Physics 112
Exam 3
Summer 2010
41. 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?
A) 2.5m
B) 5.0m
C) 6.25 m
D) 25.0 m
E) none of these
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.
42. The speed of light in a certain substance is 89% of its value in water. What is the
index of refraction of this substance? (The index of refraction of water is 1.33)
A) 1.00
B) 1.18
C) 1.33
D) 1.49
E) 1.52
Solution:
We find the index of refraction from n 
n
c
c
1.33

 water 
 n  1.49
v 0.89v water 0.89 0.89
43. A ray of light is refracted through three different materials. Rank the materials
according to their index of refraction.
A) n1  n2  n3
B) n1  n3  n2
C) n2  n1  n3
D) n2  n3  n1
E) n3  n2  n1
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 = smaller n material to larger n material; bends away from normal = larger n material to
smaller n material).
From the first material to the second material the ray bends toward the normal, thus n1 < n2.
From the second material to the third material the ray bends away from the normal, thus 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 and n1 < n3.
Thus, the overall ranking of indices of refraction is: n1 < n3 < n2.
Physics 112
Exam 3
Summer 2010
44. A substance has an index of refraction of 1.46. Light is passing through it at 53.0°.
At what angle will it leave into the air?
A) It will not leave.
B) 59.1°
C) 53.0°
D) 43.2°
E) 33.2°
Solution:
The critical angle for the total internal refractions is given by the equation:
 c  43.2   53.0  .
sin  c  n2 n1 
sin  c  1 1.46 
45. Is it possible to see a virtual image?
A) No, since the rays that seem to emanate from a virtual image do not in fact emanate
from the image.
B) No, since virtual images do not really exist.
C) Yes, the rays that appear to emanate from a virtual image can be focused on the retina
just like those from an illuminated object.
D) Yes, since almost everything we see is virtual because most things do not themselves
give off light, but only reflect light coming from some other source.
E) Yes, but only indirectly in the sense that if the virtual image is formed on a sheet of
photographic film, one could later look at the picture formed.
Solution:
Yes, the rays coming from the real object can appear to emanate from a virtual image,
and can be focused on the retina.
46. 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? Is the image real or virtual? Is the image
inverted or upright?
A) At the focus. It is real and inverted.
B) At the focus. It is real and upright.
C) At the focus. It is virtual and inverted.
D) At the center of curvature. It is real and inverted.
E) At the center of curvature. It is virtual and inverted.
Solution:
a) With di  do , we locate the object from
 1  1 1
   ;
 d o   di  f
Physics 112
Exam 3
Summer 2010
 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.
47. A single concave spherical mirror produces an image which is
A) always virtual
B) always real
C) real only if the object distance is less than f
D) real only if the object distance is greater than f
E) real only if the object distance is greater than 2f
48. An object is placed between a converging lens and its focal point. The image formed
is
A) virtual and erect
B) virtual and inverted
C) real and erect
D) real and inverted
E) image is not formed
49. A glass converging lens has one flat side and another with a radius of 10 cm. What is
the focal length of the lens? ( The index of refraction of the glass in the lens is 1.50)
A) 10 cm
B) 20 cm
C) 30 cm
D) 40 cm
E) 50 cm
Solution:
R1  
R2  10cm
n  1.5
f ?
 1
1
1 

 n  1 
f
R
R
2 
 1
1
1 
1

 1.5  1 0 

f
 10cm  20cm
f  20cm
Physics 112
Exam 3
Summer 2010
50. An object is placed in front of converging lens with focal distance 15 cm. What is the
distance between the object and the lens if magnification is 3?
A) 5 cm
B) 10 cm
C) 15 cm
D) 20 cm
E) 25 cm
Solution:
d
m   i  d i  md 0
d0
1
1 1
  
d0 di
f
1

d 0  f 1   ;
 m
1
1
1

 
d 0 md 0
f
1  1 1
1    
d0  m  f
 1
d 0  15cm 1   
 3
d 0  10cm .
51. In a Young's double slit experiment, if the separation between the two slits is 0.050
mm and the distance from the slits to a screen is 2.5 m, find the spacing between the
first-order and second-order bright fringes for light with wavelength of 600 nm.
A) 1.0 cm
B) 2.0 cm
C) 3.0 cm
D) 4.0 cm
E) 5.0 cm
Solution:
Given:
d  0.050mm
  600nm
x  2.5m
y2  y1  ?
Conditions for constrictive interference:
From geometry:
Because  / d  1  sin  m  tan  m .
Combining these equations we have:
y2  y1  x / d
y2  y1  3.0cm
d sin  m  m
tan  m  y m x

