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 lens is to be constructed out of glass so that it has a focal length of 40
R2 = 
cm. What is the radius of curvature of the curved surface?
1 1 
1
1
 1
 (n  1)     (1.5  1) 
 ;
f
 40 cm  
 R1 R2 
1
1
 (0.5)   ;
40 cm
 R1 
R1  0.5(40 cm);
R1 = ?
n = 1.5
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
 (n  1)     (1.5  1)   ; R  f ;
f
R R
R
R2
R1
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.
R1
1 1 
1
1
 1
 (n  1)     (1.54  1) 
 ;
f
 -12 cm  
 R1 R2 
1
 1 
 (1.54  1) 
;
f
 12 cm 
1
0.540

;
f 12 cm
R2 = 
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?
 20 cm + 12 cm 
1
1 
 1
 (1.5  1) 

  (0.5) 
;
f
 12 cm 20 cm 
 (12 cm)(-20 cm) 
497
f = 30 cm
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
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 )
 10 cm + 15 cm 
1
1 
 1
 (1.5  1) 

  (0.5) 
;
f
 15 cm 10 cm 
 (10 cm)(15 cm) 
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 = ?
 30 cm + 20 cm 
1
1 
 1
 (1.54  1) 

  (0.54) 
;
f
 20 cm -30 cm 
 (20 cm)(-30 cm) 
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?
R2 = 
1 1 
1
 (n  1)    ;
f
 R1 R2 
1
1
 (0.52)   ;
400 mm
 R1 
 1 1
1
 (1.52  1)    ;
400 mm
 R1  
f = 400 mm
n = 1.52
R1  0.52(400 mm);
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 = ?
 60 cm - 20 cm 
1
1 
 1
 (1.5  1) 

  (0.5) 
;
f
 -20 cm 60 cm 
 (-20 cm)(60 cm) 
f = -60 cm
The lens is a diverging lens, since the focal length is negative.
498
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
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?
 -70 cm + 25 cm 
1
1 
 1
 (1.54  1) 

  (0.54) 
;
f
 25 cm -70 cm 
 (25 cm)(-70 cm) 
f = +72.0 cm
The lens is a 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 25 cm.
What are the nature, size, and location of the image formed?
q
pf
(35 cm)(25 cm)

;
p f
35 cm - 25 cm
M
q = 87.5 cm
F
2F
q y '
qy ' (87.5 cm)(7 m)
 ; y' 

p
y
p
35 cm
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?
q
pf
(30 cm)(12 cm)

;
p f
30 cm - 12 cm
M
q = 20 cm
F
2F
q y '
qy ' (20 cm)(8 m)
 ; y' 

p
y
p
30 cm
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.
499
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
36-12. (Cont.). Find object distance for given values:
p
qf
(40 cm)(15 cm)

;
q f
-40 cm - 15 cm
F
p = 10.9 cm
2F
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.)
q
(12 cm)(20 cm)
;
12 cm - 20 cm
q = -30 cm; M 
q y '
qy ' (30 cm)(50 mm)
 ; y' 

p
y
p
12 cm
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?
f 
pq
(30 cm)(60 cm)

;
p  q 30 cm + 60 cm
f = +20 cm
F
2F
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?
q
pf
(60 cm)(18 cm)

;
p f
60 cm - 18 cm
q = 25.7 cm
q = +25.7cm, real, inverted, and diminished.
Note: Show that the magnification of the image is M = - 0.428.
500
F
2F
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
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?
q
(40 mm)(60 mm)
;
40 cm - 60 cm
q = -120 mm; y ' 
qy ' (120 mm)(6 cm)

p
40 mm
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?
q
pf
(4 cm)(-24 cm)

;
p  f 4 cm - (-24 cm)
M
q = -3.43 cm
q y '
qy ' (3.43 cm)(6 m)
 ; y' 

p
y
p
4 cm
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.)
p = 15 cm, 20 cm, 30 cm, 40 cm, and 60 cm;
q
pf
;
p f
M
y ' q

;
y
p
f = 20 cm, y = 6 cm
y' 
qy
p
(a) q 
(15 cm)(20 cm)
= -60 cm, virtual
15 cm - 20 cm
y' 
(60 cm)(6 cm)
= 24 cm, erect
(15 cm)
(b) q 
(20 cm)(20 cm)
= , no image
20 cm - 20 cm
y' 
()(6 cm)
= , no image
(20 cm)
y' 
(60 cm)(6 cm)
= -12 cm, inverted
(30 cm)
(c) q 
(30 cm)(20 cm)
= 60 cm, real
30 cm - 20 cm
501
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)
= 40 cm, real
40 cm - 20 cm
y' 
(40 cm)(6 cm)
= -6 cm, inverted
(40 cm)
(e) q 
(60 cm)(20 cm)
= 30 cm, real
60 cm - 20 cm
y' 
(30 cm)(6 cm)
= -3 cm, inverted
(60 cm)
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
(45 cm)(90 cm)

;
p  q 45 cm + 90 cm
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?
M
q
 2;
p
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 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 be held to an
object to produce an erect image three times the size of the object?
M
q
 3;
p
-3(p – f) = f;
q  3 p; q 
-3p + 3f = f;
p=
2
502
3
pf
;
p f
 3p 
f  2 3 (27 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 from a specimen produces an erect image that is twice
the object size. What is the focal length of the lens?
M
q
 2;
p
-2(p – f) = f;
pf
;
p f
q  2 p; q 
-2p + 2f = f;
2p 
f = 2p = 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)
 104 cm;
65 cm - 40 cm
M
q (104 cm)

