Physics 112
Homework 9 (Ch23)
Due July 17
1. Suppose that you want to take a photograph of yourself as you look at your image in a flat
mirror 2.5 m away. For what distance should the camera lens be focused?
Solution
For a flat mirror the image is as far behind the mirror as the object is in front, so the
distance from object to image is
do  di  2.5m  2.5m  5.0m.
2. A person whose eyes are 1.68 m above the floor stands 2.20 m in front of a vertical plane
mirror whose bottom edge is 43 cm above the floor (see figure below) . What is the
horizontal distance x to the base of the wall supporting the mirror of the nearest point on the
floor that can be seen reflected in the mirror?
Solution
The angle of incidence is the angle of reflection.
Thus we have
tan 
( H  h) h
 ;
L
x
Mirror
H
x  hL /( H  h)
x  0.43m 2.20m  / 1.68m  0.43m 


which gives x  0.76m  76cm.
h
x
L
3. 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?
Solution
Only concave mirror can produce magnified image.
We find the image distance from the magnification:
m
hi  di

;
ho
do
 d i  md 0 ;
 d i  4.5  2.2cm ; which gives di   9.90cm.
We find the focal length from
 1  1
 
 d o   di
 1
 ;
 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.
r  2 f  2d 0
m
4.5
 2  2.20cm 
 5.7cm
m 1
4.5  1
Physics 112
Homework 9 (Ch23)
Due July 17
4. The speed of light in a certain substance is 89% of its value in water. What is the index of
refraction of this substance?
Solution
We find the index of refraction from
c
v ;
n
0.89 vwater 
0.89c c
 ,
1.33 n
which gives n  1.49.
5. 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?
Solution
We find the angle of refraction in the water from
n1 sin 1  n2 sin  2 ;
1.33 sin 42.5    sin 2 , which gives  2 
64.0.
6. In searching the bottom of a pool at night, a watchman shines a narrow beam of light from
his flashlight, 1.3 m above the water level, onto the surface of the water at a point 2.7 m from
the edge of the pool (see figure below). Where does the spot of light hit the bottom of the
pool, measured from the wall beneath his foot, if the pool is 2.1 m deep?
Solution
We find the angle of incidence from the distances:
tan1 
L1  2.7 m 

 2.076, so 1  64.3.
h1 1.3m 
For the refraction from air into water, we have
nair sin 1  nwater sin  2 ;
1.00 sin 64.3   sin 2 , which gives  2  42.6.
We find the horizontal distance from the edge of the pool from
L2  L1  h2 tan  2 
 2.7m   2.1m  tan 42.6  . m.
h2
nair
1
h1
2
L1
L2
nwater
Physics 112
Homework 9 (Ch23)
Due July 17
7. A beam of light in air strikes a slab of glass (n  1.52) and is partially reflected and partially
refracted. Find the angle of incidence if the angle of reflection is twice the angle of
refraction.
Solution
The angle of reflection is equal to the angle of incidence:
 ref1  1  2 2 .
For the refraction we have
nair sin1  nglass sin  2 ;
1.00 sin 2 2  1.52 sin 2 .
We use a trigonometric identity for the left-hand side:
sin 2 2  2sin  2 cos 2  1.52  sin  2 , or cos 2  0.760, so  2  40.5.
Thus the angle of incidence is 1  2 2  81.0.
8. Where should an object be placed in front of a concave mirror so that it produces an image at
the same location as the object? (b) Is the image real or virtual? (c) Is the image inverted or
upright? (d) What is the magnification of the image?
Solution
(a) With di  d o , we locate the object from
 1  1 1
   ;
 d o   di  f
 1   1  1
      , which gives d o  2 f  r.
 do   do  f
The object should be placed at the center of curvature.
(b) Because the image is in front of the mirror, d i  0, it is real.
(c) The magnification is
m
 di  d o

 1.
do
do
Because the magnification is negative, the image is inverted.
(d) As found in part (c), m  1.
Physics 112
Homework 9 (Ch23)
Due July 17
9. A 4.5-cm-tall object is placed 28 cm in front of a spherical mirror. It is desired to produce a
virtual image that is upright and 3.5 cm tall. (a) What type of mirror should be used? (b)
Where is the image located? (c) What is the focal length of the mirror? (d) What is the radius
of curvature of the mirror?
Solution
(a) To produce a smaller image located behind the surface of the mirror requires a convex
mirror.
(b) We find the image distance from the magnification:
m
hi  di

;
ho
do
 3.5cm    di , which gives
 4.5cm   28cm 
di   21.8cm.
As expected, di  0. The image is located 22 cm behind the surface.
(c) We find the focal length from
 1  1 1
    ;
 d o   di  f
 1  
 1
1


  , which gives f   98cm.
  28cm      21.8cm   f
(d) The radius of curvature is
r  2 f  2   98cm   196cm.