Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
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 plano-convex lens lens is to be construc constructed ted out of glass glass so that that it has a focal focal length length of 40 40 R 2 = ∞
cm. What is the radius radius of curvature of the curved surface? surface? 1 f
1 1 1 1 = (n − 1) + = (1.5 − 1) + ; R R 40 cm ∞ 1 2
1 = (0.5) ; 40 cm R1 1
R1
= 0.5(40 cm);
R 1 = ? n = 1.5
R1 = 20.0 cm
36-2. If one uses uses a glass, glass, double-c double-convex onvex lens lens to obtain obtain a focal focal length length of 30 30 cm, what what must be the curvature of each convex surface? 1 f
[ R1 = R2 = R ]
1 1 2 = (n − 1) + = (1.5 − 1) ; R = R R R
R2
R1
f ; R = 30 cm
36-3. The curved curved surface surface of a plano-conc plano-concave ave lens lens has has a radius of –12 cm. What is the the focal focal length if the lens is made from a material with a refractive index of 1.54. 1 f 1 f
1 1 1 1 1) + = (n − 1) = (1.54 − 1) + ; ∞ R R -12 cm 1 2 1 ; = (1.54 − 1) −12 cm
R1 R2 =
1 f
=
0. 0.540
−12 cm
;
∞
f = -22.2 cm
36-4. A converging converging meniscu meniscuss lens (n (n = 1.5) has a concave concave surface surface whose whose radius radius is –20 cm and and a convex surface whose radius is 12 cm. What is the focal length? 1 1 = (0.5) −20 cm + 12 cm ; = (1.5 − 1) + (12 cm)(-20 cm) f 12 cm −20 cm 1
241
f = 30 cm
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
36-5. A converging converging lens such such as shown in Fig. Fig. 36-8a is made made of glass. glass. The first first surfac surfacee 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 )
10 cm + 15 cm 1 1 = (1.5 − 1) + = ( 0 . 5 ) (10 cm)(15 cm) ; f 15 cm 10 cm 1
f = 12 cm
36-6. A meniscus meniscus lens lens has a convex surfac surfacee whose radius radius is is 20 cm and and a concave concave surface 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: 1.54: Find: f = ?
−30 cm + 20 cm 1 1 ( 0 . 5 4 ) = (1.54 − 1) + = (20 cm)(-30 cm) ; f 20 cm -30 cm 1
f = 1.11 m
36-7. A plano-convex plano-convex lens lens is ground from from crown crown glass glass (n = 1.52). 1.52). What What should should be the the radius radius of the curved surface if the desired focal length is to be 400 mm? 1 f
1 1 = (n − 1) + ; R1 R2
1 = (0.52) ; 400 mm R1 1
1 1 = (1.52 − 1) + ; 400 mm R1 ∞ 1
R 2 = ∞ f = 400 mm n = 1.52
R1
= 0.52(400 mm);
R1 = 208 mm
36-8. The magnitude magnitudess of the concave concave and and convex surfaces surfaces of a glass lens are are 200 and 600 600 mm, respectively. What is the focal length? Is it diverging or converging? Given: R1 = -20 -20 cm , R2 = +60 cm , n = 1.5: 1.5: Find: f = ?
60 cm - 20 cm 1 1 ; = (1.5 − 1) + = (0.5) f 2 0 c m 6 0 c m ( 2 0 c m ) ( 6 0 c m ) 1
f = -60 cm
The lens is a diverging lens, diverging lens, since the focal length is negative.
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Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
36-9. A plastic plastic lens lens (n = 1.54) 1.54) has a convex surface surface of radius radius 25 cm and and a concave concave surface surface of –70 cm. What is the focal length? Is it diverging or converging?
-70 cm + 25 cm 1 1 ; = (1.54 − 1) + = (0.54) f 2 5 c m 7 0 c m ( 2 5 c m ) ( 7 0 c m ) 1
f = +72.0 cm
The lens is a converging lens, converging lens, since the focal length is positive.
Images Formed by Thin Lenses 36-10. A 7-cm tall pencil is placed 35 cm from a thin converging lens of focal length length 25 cm. What are the nature, size, and location of the image formed? q=
M
pf p − f
=
−q p
=
=
(35 cm)(25 cm) 35 cm - 25 cm
y' y
; y' =
;
q = 87.5 cm
F
2F
− qy ' −(87.5 cm)(7 m) = p
35 cm
q = +87.5 +87.5 cm, y’ = -17.5 cm , real, inverted, and enlarged.