;
p
(65 cm)
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 =  ]
R1
1 1 
1
1
 1
 (n  1)     (1.54  1) 
 ;
f
 -20 cm  
 R1 R2 
1
 1 
 (1.54  1) 
;
f
 20 cm 
1
0.540

;
f 20 cm
R2 = 
f = -37.4 cm
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 1 
1
 (n  1)    ;
f
 R1 R2 
1
1 
 1
 (n  1) 

;
79 cm
 40 cm 30 cm 
1
 10 cm 
 (n  1) 
; n  1  1.52;
2 
79 cm
 -1200 cm 
503
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
(15 cm)(20 cm)

;
p f
15 cm - 20 cm
Object
q = -60 cm
F
The image is erect, virtual, and enlarged.
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
2F
qf
(80 cm)(20 cm)
p

;
q f
80 cm - 20 cm
p = 26.7 cm
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.
f 
pq
(36 cm)(18 cm)

;
p  q 36 cm + 18 cm
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?
q
pf
(2000 mm)(55 mm)

;
p f
2000 mm - 55 mm
q = 56.6 mm
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Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
*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?
M
q
1
 ;
p
4
p  4q  4(80 mm);
p = 240 mm
*36-32. How far from a diverging lens should an object be placed in order that its image be onefourth the size of the object? The focal length is -35 cm.
M
q 1
p
pf
 ; q
; q
;
p 4
4
p f
-(p – f) = 4f;
-p + f = 4f;
p
pf

4
p f
p = -3f=-3(-35 cm);
F
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 1 
1
 (n  1)    ;
f
 R1 R2 
 R  20 cm 
1
 0.5  2
;
8 cm
(20
cm)
R
2 

 1
1
1 
 (1.5  1) 
 ;
8 cm
 20 cm R2 
2.5R2  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)
 20 cm;
20 cm - 10 cm
q2 
p2  60 cm - 20 cm = 40 cm
(40 cm)(15 cm)
 24 cm;
40 cm - 15 cm
505
q = 24.0 cm, real image
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
60 cm
*36-34. (Cont.) q1 = 20 cm; q2 = 24 cm
M = M1 x M2
 q  q  q q
M   1  2   1 2
 p1  p2  p1 p2
M
(20 cm)(24 cm)
 0.6;
20 cm)(40 cm)
f2 = 15 cm
f1 = 10 cm
cm
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 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.)
q1 
(75 cm)(25 cm)
 37.5 cm;
75 cm - 25 cm
p2  60 cm - 20 cm = 40 cm
q2 
(12.5 cm)(-25 cm)
 8.33 cm;
12.5 cm -(-25 cm)
M
M  M1 x M 2 
q1q2 (37.5 cm)(-8.33 cm)

;
p1 p2
(75 cm)(12.5 cm)
( q1 )( q2 )
p1 p2
M = -0.333
q = -8.33 cm left of lens 2; virtual, inverted, diminished; M = -0.333
506
Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
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 = ;
(b) q 
(c) M 
1 1 1
  ;
 q f
q = f = 50 mm
pf
(500 mm)(50 mm)

;
p f
500 mm - 50 mm
q (55.6 mm)

;
p
500 mm
q = 55.6 mm for object at 500 mm.
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?
Let p = x and q = (30 cm – x); q 
30 cm - x 
x(5 cm)
; (30  x)( x  5)  5 x ;
x  5 cm
30x – 150 - x2 + 5x = 5x;
30 cm
pf
;
p f
x
30 cm - x
x2 –30x + 150 = 0; ( Solve the quadratic equation for x)
Two positions are: x = 6.34 cm and 23.7 cm from object.
507
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
y ' (60 cm)

;
y
0.5 cm
q
 120;
p
p
q – f = 120f;
M  120
q
qf
; p
;
120
q f
121f = q;
p
f 
q
qf

;
120 q  f
q
600 cm

;
121
121
q
600 cm

;
120
120
1
f

120 q  f
f = 4.96 cm
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?
x
p1 = 6000 cm, f1 = 90 cm, f2 = 5 cm;
60 m
(a)
q
(6000 cm)(90 cm)
 91.37 cm
6000 cm - 90 cm
p2 = x – 91.37 cm;
x  91.3 cm 
p2 
q2 f 2
q2  f 2
(25 cm)(5 cm)
;
-25 cm - 5 cm
90 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 cm.
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Physics, 6th Edition
Chapter 36. Lenses and Optical Instruments
*36-39. (Cont.)
(b) p2 = 95.54 cm – 91.37 cm;
M
q1q2 (91.37 cm)( 25 cm)

;
p1 p2 (6000 cm)(4.17 cm)
y’ = (-0.0914)(30 cm);
p2 = 4.17 cm;
M = -0.0914;
M = M1 x M2
y’ = My
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?
q1 
p2 
*36-6.
x
(4000 cm)(30 cm)
 30.23 cm;
4000 cm - 30 cm
q2 f 2
(25 cm)(-2.5 cm)

;
q2  f 2 -25 cm - (-2.5 cm)
p2 = x – 30.23 cm;
M
4000 cm
p2 = -2.78 cm
-25 cm
x = 30.23 cm + p2 = 30.23 cm +(-2.78 cm);
x = 27.5 cm
q1q2 (30.23 cm)(-25 cm)

;
p1 p2 (4000 cm)(2.78 cm)
M = -0.068
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;
p1 
q1 f1
(235 mm)(19 mm)

;
q1  f1 235 mm - 19 mm
509
q = 20.7 mm