1
Chapter 23
3.
(II) Two mirrors meet at a 135° angle, Fig. 23–47. If light rays strike one
mirror at 40° as shown, at what angle  do they leave the second mirror?

3. From the triangle formed by the mirrors and the
first reflected ray, we have
      180;

40  135    180, which gives   5.
8.

(I) How far from a concave mirror (radius 23.0 cm) must an object be placed if
its image is to be at infinity?
9. The ball is a convex mirror with a focal length
r   4.5cm 
f  
 2.25cm.
2
2
We locate the image from
 1  1 1
   ;
 d o   di  f

 1
1
1
, which gives di   2.09cm.

 
  30.0cm    di    2.25cm 
The image is 2.09 cm behind the surface, virtual.
The magnification is
   2.09cm 
d
m i 
  0.070.
do
 30.0cm 
Because the magnification is positive, the image is upright.
2
11.
(II) A dentist wants a small mirror that, when 2.20 cm from a tooth, will
produce a 4.5  upright image. What kind of mirror must be used and what must
its radius of curvature be?
11. We find the image distance from the magnification:
h d
m i  i ;
ho
do
 di
 4.5 
, which gives di   9.90cm.
 2.20cm 
We find the focal length from
 1  1 1
   ;
 d o   di  f

 
 1
1
1


  , which gives f  2.83cm.
  2.20cm      9.90cm   f
Because the focal length is positive, the mirror is concave with a radius of
r  2 f  2  2.83cm  5.7cm.
14.
(II) You are standing 3.0 m from a convex security mirror in a store. You
estimate the height of your image to be half of your actual height. Estimate the
radius of curvature of the mirror.
14. We find the image distance from the magnification, noting an upright image as in Fig. 23-46.
  3.0m   1.5m.
h  di
 di
m   0.5  i 

, which yields di 
ho
do
 3.0m 
  0.5
The thin lens equation gives the inverse focal length

1  1  1  1  
1
    

 , so f   3.0 m.
f  d o   di    3.0 m     1.5m  
The radius of the mirror is r  2 f  2  3.0m    6.0m.
16.
(II) The image of a distant tree is virtual and very small when viewed in a
curved mirror. The image appears to be 18.0 cm behind the mirror. What kind
of mirror is it, and what is its radius of curvature?
16. We take the object distance to be ∞, and find the focal length from
 1  1 1
   ;
 d o   di  f

 1
1

  , which gives f  18.0cm.
  18.0cm   f
Because the focal length is negative, the mirror is convex.
The radius is
1
 

3
r  2 f  2  18.0cm   36.0cm.
24.
(I) The speed of light in ice is 2.29  108 m s . What is the index of refraction of ice?
24. We find the index of refraction from
c
v ;
n
3.00  108 m s 

8
2.29  10 m s 
, which gives n  1.31.
n
25.
(II) The speed of light in a certain substance is 89% of its value in water. What is the index of
refraction of this substance?
25. We find the index of refraction from
c
v ;
n
0.89c c
0.89 vwater 
 , which gives n  1.49.
1.33 n
27.
(I) A diver shines a flashlight upward from beneath the water at a 42.5° angle to the vertical. At
what angle does the light leave the water?
27. We find the angle of refraction in the water from
n1 sin 1  n2 sin  2 ;
1.33 sin 42.5    sin 2 ,
30.
which gives  2  64.0.
(II) An aquarium filled with water has flat glass sides whose
index of refraction is 1.52. A beam of light from outside the
aquarium strikes the glass at a 43.5° angle to the perpendicular
(Fig. 23–49). What is the angle of this light ray when it enters
(a) the glass, and then (b) the water? (c) What would be the
refracted angle if the ray entered the water directly?
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30. (a) We find the angle in the glass from the refraction
at the air–glass surface:
n1 sin 1  n2 sin  2 ;
1.00 sin 43.5    sin 2 , which gives  2 
26.9.
(b) Because the surfaces are parallel, the refraction angle
from the first surface is the incident angle at the second
surface. We find the angle in the water from the refraction
at the glass–water surface:
n2 sin  2  n3 sin  3 ;
1.52 sin 26.9   sin3 ,
which gives  3  31.2.
(c) If there were no glass, we would have
n1 sin 1  n3 sin  3;
1.00 sin 43.5   sin3,
which gives  3  31.2.
Note that, because the sides are parallel,  3 is independent of the presence of the glass.
39.
(II) A beam of light is emitted 8.0 cm beneath the surface of a liquid and strikes the surface 7.0
cm from the point directly above the source. If total internal reflection occurs, what can you say
about the index of refraction of the liquid?
39. We find the angle of incidence from the distances:
L  7.0cm 
tan 1  
 0.875, so 1  41.2.
h 8.0cm 
For the maximum incident angle for the refraction from liquid into air, we have
nliquid sin1  nair sin 2 ;
nliquid sin 1max  1.00  sin 90, which gives sin 1max 
1
nliquid
.
Thus we have
sin 1  sin 1max 
1
nliquid
sin 41.2  0.659 
47.
;
1
nliquid
, or nliquid  1.5.
(II) A stamp collector uses a converging lens with focal length 24 cm to view a
stamp 18 cm in front of the lens. (a) Where is the image located? (b) What is the
magnification?
47. (a) We locate the image from
 1  1 1
    ;
 d o   di  f
 1  1
1
, which gives di   72cm.

