Solucionario Capitulo 36 - Paul E. Tippens

Chapter 36. Lenses and Optical Instruments Physics, 6th Edition

Chapter 36. Lenses and Optical Instruments

Focal Length and the Lensmaker’s Equation (Assume n = 1.50 unless told otherwise.)

36-1. A plano-convex lens is to be constructed out of glass so that it has a focal length of 40

cm. What is the radius of curvature of the curved surface?

1 2

1 1 1 1 1( 1) (1.5 1) ;40 cm

nf R R

11

1 1(0.5) ; 0.5(40 cm);40 cm

RR

R1 = 20.0 cm

36-2. If one uses a glass, double-convex lens to obtain a focal length of 30 cm, what must be

the curvature of each convex surface? [ R1 = R2 = R ]

1 1 1 2( 1) (1.5 1) ; ;n R ff R R R

R = 30 cm

36-3. The curved surface of a plano-concave lens has a radius of –12 cm. What is the focal

length if the lens is made from a material with a refractive index of 1.54.

1 2

1 1 1 1 1( 1) (1.54 1) ;-12 cm

nf R R

497

R2R1

R2 =

R1 = ?

n = 1.5

R2 =

R1


1 1 1 0.540(1.54 1) ; ;12 cm 12 cmf f

f = -22.2 cm

36-4. A converging meniscus lens (n = 1.5) has a concave surface whose radius is –20 cm and a

convex surface whose radius is 12 cm. What is the focal length?

1 1 1 20 cm + 12 cm(1.5 1) (0.5) ;12 cm 20 cm (12 cm)(-20 cm)f

f = 30 cm

36-5. A converging lens such as shown in Fig. 36-8a is made of glass. The first surface has a

radius of 15 cm and the second surface has a radius of 10 cm. What is the focal length?

Both surfaces are convex, and therefore positive: ( R1 = +15 cm, R2 = + 10 cm, n = 1.5 )

1 1 1 10 cm + 15 cm(1.5 1) (0.5) ;15 cm 10 cm (10 cm)(15 cm)f

f = 12 cm

36-6. A meniscus lens has a convex surface whose radius is 20 cm and a concave surface

whose radius is – 30 cm. What is the focal length if the refractive index is 1.54.

Given: R1 = 20 cm, R2 = -30 cm, n = 1.54: Find: f = ?

1 1 1 30 cm + 20 cm(1.54 1) (0.54) ;20 cm -30 cm (20 cm)(-30 cm)f

f = 1.11 m

36-7. A plano-convex lens is ground from crown glass (n = 1.52). What should be the radius of

the curved surface if the desired focal length is to be 400 mm?

498

R2 =

f = 400 mm

n = 1.52


1 2 1

1 1 1 1 1 1( 1) ; (1.52 1) ;400 mm

nf R R R

11

1 1(0.52) ; 0.52(400 mm);400 mm

RR

R1 = 208 mm

36-8. The magnitudes of the concave and convex surfaces of a glass lens are 200 and 600 mm,

respectively. What is the focal length? Is it diverging or converging?

Given: R1 = -20 cm, R2 = +60 cm, n = 1.5: Find: f = ?

1 1 1 60 cm - 20 cm(1.5 1) (0.5) ;-20 cm 60 cm (-20 cm)(60 cm)f

f = -60 cm

The lens is a diverging lens, since the focal length is negative.

36-9. A plastic lens (n = 1.54) has a convex surface of radius 25 cm and a concave surface of

–70 cm. What is the focal length? Is it diverging or converging?

1 1 1 -70 cm + 25 cm(1.54 1) (0.54) ;25 cm -70 cm (25 cm)(-70 cm)f

f = +72.0 cm

The lens is a converging lens, since the focal length is positive.

Images Formed by Thin Lenses

499


36-10. A 7-cm tall pencil is placed 35 cm from a thin converging lens of focal length 25 cm.

What are the nature, size, and location of the image formed?

(35 cm)(25 cm) ;35 cm - 25 cm

pfqp f

q = 87.5 cm

' ' (87.5 cm)(7 m); '35 cm

q y qyM yp y p

q = +87.5 cm, y’ = -17.5 cm, real, inverted, and enlarged.

36-11. An object 8 cm high is placed 30 cm from a thin converging lens of focal length 12 cm.

What are the nature, size, and location of the image formed?

(30 cm)(12 cm) ;30 cm - 12 cm

pfqp f

q = 20 cm

' ' (20 cm)(8 m); '30 cm

q y qyM yp y p

q = +20.0 cm, y’ = -5.33 cm, real, inverted, and diminished.

36-12. A virtual, erect image appears to be located 40 cm in front of a lens of focal length 15

cm. What is the object distance? The focal length must be positive (converging), since

real objects cannot produce erect virtual images farther away from lens than the object.