36-11. An object 8 cm high is placed placed 30 cm from a thin converging lens lens of focal length 12 cm. What are the nature, size, and location of the image formed? q=
M
pf p − f
=
−q p
=
=
(30 cm)(12 cm) 30 cm - 12 cm
y' y
; y' =
;
q = 20 cm
F
2F
− qy ' −(20 cm)(8 m) = p
30 cm
q = +20.0 +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.
243
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments 36-12. (Cont.). Find object distance for given values: p =
qf q − f
(−40 cm)( cm)(15 15 cm) cm)
=
-40 cm - 15 cm
;
F
p = 10.9 cm
2F
Note: Note: You should show that the magnification mag nification is +2.67 +2.67..
36-13. A 50 mm tall 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? image? (See drawing for Prob. 36-12 above.) q=
(12 cm)(20 cm) 12 cm - 20 cm
;
q = -30 -30 cm; M
=
−q p
=
y' y
; y' =
3 0 cm)(50 mm) − qy ' −(−30 = p
12 cm
q = -30.0 -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 op posite side of the lens. What is the focal length of the lens? f =
pq p + q
=
(30 cm)(60 cm) 30 cm + 60 cm
;
f = +20 cm
F
2F
The focal length is positive, so the lens is converging. f = 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? distance? Is the image real or virtual? q=
pf p − f
=
(60 cm)(18 cm) 60 cm - 18 cm
;
q = 25.7 cm
q = +25.7cm, +25.7cm, real, inverted, and diminished. Note: Show that the magnification of the image is M = M = - 0.428.
244
F
2F
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
36-16. A plano-convex lens is is held 40 mm from a 6-cm 6-cm object. What are the nature and location of the image formed if the focal length is 60 mm? q=
(40 mm)(60 mm) 40 cm - 60 cm
q = -120 -120 mm; y ' =
;
−qy ' −(−120 mm)(6 cm) = p
40 mm
q = -120 -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? q=
M
pf p − f
=
−q p
=
=
(4 cm)(-24 cm) 4 cm - (-24 cm)
y' y
; y' =
;
q = -3.43 cm
3.43 cm)(6 m) − qy ' −(−3. = p
4 cm
q = -3.43 cm, y’ = 5.14 cm , virtual, erect, and diminished.
36-18. The focal length length of a convergi converging ng lens is is 200 mm. An object object 60 mm high is is mounted 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.) p = 15 cm, 20 cm, 30 cm, 40 cm, and 60 cm; q=
pf p − f
(a) q =
(15 cm)(20 cm)
(b) q =
(20 cm)(20 cm)
(c) q =
15 cm - 20 cm
20 cm - 20 cm
(30 cm)(20 cm) 30 cm - 20 cm
;
M
=
y ' y
=
−q p
;
= -60 cm, virtual y ' =
= ∞, no image
y ' =
y ' =
= 60 cm, real
245
f = 20 cm, y = 6 cm y'=
− qy p
cm)(6 cm) −(−60 cm) (15 cm)
−(∞)(6 cm) (20 cm)
= ∞, no image
−(60 cm)(6 cm) (30 cm)
= 24 cm, erect
= -12 cm, inverted
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments 36-18. (Cont.) Images formed by converging mirror, f = 20 cm, y = 6 cm. (d) q =
(40 cm)(20 cm)
(e) q =
(60 cm)(20 cm)
40 cm - 20 cm
60 cm - 20 cm
= 40 cm, real
y ' =
= 30 cm, real
y ' =
−(40 cm)(6 cm) (40 cm)
−(30 cm)(6 cm) (60 cm)
= -6 cm, inverted
= -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? f =
pq p + q
=
(45 cm)(90 cm) 45 cm + 90 cm
; f = +30 cm
Magnification *36-20. An object is located 20 cm from from a converging lens. If the the magnification is –2, what is the image distance? M
=
−q p
= −2;
q = 2 p = 2(20 cm);
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 is the magnification? M =
−q −(−10 cm) ; = p
20 cm
M = +½ F
*36-22. A magnifying glass has a focal length of 27 cm. How close must this glass glass be held to an object to produce an erect image three times the size of the object? ob ject? M
