 1 1

/f4 4

fdE A t R t t tπππ= = = =

Similarly, for the second f number, the same exposure results when the proper exposure time isused. Therefore, the exposure equations for both f numbers are set equal to each other. We thensolve for the second exposure time for an f number of 8.

( ) ( )

( )

2 21 2

1 2

2 22 1 1 2

22

22 12 2 2

1

/f /f4 4

f f18 s

f 250 8.9 × 10 s(1.7)f

f ft t

t t

tt

π π

−

=

=

⎛ ⎞⎜ ⎟⎝ ⎠= = =

34-88. (a) Since the image must be focused on the retina, the focal length of the eye is equal to theimage distance 2.2 cm for an object placed at infinity.(b) For an object placed at 25 cm from the eye, the focal length must be1 1 1 1 1

25 cm 2.2 cm2.0 cm

f s sf

= + = +′

=

34-89. (a)1 2

2 1

2

1 1 1

1 1 1 1 12.2 cm 2.0 cm

22 cm

f f f

f f ff

+ =

= − = −

= −This is a diverging lens.

CHAPTER 34

744

(b) ( )

( ) ( )

1 2 1 2

2 1

2 1

2

1 1 1 1 11

1 1 1

1 1 1 1 11 22 cm 1.33 1 0.80 cm

0.72 cm

n nnf R R R Rn nR f R

R f n RR

⎛ ⎞ − −= − + = +⎜ ⎟⎝ ⎠

− −= −

= − = −− − − −

= −34-90. The focal length of the contact lens in Problem 34-66 is 35 cm. The lens is convex and

converging. This lens is designed for a far-sighted person. The number of diopters is the inverseof the focal length in meters, so 1/0.35 m = +2.9 diopters.

†34-91. The angular magnification is given by25 cm

fθθ

′=

where f can be determined from the lens formula from the case when the object distance isinfinite, then f = s and θ′ = h/f = h/s. For an image distance s′ = −25 cm, s is given by

( )

1 1 125 cm

25 cm25 cm

s ff

sf

= +

=+

Then,( )

( )25 cm 25 cm 25 cm 1

25 cm25 cm

hh ffs f fhf

θθ θ

′ += = = = +

+34-92. The net angular magnification is, according to Equation 34.33,

ocular

25 cm 550s sMs f s

θθ

′ ′ ′= = =

where s is found from the lens formula.1 1 1

0.40 cm 22.4 cm0.407 cm

ss

= −

=The ocular focal length is then

ocular

ocular

25 cm 550

25 cm 25 cm 22.4 cm 2.5 cm550 550 0.407 cm

sf s

sfs

′=

′ ⎛ ⎞= = =⎜ ⎟⎝ ⎠

Then, then the ocular magnification is

ocular25 cm 102.5 cm

M = =

†34-93. The magnification is given by Equation 34.35, except that the objective and ocular are reversed.ocular

objective

2.5 cm 1160 cm 64

fMf

θθ

′= = = =

CHAPTER 34

745

34-94. objective

ocular

90.0 cm 721.25 cm

fM

fθθ

′= = = =

34-95. objective

ocular

1680 cm 1344 13401.25 cm

fM

fθθ

′= = = = =

34-96. Since the light collected is proportional to the area of the mirror, we simply take the ratio of thenew proposed mirror to that of the existing six mirrors.

( )2

2

(6.5 m)enhancement = 2.26 1.8 m

=

†34-97. The telescope designed by Galileohas the focal lengths of the objectiveand ocular lenses aligned at the samepoint to the right of the lenses asshown. A image formed by theconverging ocular lens becomes avirtual object for the divergingobjective lens.

The objective lens then refracts the rays from the virtual object, producing an upright image thatappears to be at infinity. The original rays come from infinity at an angle θ and exit the system atan angle θ′. Therefore, the angular magnification is

ocular

objective

fMf

θθ

′= = −

The negative sign indicates that the image is erect, since focular is negative.34-98. The angles are shown in the drawing. The first

angle of reflection is 60°. The ray forms oneside of a triangle that is opposite the 145°angle. The angle on the lower left must be 30°since it must be added to the angle of reflectionto give 90°. Since the three angle of a trianglemust add to 180°, the third angle of the triangleis 5°. This means that the second angle ofincidence must be 85°. Since the angle ofincidence is equal to the angle of reflection, thefinal reflected angle is also 85°.