36-12. (Cont.). Find object distance for given values:

500

2FF

2FF

2FF


( 40 cm)(15 cm) ;-40 cm - 15 cm

qfpq f

p = 10.9 cm

Note: You should show that the magnification is +2.67.

36-13. A 50 mm tall object is placed 12 cm from a converging lens of focal length 20 cm. What

are the nature, size, and location of the image? (See drawing for Prob. 36-12 above.)

(12 cm)(20 cm) ;12 cm - 20 cm

q q = -30 cm;

' ' ( 30 cm)(50 mm); '12 cm

q y qyM yp y p

q = -30.0 cm, y’ = +125 mm, virtual, erect, and enlarged.

36-14. An object located 30 cm from a thin lens has a real, inverted image located a distance of

60 cm on the opposite side of the lens. What is the focal length of the lens?

(30 cm)(60 cm) ;30 cm + 60 cm

pqfp q

f = +20 cm

The focal length is positive, so the lens is converging.

f = +20 cm, converging

36-15. A light source is 600 mm from a converging lens of focal length 180 mm. Construct the

image using ray diagrams. What is the image distance? Is the image real or virtual?

(60 cm)(18 cm) ;60 cm - 18 cm

pfqp f

q = 25.7 cm

501

2FF

2FF


q = +25.7cm, real, inverted, and diminished.

Note: Show that the magnification of the image is M = - 0.428.

36-16. A plano-convex lens is held 40 mm from a 6-cm object. What are the nature and location

of the image formed if the focal length is 60 mm?

(40 mm)(60 mm) ;40 cm - 60 cm

q q = -120 mm;

' ( 120 mm)(6 cm)'40 mm

qyyp

q = -120 mm, y’ = +18.0 cm, virtual, erect, and enlarged.

36-17. An object 6 cm high is held 4 cm from a diverging meniscus lens of focal length –24 cm.

What are the nature, size, and location of the image?

(4 cm)(-24 cm) ;4 cm - (-24 cm)

pfqp f

q = -3.43 cm

' ' ( 3.43 cm)(6 m); '4 cm

q y qyM yp y p

q = -3.43 cm, y’ = 5.14 cm, virtual, erect, and diminished.

36-18. The focal length of a converging lens is 200 mm. An object 60 mm high is mounted on a

movable track so that the distance from the lens can be varied. Calculate the nature, size,

and location of the image formed for the following object distances: (a) 150 mm, (b) 200

mm, (c) 300 mm, (d) 400 mm, (e) 600 mm. (Figures not drawn to conserve space.)

502


p = 15 cm, 20 cm, 30 cm, 40 cm, and 60 cm; f = 20 cm, y = 6 cm

;pfqp f

' ; 'y q qyM yy p p

(a)

(15 cm)(20 cm)15 cm - 20 cm

q = -60 cm, virtual

( 60 cm)(6 cm)'(15 cm)

y

= 24 cm, erect

(b)

(20 cm)(20 cm)20 cm - 20 cm

q = , no image

( )(6 cm)'(20 cm)

y

= , no image

(c)

(30 cm)(20 cm)30 cm - 20 cm

q = 60 cm, real

(60 cm)(6 cm)'(30 cm)

y

= -12 cm, inverted

36-18. (Cont.) Images formed by converging mirror, f = 20 cm, y = 6 cm.

(d)

(40 cm)(20 cm)40 cm - 20 cm

q = 40 cm, real

(40 cm)(6 cm)'(40 cm)

y

= -6 cm, inverted

(e)

(60 cm)(20 cm)60 cm - 20 cm

q = 30 cm, real

(30 cm)(6 cm)'(60 cm)

y

= -3 cm, inverted

36-19. An object 450 mm from a converging lens forms a real image 900 mm from the lens.

What is the focal length of the lens?

503


(45 cm)(90 cm) ;45 cm + 90 cm

pqfp q

f = +30 cm

Magnification

*36-20. An object is located 20 cm from a converging lens. If the magnification is –2, what is the

image distance?

2; 2 2(20 cm);qM q pp

q = 40 cm, real, inverted, and enlarged.

*36-21. A pencil is held 20 cm from a diverging lens of focal length –10 cm. What is the

magnification?

( 10 cm) ;20 cm

qMp

M = +½

*36-22. A magnifying glass has a focal length of 27 cm. How close must this glass be held to an

object to produce an erect image three times the size of the object?

3; 3 ; ; 3q pf pfM q p q pp p f p f

-3(p – f) = f; -3p + 3f = f; p = 2 2

3 3 (27 cm);f p = 18.0 cm

504

F


*36-23. A magnifying glass held 40 mm from a specimen produces an erect image that is twice

the object size. What is the focal length of the lens?