dy m x  m 


y m  m x d 

y 2  y1  600  10 9 m 2.5m / 0.050  10 3 m  3.0  10 2 m
Physics 112
Exam 3
Summer 2010
52. The separation between adjacent maxima in a double-slit interference pattern using
monochromatic light is
A) greatest for red light
B) greatest for green light
C) greatest for blue light
D) the same for all colors of light
E) greatest for red light or for blue light depending on the distance between the slits
Solution:
Conditions for constrictive interference: d sin  m  m .
For given index m, angle  m is increasing with increasing wavelength  .
Maximum value of  m is for the maximum value of  , which is for red light.
53. In order to obtain a good single slit diffraction pattern, the slit width could be:
A) /100
B) /10
C) 
D) 10
E) 100
Solution:
D sin  m  m
m  1,2,3,...
If D  10 , than sin  m  0.1m and one can observe several dark and bright fringes.
54. Monochromatic light falls on a slit that is 2.20  10 3 mm wide. If the angle between
the first dark fringes on either side of the central maximum is 35.0° (dark fringe to
dark fringe), what is the wavelength of the light used?
A) 466nm
B) 552nm
C) 585nm
D) 610nm
E) 662nm
Solution:
The angle from the central maximum to the first minimum is 1  35 / 2  17.5 .
We find the wavelength from
  D sin 1
D sin  m  m 
  2.20  10 mm sin 17.5 ,
3

m  1,2,3,...
which gives
  6.62  10 7 m  662nm
Physics 112
Exam 3
Summer 2010
55. What is the highest spectral order that can be seen if a grating with 6000 lines per cm
is illuminated with 633-nm laser light? Assume normal incidence.
A) 2
B) 3
C) 4
D) 5
E) 6
Solution:
d sin  m  m
m  0,1,2,...
The maximum angle is 90°, so we have


 2
1
 10 m / cm 
m  
 6000 lines / cm 


  90 and m 
 d .
d sin 90 

 633 10 m  2.6 .
9
Thus two orders can be seen on each side of the central white line.
56. A lens appears greenish yellow (   570 nm is strongest) when white light reflects
from it. What minimum thickness of coating (n  1.25) do you think is used on such a
glass (n  1.52) lens?
A) 180nm
B) 205nm
C) 228nm
D) 350nm
E) 420nm
Solution:
There are a phase shifts (  / 2 ) on both boundaries of the coating film. The total phase
shift is  . Constrictive interference should be observed if 2t  m film with minimum
m=1.
 film  570nm
t min 


 228nm
2
2n 2 1.25
Physics 112
Exam 3
Summer 2010
57. A binary star system in the constellation Orion has an angular separation of 10-5 rad.
If the wavelength of the light from the system is 500 nm, what is the smallest aperture
(diameter) telescope that can just resolve the two stars?
A) 0.50 cm
B) 0.61 cm
C) 5.0 cm
D) 6.1 cm
E) 7.0 cm
Solution:
  1  sin   
D  1.22 sin  1.22 
D sin   1.22 
9
D  1.22500  10 m 10 5  6.1  10 2 m  6.1cm
58. A magnifying glass with a focal length of 7.5cm is used to read print placed at a
distance 5.0 cm. Calculate the angular magnification.
A) 1
B) 2
C) 3
D) 4
E) 5
Solution:
M
 ' 25cm 25cm


5

d0
5.0cm
59. A person’s eye is corrected by a  4.00-diopter lens. What is this eye’s far point
without glasses? (Assume that the distance between the eye and the lens can be
neglected.)
A) 20cm
B) 25cm
C) 30cm
D) 35cm
E) 40cm
Solution:
We find the far point by finding the image distance for an object at infinity ( d 0   ):
1
1
1
1 1

 P

 4.00 D 
d i  0.250m  25.0cm
d0 di
f
 di
Physics 112
Exam 3
Summer 2010
60. An astronomical telescope has its two lenses spaced 90.0 cm apart. If the objective
lens has a focal length of 89.0 cm, what is the magnification of this telescope?
Assume a relaxed eye.
A) -45
B) -80
C) -90
D) -110
E) -120
Solution:
For both object and image far away, we find the focal length of the eyepiece from the
separation of the lenses:
L  fo  fe ;
 f e  L  f 0  90.0cm  89.0cm  1.0cm
The magnification of the telescope is given by
M 
fo
90.0cm

 90 .
fe
1.0cm
Physics 112
Exam 3
Summer 2010
Record Sheet
You may fill in this sheet with your choices, detach it and take it with you after the exam
for comparison with the posted answers
41
B) 5.0m
51
C) 3.0 cm
42
D) 1.49
52
A) greatest for red light
43
B) n1  n3  n2
53
D) 10
44
A) It will not leave.
54
E) 662nm
45 C) Yes, the rays that
55
A) 2
appear to emanate from a
virtual image can be focused
on the retina just like those
from an illuminated object.
46
D) At the center of
curvature. It is real and
inverted.
47
D) real only if the object
distance is greater than f
56
C) 228nm
48
A) virtual and erect
58
E) 5
49
B) 20 cm
59
B) 25cm
50
B) 10 cm
60
C) -90
57
D) 6.1 cm