  
18cm
d
24cm

  i
The negative sign means the image is 72 cm behind the lens (virtual).
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(b) We find the magnification from
  72cm    4.0.
d
m i 
do
18cm 
49.
(II) An 80-mm-focal-length lens is used to focus an image on the film of a
camera. The maximum distance allowed between the lens and the film plane is
120 mm. (a) How far ahead of the film should the lens be if the object to be
photographed is 10.0 m away? (b) 3.0 m away? (c) 1.0 m away? (d) What is the
closest object this lens could photograph sharply?
49. (a) We find the image distance from
 1  1 1
    ;
 d o   di  f

 1
1
1
, which gives di  81mm.

  
3
10.0

10
mm
d
80mm

  i
(b) For an object distance of 3.0 m, we have

 1
1
1
, which gives di  82mm.

  
3
 3.0  10 mm   di  80mm
(c) For an object distance of 1.0 m, we have

 1
1
1
, which gives di  87 mm.

  
3
 1.0  10 mm   di  80mm
(d) We find the smallest object distance from
 1   1 
1
, which gives di  240mm  24cm.

  

d
120mm
80mm


omin


53.
(II) (a) How far from a 50.0-mm-focal-length lens must an object be placed if its
image is to be magnified 2.00  and be real? (b) What if the image is to be
virtual and magnified 2.00 ?
53. We can relate the image and object distance from the magnification:
d
d
m  i , or do  i .
do
m
We use this in the lens equation:
 1  1 1
    ;
 d o   di  f
m  1  1
       , which gives di  1  m  f .
 di   di  f
(a) If the image is real, di  0. With f  0, we see that m  1; thus m   2.00. The image
distance is
di  1    2.00  50.0mm   150mm.
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The object distance is
 d  150mm 
do  i 
 75.0mm.
m
  2.00
(b) If the image is virtual, di  0. With f  0, we see that m  1; thus m   2.00. The
image distance is
di  1    2.00  50.0mm    50mm.
The object distance is
 d    50mm 
do  i 
 25.0mm.
m
  2.00
55.
(II) (a) A 2.00-cm-high insect is 1.20 m from a 135-mm-focal-length lens. Where
is the image, how high is it, and what type is it? (b) What if f  135 mm?
55. (a) We find the image distance from
 1  1 1
    ;
 d o   di  f

 1
1
1
, which gives di  152mm  real,behind thelens .

  
3
 1.20  10 mm   di  135mm
We find the height of the image from
h
d
m i  i ;
ho
do
 152 mm 
hi

, which gives hi   0.254cm  inverted  .
2.00cm 1.20  103 mm 
(b) We find the image distance from
 1  1 1
    ;
 d o   di  f

 1
1
1
,

  
3
 1.20  10 mm   di   135mm 
which gives di  121mm  virtual,in front of thelens .
We find the height of the image from
h
d
m i  i ;
ho
do
  121mm 
hi

, which gives hi   0.202cm  upright  .
 2.00cm  1.20  103 mm 