2; 2 ; ; 2q pf pfM q p q pp p f p f

-2(p – f) = f; -2p + 2f = f; f = 2p = 2(40 mm); f = 80.0 mm

*36-24. What is the magnification of a lens if the focal length is 40 cm and the object distance is

65 cm?

(65 cm)(40 cm) (104 cm)104 cm; ;65 cm - 40 cm (65 cm)

qq Mp

M = -1.60

Challenge Problems

36-25. The radius of the curved surface in a plano-concave lens is 20 cm. What is the focal

length if n = 1.54? [ R1 = - 20 cm, R2 = ]

1 2

1 1 1 1 1( 1) (1.54 1) ;-20 cm

nf R R

1 1 1 0.540(1.54 1) ; ;20 cm 20 cmf f

f = -37.4 cm

505

R2 =

R1


36-26. A thin meniscus lens if formed with a concave surface of radius –40 cm and a convex

surface of radius +30 cm. If the resulting focal length is 79.0 cm, what was the index of

refraction of the transparent material?

1 2

1 1 1 1 1 1( 1) ; ( 1) ;79 cm 40 cm 30 cm

n nf R R

2

1 10 cm( 1) ; 1 1.52;79 cm -1200 cm

n n n = 2.52

36-27. A converging lens has a focal length of 20 cm. An object is placed 15.0 cm from the lens.

find the image distance and the nature of the image. The object is inside the focus.

(15 cm)(20 cm) ;15 cm - 20 cm

pfqp f

q = -60 cm

The image is erect, virtual, and enlarged.

36-28. How far from a source of light must a lens be placed if it is to form an image 800 mm

from the lens? The focal length is 200 mm.

(80 cm)(20 cm) ;80 cm - 20 cm

qfpq f

p = 26.7 cm

506

Image

Object

2FF

2FF


36-29. A source of light 36 cm from a lens projects an image on a screen 18.0 cm from the lens.

What is the focal length of the lens? Is it converging or diverging.

(36 cm)(18 cm) ;36 cm + 18 cm

pqfp q

f = 12 cm, converging.

36-30. What is the minimum film size needed to project the image of a student who is 2 m tall?

Assume that the student is located 2.5 m from the camera lens and that the focal length is

55.0 mm?

(2000 mm)(55 mm) ;2000 mm - 55 mm

pfqp f

q = 56.6 mm

*36-31. When parallel light strikes a lens, the light diverges, apparently coming from a point 80

mm behind the lens. How far from an object should this lens be held to form an image

one-fourth the size of the object?

1 ; 4 4( 80 mm);4

qM p qp

p = 240 mm

*36-32. How far from a diverging lens should an object be placed in order that its image be one-

fourth the size of the object? The focal length is -35 cm.

507F


1 ; ; ; 4 4 4

q p pf p pfM q qp p f p f

-(p – f) = 4f; -p + f = 4f; p = -3f=-3(-35 cm); p = 105 cm

*36-33. The first surface of a thin lens has a convex radius of 20 cm. What should be the radius

of the second surface to produce a converging lens of focal length 8.00 cm?

1 2 2

1 1 1 1 1 1( 1) ; (1.5 1) ;8 cm 20 cm

nf R R R

22 2

2

1 20 cm 0.5 ; 2.5 0.5 10 cm;8 cm (20 cm)

R R RR

R2 = +5.00 cm

*36-34 Two thin converging lenses are placed 60 cm apart and have the same axis. The first lens

has a focal length 10 cm, and the second has a focal length of 15.0 cm. If an object 6.0

cm high is placed 20 cm in front of the first lens, what are the location and size of the

final image? Is it real or virtual? ( See figure on following page.)

1 2(20 cm)(10 cm) 20 cm; 60 cm - 20 cm = 40 cm20 cm - 10 cm

q p

2(40 cm)(15 cm) 24 cm;40 cm - 15 cm

q q = 24.0 cm, real image

508


*36-34. (Cont.) q1 = 20 cm; q2 = 24 cm

M = M1 x M2

1 2 1 2

1 2 1 2

q q q qMp p p p

(20 cm)(24 cm) 0.6; ' (0.6)(6 cm); ' 3.6 cm;20 cm)(40 cm)

M y My y

q = 24 cm beyond lens 2, y’ = 3.6 cm, real, erect, and diminished.

*36-35. A converging lens of focal length 25 cm is placed 50 cm in front of a diverging lens

whose focal length is –25 cm. If an object is placed 75 cm in front of the converging lens,

what is the location of the final image? What is the total magnification. Is the image real

or virtual? (Take image of 1st as object for 2nd.)

