Physics 112
Homework 10 (Ch23 &24)
Due July 22
1. A sharp image is located 78.0 mm behind a 65.0-mm-focal-length converging lens. Find the
object distance (a) using a ray diagram, (b) by calculation.
Solution
(a) From the ray diagram, the object distance is about six focal lengths, or 390 mm.
F I
O
F
(b) We find the object distance from equation:
1
1 1
  
d o di f
 1  

1
1
,
 

 do   78.0mm  65.0mm
which gives do  390mm  39.0cm.
2. 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?
Solution
(a) We locate the image from equation:
1
1 1
  
d o di f
 1  1
1
,

  
 18cm   di  24cm
which gives di   72cm.
The negative sign means the image is 72 cm behind the lens (virtual).
(b) We find the magnification from equation:
m
  72cm    4.0.
di

do
18cm 
3. An object is located 1.5 m from an 8.0-D lens. By how much does the image move if the
object is moved (a) 1.0 m closer to the lens, and (b) 1.0 m farther from the lens?
Solution
We find the image distance from equation:
1
1 1
  
d o di f
1
1
  8.0D,
1.5m di
which gives di  0.1364 m.
(a) With d o  0.5m, the new image distance is determined by
1
1
  8.0D,
0.5m di
which gives di  0.1667 m.
The image has moved 0.1667 m  0.1364m  0.0303m or 3.0 cm away from the lens.
Physics 112
Homework 10 (Ch23 &24)
Due July 22
(b) With d o  2.5m, the new image distance is determined by
1
1
  8.0D,
2.5m di
which gives di  0.1316 m.
The image has moved 0.1316m  0.1364m   0.0048m or 0.5 cm toward the lens.
4. 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?
Solution
(a) We find the image distance from equation:

 1
1
1
,

  
3
 1.20  10 mm   di  135mm
1
1 1
  
d o di f
which gives di  152mm  real,behind thelens .
We find the height of the image from equation:
m
hi
d
 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 equation:
1
1 1
  
d o di f

 1
1


3
 1.20  10 mm   di

1
, which gives

  135mm 
di  121mm  virtual,in front of thelens .
We find the height of the image from equation:
m
hi
d
 i ;
ho
do
  121mm 
hi

, which gives hi   0.202cm  upright .
 2.00cm  1.20  103 mm 
5. Both surfaces of a double convex lens have radii of 31.0 cm. If the focal length is 28.9 cm,
what is the index of refraction of the lens material?
Solution
R  R1  R2  31.0cm
We find the index from the lensmaker’s equation:
1
1
 1
 1
1
1 
1 
R
 2
n  1   f     1   f   1 
 n  1   
2f
f
 R
 R1 R2 
  R1 R2 
31cm
n  1
n  1.54.
2  28.9cm
Note! We using different sign convention than the Giancoli book.
Physics 112
Homework 10 (Ch23 &24)
Due July 22
6. A prescription for a corrective lens calls for 1.50 D. The lensmaker grinds the lens from a
“blank” with n  1.56 and a preformed convex front surface of radius of curvature of 40.0 cm.
What should be the radius of curvature of the other surface?
Solution
We find the radius from the lensmaker’s equation:
1
1
1 
 n  1   
f
 R1 R2 
1
1
1
P



R2 R1 n  1
1
1
1



f n  1 R1 R2
1
1
P 
1.5 
 1
R2   
R2  

 5.6m ; R2  5.6m

 0.400m 1.56  1
 R1 n  1
Note! We using different sign convention than the Giancoli book.
7. A ray of light is refracted through three different materials (see figure below). Rank the
materials according to their index of refraction, least to greatest.
Solution
By looking at the direction and the relative amount that the light rays bend at each interface, we can infer
the relative sizes of the indices of refraction in the different materials (bends toward normal = faster
material to slower material or smaller n material to larger n material; bends away from normal = slower
material to faster material or larger n material to smaller n material). From the first material to the second
material the ray bends toward the normal, thus it slows down and n1 < n2. From the second material to the
third material the ray bends away from the normal, thus it speeds up and n2 > n3. Careful inspection
shows that the ray in the third material does not bend back away from the normal as far as the ray was in
the first material, thus the speed in the first material is the faster than in the third material and n1 < n3.
Thus, the overall ranking of indices of refraction is: n1 < n3 < n2.
8. Monochromatic light falls on two very narrow slits 0.048 mm apart. Successive fringes on a
screen 5.00 m away are 6.5 cm apart near the center of the pattern. Determine the wavelength
and frequency of the light.
Solution
We find the location on the screen from equation: y  L tan .
For small angles, we have:
sin  tan ,
 m  mL
which gives
y  L
.

d
 d 
Physics 112
Homework 10 (Ch23 &24)
Due July 22
For adjacent fringes, m  1, so we have
y 
L m
;
d
 



dy
0.048 10 3 m 6.510 2 m
 
; which gives   6.2  107 m.
Lm
5.00m 1
The frequency is
 3.00  10 m s  
f  
  6.24  10 m 
8
c
7
4.8  1014 Hz.
9. In a double-slit experiment, it is found that blue light of wavelength 460 nm gives a secondorder maximum at a certain location on the screen. What wavelength of visible light would
have a minimum at the same location?
Solution
For constructive interference of the second order for the blue light, we have
d sin  mb   2  460nm   920nm.
For destructive interference of the other light, we have
d sin    m  12   , m  0, 1, 2, 3, ... .
When the two angles are equal, we get
920 nm   m  12   , m  0, 1, 2, 3, ... .
For the first three values of m, we get
920 nm   0  12   , which gives   1.84  103 nm;
920 nm  1 
  , which gives   613nm;
920 nm   2  12   , which gives   368nm.
1
2
The only one of these that is visible light is 613 nm.
10.
A light beam strikes a piece of glass at a 60.00° incident angle. The beam contains two
wavelengths, 450.0 nm and 700.0 nm, for which the index of refraction of the glass is 1.4820
and 1.4742, respectively. What is the angle between the two refracted beams?
Solution
We find the angles of refraction in the glass from
n1 sin 1  n2 sin  2 ;
1.00 sin 60.00  1.4820 sin 2,450 , which gives  2,450  35.76;
1.00 sin 60.00  1.4742 sin 2,700 , which gives 2,700  35.98.
Thus the angle between the refracted beams is
 2,700   2,450  35.98  35.76  0.22.