34-99. (a) 12 1

2

12

sin sin

1.0sin sin 50 30.7 311.5

nn

θ θ

θ −

=

⎛ ⎞= ° = ° = °⎜ ⎟⎝ ⎠

(b) 13

1.5sin sin 30.7 501.0

θ − ⎛ ⎞= ° = °⎜ ⎟⎝ ⎠

foc

fobj

virtualobject

θ

θ′

30°

60°5°

85°

CHAPTER 34

746

34-100. The situation is shown in the drawing. The refracted rayexits the water at an angle θ2 given by

12 1

2

12

sin sin

1.33sin sin 40 58.7 591.0

nn

θ θ

θ −

=

⎛ ⎞= ° = ° = °⎜ ⎟⎝ ⎠

The distance from the corner to the point where the rayexits the water is L, which can be found from

1

1

tan

tan

Ld

L d

θ

θ

=

=The apparent depth d′ can then be determined.

( )2

1

2 2

tan

2.0 m tan 40 tan 1.02 m = 1.0 mtan tan tan 58.7

Ld

dLd

θ

θθ θ

=′

°′ = = = =

°†34-101. Consider the drawing. The refracted ray is parallel to the base

of the equilateral triangle, so θ2 + 60° = 90°, so θ2 = 30°. ApplySnell’s Law to find the incident angle.

21 2

1

11

sin sin

1.66sin sin 30 561.0

nn

θ θ

θ −

=

⎛ ⎞= ° = °⎜ ⎟⎝ ⎠

34-102. ( )21 2

1

sin sin 72= 1.0 1.34sin sin 45

n n θθ

°= =°

†34-103. This is similar to Example 4. A ray of light leaving the bottomof the beaker undergoes two refractions, one at the water-oilinterface, and one at the oil-air interface as shown in thedrawing. The thickness of each layer is 10 cm. If one were inthe oil looking at the refracted ray, the apparent depth would be

( )2

1

1.4810.0 cm 11.13 cm1.33

nd dn

′ = = =

So, the apparent depth would be greater than the actual depth.To find the apparent depth one sees looking from above thecontainer in air, we simply reapply the above equation.

( )3

2

1.0011.13 cm 7.5 cm1.48

nd dn

′′ ′= = =

34-104. There are two refractions. The first is at the water–glass interface.1

2 12

sin sin nn

θ θ=

The second is at the glass–air interface.

θ1

θ2

L

d′d

θ1 θ2

60°

θ1

θ2

θ3

d′′

d′

d

CHAPTER 34

747

2 2 1 13 2 1 1

3 3 2 3

13

sin sin sin sin

1.33sin sin 45.0 701.00

n n n nn n n n

θ θ θ θ

θ −

⎛ ⎞⎛ ⎞= = = ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎛ ⎞= ° = °⎜ ⎟⎝ ⎠

†34-105. The drawing shows the relevant angles for thesituation.The angle α is related to the angle of incidence θ asfollows.180 90 (90 ) 0α θα θ

° − ° − − ° − ==

The requirement for total internal reflection is that

critical

1 1critical

1sin

1 1sin sin 38.68 38.71.60

n

n

θ

θ − −

=

⎛ ⎞ ⎛ ⎞= = = ° = °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Total internal reflection will occur when θ ≥ θcritical. Therefore, the requirement is that α ≥ 38.7°.34-106. If we assume that the light emits light in all directions, then the dark portions of the water are

produced by total internal reflection (because the light does not escape the pool). The criticalangle for the water–air interface is

1 1critical

1 1sin sin 48.75 ,1.33n

θ − −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

which determines whether light is reflected or transmitted. The radius of the circle 2.0 metersabove the light is then

critical

critical

tan .2.0 m

(2.0 m) tan (2.0 m) tan (48.75 ) 2.3 m

r

r

θ

θ

=

= = ° =34-107.