1

(75 cm)(25 cm) 37.5 cm;75 cm - 25 cm

q

2 60 cm - 20 cm = 40 cmp

1 22 1 2

1 2

(12.5 cm)(-25 cm) ( )( )8.33 cm; x 12.5 cm -(-25 cm)

q qq M M Mp p

1 2

1 2

(37.5 cm)(-8.33 cm) ;(75 cm)(12.5 cm)

q qMp p

M = -0.333

509

60 cm

f1 = 10 cm cm

f2 = 15 cm


q = -8.33 cm left of lens 2; virtual, inverted, diminished; M = -0.333

Critical Thinking Problems

*36-36. A camera consists of a converging lens of focal length 50 mm mounted in front of a

light-sensitive film as shown in Fig. 36-19. When photographing infinite objects, how far

should the lens be from the film? What is the image distance when photographing object

500 mm from the lens? What is the magnification?

(a) p = ;

1 1 1 ;q f

q = f = 50 mm

(b)

(500 mm)(50 mm) ;500 mm - 50 mm

pfqp f

q = 55.6 mm for object at 500 mm.

(c)

(55.6 mm) ;500 mm

qMp

M = -0.111, inverted, real, and diminished.

*36-37. An object is placed 30 cm from a screen. At what points between the object and the

screen can a lens of focal length 5 cm be placed to obtain an image on the screen?

51030 cm - x

x

30 cm


Let p = x and q = (30 cm – x); ;pfq

p f

(5 cm)30 cm - ; (30 )( 5) 55 cm

xx x x xx

;

30x – 150 - x2 + 5x = 5x; x2 –30x + 150 = 0; ( Solve the quadratic equation for x)

Two positions are: x = 6.34 cm and 23.7 cm from object.

*36-38. A simple projector is illustrated in Fig. 36-20. The condenser provides even illumination

of the film by the light source. The frame size of regular 8-mm film is 5 x 4 mm. An

image is to be projected 600 x 480 mm on a screen located 6 m from the projection lens.

What should be the focal length of the projection lens? How far should the film be from

the lens?

' ( 60 cm) ; 1200.5 cm

yM My

1120; ; ; ; 120 120 120

q q qf q qf fM p pp q f q f q f

511


q – f = 120f; 121f = q;

600 cm ;121 121qf

f = 4.96 cm

600 cm ;120 120

qp p = 5.00 cm

*36-39. A telescope has an objective lens of focal length 900 mm and an eyepiece of focal

length 50 mm. The telescope is used to examine a rabbit 30 cm high at a distance of 60

m. What is the distance between the lenses if the final image is 25 cm in front of the

eyepiece? What is the apparent height of the rabbit as seen through the telescope?

p1 = 6000 cm, f1 = 90 cm, f2 = 5 cm;

(a)

(6000 cm)(90 cm) 91.37 cm6000 cm - 90 cm

q

p2 = x – 91.37 cm;

2 22

2 2

q fpq f

( 25 cm)(5 cm)91.3 cm ; 91.37 cm 4.167;-25 cm - 5 cm

x x

x = 95.54 cm

The two lens must be separated by a distance of 95.5 cm.

*36-39. (Cont.) (b) p2 = 95.54 cm – 91.37 cm; p2 = 4.17 cm; M = M1 x M2

1 2

1 2

(91.37 cm)( 25 cm) ;(6000 cm)(4.17 cm)

q qMp p

M = -0.0914; y’ = My

512

5 cm90 cm

60 m

25 cm

x


y’ = (-0.0914)(30 cm); y’= -2.74 cm, virtual, inverted, and diminished

*36-40. The Galilean telescope consists of a diverging lens as the eyepiece and a converging

lens as the objective. The focal length of the objective is 30 cm, and the focal length of

the eyepiece is –2.5 cm. An object 40 m away from the objective has a final image

located 25 cm in front of the diverging lens. What is the separation of the lenses? What

is the total magnification?

1

(4000 cm)(30 cm) 30.23 cm;4000 cm - 30 cm

q

2 22

2 2

( 25 cm)(-2.5 cm) ;-25 cm - (-2.5 cm)

q fpq f

p2 = -2.78 cm

p2 = x – 30.23 cm; x = 30.23 cm + p2 = 30.23 cm +(-2.78 cm); x = 27.5 cm

1 2

1 2

(30.23 cm)(-25 cm) ;(4000 cm)(2.78 cm)

q qMp p

M = -0.068

*36-6. The focal length of the eyepiece of a particular microscope is 3.0 cm, and the focal

length of the objective lens is 19 mm. The separation of the two lenses is 26.5 cm, and

the final image formed by the eyepiece is at infinity. How far should the objective lens

be placed from the specimen being studied?

x = 256 mm; f1 = 19 mm; f2 = 30 mm; p2 = f2 = 30 mm;

q1 = 265 mm – 30 mm = 235 mm;

513

4000 cm

-25 cm

x


1 11

1 1

(235 mm)(19 mm) ;235 mm - 19 mm

q fpq f

q = 20.7 mm

514

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Solucionario Capitulo 36 - Paul E. Tippens