=
−q p
-3( p p – f) = f;
= +3;
q = −3 p; q =
-3p + 3f = f;
p=
2
246
3
pf p− f
;
− 3p =
f = 2 3 (27 cm); cm);
pf p − f p = 18.0 cm
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
*36-23. A magnifying glass held 40 mm mm from a specimen produces an erect image that is twice twice the object size. What is the focal length of the lens? M
=
−q p
-2( p p – f) = f;
= +2;
pf
q = −2 p; q =
-2p + 2f = f;
p− f
;
− 2p=
f = 2p = 2(40 2(40 mm);
pf
p − f
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? q=
(65 cm)(40 cm) 65 cm - 40 cm
= 104 cm;
M =
−q −(104 cm) ; = p
(65 cm)
M = -1.60 -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 if n = 1.54? [ R1 = - 20 cm, R2 = ∞ ] 1 f 1 f
R1
1 1 1 + 1 ; 1) + ( 1 . 5 4 1 ) = (n − 1) = − -20 cm ∞ R1 R2 1 ; = (1.54 − 1) −20 cm
1 f
=
0. 0.540
−20 cm
;
R2 =
∞
f = -37.4 cm
36-26. A thin meniscus lens if formed formed with a concave surface of radius –40 cm and a convex surface of radius +30 cm. If the resulting focal focal length is 79.0 cm, what was the index of refraction of the transparent material? 1 f
1 1 = (n − 1) + ; R1 R2
1 1 = ( n − 1) + ; 79 cm −40 cm 30 cm 1
−10 cm = (n − 1) ; 79 cm -1200 cm2 1
247
n − 1 = 1. 1.52;
n = 2.52
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
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. q=
pf p − f
=
(15 cm)(20 cm) 15 cm - 20 cm
;
Object q = -60 cm F
The image is erect, virtual, and enlarged. en larged.
2F
Image
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. F p =
qf q − f
=
(80 cm)(20 cm) 80 cm - 20 cm
2F
;
p = 26.7 cm
36-29. A source of light 36 cm from a lens projects projects an image on a screen 18.0 cm from from the lens. What is the focal length length of the lens? Is it converging or diverging. f =
pq p + q
=
(36 cm)(18 cm) 36 cm + 18 cm
;
f = 12 cm, converging.
36-30. What is the minimum minimum film size needed to project project the image of a student who is 2 m tall? tall? Assume that the student is located 2.5 m from the camera lens and that the focal length is 55.0 mm? q=
pf p − f
=
(2000 mm)(55 mm) 2000 mm - 55 mm
;
q = 56.6 mm
248
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
*36-31. When parallel light strikes a lens, the light d iverges, 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?
=
M
−q p
=+
1 4
;
p = − 4q = − 4(− 80 mm);
p = 240 mm
*36-32. How far from a diverging lens should an object be placed placed in order that its image be onefourth the size of the object? The focal length is -35 cm. M
=
−q p
=
1 4
; q=
-( p p – f) = 4f;
−p 4
; q=
-p + f = 4f;
pf p− f
−p
;
=
4
pf
p − f
p = -3f=-3(-35 -3f=-3(-35 cm);
F p = 105 cm
*36-33. The first surface of a thin lens has h as 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 f
1 1 = (n − 1) + ; R1 R2
R + 20 cm = 0 .5 2 ; 8 cm ( 2 0 c m ) R 2 1
1 1 = (1.5 − 1) + ; 8 cm 20 cm R2 1
2.5 R2
= 0.5R2 + 10 cm;
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.) q1
=
(20 cm)(10 cm)
q2
=
20 cm - 10 cm
= 20 cm;
(40 cm)(15 cm) 40 cm - 15 cm
p2
= 24 cm; 249
= 60 cm - 20 cm = 40 cm q = 24.0 cm, real image
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments *36-34. (Cont.)
60 cm
q1 = 20 cm; cm; q2 = 24 cm M = M1 x M2
−q1 − q2 p = p 1 2
M =
q1q2 p1 p2
f 2 = 15 cm
f 1 = 10 cm cm
M
=
(20 cm)(24 cm) 20 cm)(40 cm)
= +0.6;
y ' = My = (0.6)(6 cm);
y ' = 3.6 cm;
q = 24 cm beyond lens 2, y’ = 3.6 cm, real, erect, and diminished.