Mirror Object distance Image characteristicsConcave s < f

f < s < 2fs > 2f

Virtual, upright, and magnifiedReal, inverted, and magnifiedReal, inverted, and reduced

Convex all Virtual, upright, and reduced34-108. If a distant object (s ≈ ∞) is placed in front of a convex mirror, the virtual image appears at the

focal point of the mirror. Therefore, since the radius of curvature R = 2f, R = 2(1.0 m) = 2.0 m.†34-109. A convex mirror always produces virtual, upright, and reduced images, so the image is behind the

mirror. To find its location, use the mirror formula. The focal distance is equal to one-half theradius of curvature,f = 0.5R = (0.5)(−4.0 m) = −2.0 m1 1 1 1 1

2.0 m 12 m1.7 m

s f ss

= − = −′ −′ = −

The magnification of the mirror is given by Equation 34.30.

θ

90° − θ

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748

1.7 m 0.14212 m

sMs′ −= − = − =

The magnification is positive, which indicates that the image is upright. The height of the imageis

1.7 m 1.7 m 0.24 m12 m

h Mh −⎛ ⎞′ = = − =⎜ ⎟⎝ ⎠

34-110. The focal length of each mirror is one-half of the radius, or 30 cm. Light from the candle reflectsfrom the concave mirror to form an image at1 1 1 1 1

30 cm 40 cm120 cm

s f ss

= − = −′′ =

This image then becomes the object, located at 120 cm − 50 cm = 70 cm, behind the convexmirror, which forms the final image at1 1 1 1 1

30 cm 70 cm52.5 cm

s f ss

= − = −′ − −′ = −

The final image is located 52.5 cm behind the convex mirror.34-111.

Lens Object distance Image characteristicsConcave s < f

s > fVirtual, upright, and reducedVirtual, upright, and reduced

Convex s < ff < s < 2fs > 2f

Virtual, upright, and magnifiedReal, inverted, magnifiedReal, inverted, reduced

34-112. The image is formed by the intersection ofvirtual rays as shown in the diagram. Thefirst ray leaves the top of the candle andtravels parallel to the principle axis. Theray is refracted by the lens along a line thatextends backward through the focal point.The second ray passes directly through thecenter of the lens and it is not refracted.The third ray is directed toward the focalpoint on the opposite side of the lens.This ray is refracted and travels parallel to the principle axis. The intersection of the three raysproduces a virtual image between the lens and the focal point that is upright and reduced.

CHAPTER 34

749

34-113. The image is formed by the intersection ofvirtual rays as shown in the diagram. Thefirst ray leaves the top of the light bulb andtravels parallel to the principle axis. The rayis refracted by the lens along a line thatpasses through the focal point on theopposite side of the lens. The second raypasses directly through the center of the lensand it is not refracted. The intersection of thetwo rays produces a virtual image betweenthe lens and the focal point that is uprightand enlarged. The image distance is1 1 1 1 1

30 cm 15 cm30 cm

s f ss

= − = −′′ = −

34-114. Apply the lens equation three times. Assume the light bulb is to the left of the first lens. For thefirst lens, the image is produced at

1 1

1

1 1 1 1 160 cm 20 cm

30 cms f ss

= − = −′′ = −

This becomes the object for the second lens with an object distance of 30 cm + 12 cm = 42 cm.

2 2

2

1 1 1 1 160 cm 42 cm

140 cms f ss

= − = −′′ = −

This image becomes the object for the third lens. The object distance is 140 cm + 12 cm =152 cm.

3 3

3

1 1 1 1 160 cm 152 cm

99 cms f ss

= − = −′′ =

The final image is located 99 cm to the right of the right lens.The magnification is the product of the magnifications of each lens.

31 21 2 3

1 2 3

30 cm 140 99 3.320 cm 42 162

ss sM M M Ms s s

′⎛ ⎞′ ′⎛ ⎞ ⎛ ⎞ − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = − − − = − − − = −⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

34-115. (a) For an image to be formed at infinity, the object must be placed at the focal point for theocular lens. This distance is s′ =180 mm + 25 mm = 205 mm(b) The objective forms an image at a distance s′ given by1 1 1 1 1

1.9 mm 205 mm1.9 mm

s f ss

= − = −′

=which is the focal point of the objective. The angular magnification is

oc

25 25 205 mm 10825 mm 1.9 mm

sMf s

′= = =