*36-35. A converging lens of focal length length 25 cm is placed 50 cm in in front of a diverging 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 1 st as object for 2nd .) q1
=
p2 q2
=
(75 cm)(25 cm) 75 cm - 25 cm
= 37.5 cm;
= 60 cm - 20 cm = 40 cm (12.5 cm)(-25 cm) 12.5 cm -(-25 cm) M =
q1q2 p1 p2
=
= −8.33 cm;
M
= M1
(37.5 cm)(-8.33 cm) (75 cm)(12.5 cm)
;
x M2
=
(−q1 )(− q2 ) p1 p2
M = -0.333
q = -8.33 cm left of lens 2; 2; virtual, inverted, diminished; M = M = -0.333
250
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
Critical Thinking Problems *36-36. A camera consists of a converging lens of focal length length 50 mm mounted in front of a light-sensitive film 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 distance when photographing object 500 mm from the lens? What is the magnification? (a) p =
(b) q =
1
;
∞
pf
=
∞
p − f
(c) M =
1
1
q
f
+ =
;
(500 mm)(50 mm) 500 mm - 50 mm
−q −(55.6 mm) ; = p
q = f = f = 50 mm
500 mm
;
q = 55.6 mm for object at 500 at 500 mm. mm.
M = -0.111, -0.111, inverted, real, and diminished.
*36-37. An object is placed 30 cm from from a screen. At what points points between the object and the screen can a lens of focal length 5 cm be placed to obtain an image on the screen? Let p = x and q = ( 30 30 cm – x); q =
30 cm - x =
x(5 cm) x − 5 cm
pf p − f
; (30 − x )( x − 5) = 5 x ;
30 x – 150 - x - x2 + 5 x = 5x;
30 cm
;
x 30 cm - x - x
x2 –30x + 150 = 0; ( Solve the quadratic equation for x)
Two positions are: x = 6.34 cm and 23.7 and 23.7 cm from cm from object .
251
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
*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? M
M
=
−q p
=
y ' y
=
( −60 cm) 0.5 cm
= −120 120;
p=
q – f = 120f;
M = −120
;
q 120
; p=
q 120
=
q− f
f =
121f = q;
p =
qf
q
;
120
q 121
600 cm 120
;
=
=
qf q− f
600 cm 121
;
;
1 120
=
f
q − f
f = f = 4.96 cm
p = 5.00 cm
*36-39. *36-39. A telescope telescope has an objecti objective ve lens of focal focal length length 900 mm and an an eyepiece 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? x
p1 = 6000 cm, f 1 = 90 cm, f 2 = 5 cm; (a)
q=
(6000 cm)(90 cm) 6000 cm - 90 cm
p2 = x – 91.37 cm; 91.37 cm;
x − 91.3 cm =
60 m
= 91.37 cm
p2
=
90 cm
q2 f 2 q2 − f 2
(−25 cm) cm)(5 cm cm) -25 cm - 5 cm
;
x − 91.37 cm = 4.167;
5 cm 25 cm
x = 95.54 cm
The two lens must be separated by a distance of 95.5 of 95.5 cm.
252
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments *36-39. (Cont.)
(b) p2 = 95.54 cm – 91.37 cm; M =
q1q2 p1 p2
=
(91. (91.37 37 cm) cm)((−25 cm) cm) (6000 cm)(4.17 cm)
p2 = 4.17 cm; ; M = -0.0914;
M = M 1 x M 2 y’ = My
y’ = (-0.0914 )(30 )(30 cm); y’= -2.74 cm, virtual, inverted, and diminished
*36-40. *36-40. The Galilean Galilean telescope telescope consists consists of a diverging diverging lens as the the eyepiece 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? q1
p2
=
=
(4000 cm)(30 cm) 4000 cm - 30 cm
q2 f 2 q2 − f 2
=
*36-6.
q1q2 p1 p2
=
x
= 30.23 cm;
( −25 cm) cm)((-2. 2.5 5 cm) cm) -25 cm - (-2.5 cm)
p2 = x – 30.23 – 30.23 cm; M =
4000 cm
-25 cm
; p2 = -2.78 cm
x = 30.23 cm + p + p2 = 30.23 cm +(-2.78 cm); x = 27.5 cm
(30.23 cm)(-25 cm) (4000 cm)(2.78 cm)
; M = -0.068
The focal focal length length of the the eyepiece eyepiece of a particular particular microscope microscope is 3.0 3.0 cm, 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; f 1 = 19 mm;
f 2 = 30 mm;
p2 = f 2 = 30 mm;
q1 = 265 mm – 30 mm = 235 mm; p1
=
q1 f 1 q1 − f 1
=
(235 mm)(19 mm) 235 mm - 19 mm
253
;
q = 20.7 mm
