Physics 112
Exam 3
Summer 2012
41. A light beam coming from an underwater spotlight exits the water (index of refraction
is n=1.33) at an angle of 66.0° to the vertical. At what angle of incidence does it hit
the air–water interface from below the surface?
A)
B)
C)
D)
E)
33◦
43◦
53◦
63◦
66◦
Solution:
We find the incident angle in the water from:
n1 sin 1  n2 sin  2 ;
1.33 sin1   sin 66.0,
which gives 1  43.4.
42. A beam of light is emitted in a pool of water (index of refraction is n=1.33) from a
depth of 62.0 cm. Where must it strike the air–water interface, relative to the spot
directly above it, in order that the light does not exit the water? (See fig.)
A)
B)
C)
D)
E)
R>71cm
R<71cm
42cm<R<50cm
35cm<R<42cm
R<35cm
air
R
H
1
1
n
Solution:
We find the critical angle for light leaving the water:
n sin 1  sin  2 ;
1.33 sin C  sin 90,
which gives  C  48.8.
If the light is incident at a greater angle than this, it will totally reflect. We see from the diagram
that
R  H tan C   62.0cm tan 48.8  70.7cm.
43. A mirror at an amusement park shows an upright image of any person who stands
1.4 m in front of it. If the image is three times the person’s height, what is the radius
of curvature?
A)
B)
C)
D)
E)
1.4 m
2.4 m
3.2 m
4.2 m
5.4 m
Page 1 of 10
Physics 112
Exam 3
Summer 2012
Solution:
We find the image distance from the magnification:
h d
m i  i ;
ho
do
 di
3 
, which gives di   4.2 m.
1.4 m 
We find the focal length from
 1  
 1
 1  1 1
1


  , which gives f  2.1m.
   ;
 1.4 m      4.2 m   f
 d o   di  f
The radius of the concave mirror is
r  2 f  2  2.1m   4.2m.
44. Convex spherical mirrors produce images which are
A)
B)
C)
D)
E)
always real and smaller than the actual object
always virtual and smaller than the actual object
always real and larger than the actual object
always virtual and larger than the actual object
could be real or virtual and larger or smaller than the actual object, depending on the
placement of the object
45. An object is placed between a convex lens and its focal point. The image formed is
A)
B)
C)
D)
E)
virtual and erect
virtual and inverted
real and erect
real and inverted
could be real or virtual and erect or inverted
46. A convex lens has a focal length f. An object is placed between f and 2f on the axis.
The image formed is located
A)
B)
C)
D)
E)
at 2f
between f and 2f
at f
at a distance greater than 2f from the lens
at a distance smaller than f
Page 2 of 10
Physics 112
Exam 3
Summer 2012
47. 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. What is the focal length of the
mirror?
A)
B)
C)
D)
E)
-38 cm
-68 cm
-98 cm
68 cm
98 cm
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
48. A parallel beam of light from a He-Ne laser, with a wavelength 656 nm, falls on two
very narrow slits 0.060 mm apart. How far apart are the fringes in the center of the
pattern on a screen 3.6 m away?
A)
B)
C)
D)
E)
0.2 cm
0.6 cm
1.7 cm
2.8 cm
3.9 cm
Solution:
For constructive interference, the path difference is a multiple of the wavelength:
d sin  m , m  0, 1, 2, 3, ... .
We find the location on the screen from
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Physics 112
Exam 3
Summer 2012
y  L tan .
For small angles, we have sin  tan , which gives
 m  mL
y  L
.

d
 d 
For adjacent fringes, m  1, so we have
y 
 3.6 m   656  109 m  1
L m
 3.9  102 m  3.9 cm.
; 
3
d
 0.060  10 m 
49. In a double-slit experiment, it is observed that the distance between adjacent maxima
on a remote screen is 1.0 cm. What happens to the distance between adjacent
maxima when the slit separation is cut in half?
A)
B)
C)
D)
E)
It increases to 2.0 cm.
It increases to 4.0 cm.
It increases to 8.0 cm.
It decreases to 0.50 cm.
It decreases to 0.25 cm.
Solution:
m
yx
d
50. When violet light of wavelength 415 nm falls on a single slit, it creates a central
diffraction peak that is 9.20 cm wide on a screen that is 2.55 m away. How wide is
the slit?
A)
B)
C)
D)
E)
23 cm
2.3 cm
5.2 mm
5.2m
23m
Solution:
We find the angle to the first minimum from the distances:
 9.20cm   0.0180  sin , because the angle is small.
tan1min  12
1min
 255cm 
We find the slit width from
D sin 1min  m ;
D  0.0180   1  415  109 m  , which gives D  2.30  105 m  0.0230mm.
Page 4 of 10
Physics 112
Exam 3
Summer 2012
51. Which color of visible light would give the best resolution in a microscope?
A)
B)
C)
D)
E)
Red
Yellow
Green
Blue
All colors give the same resolution
Solution:
To get the best resolution from a microscope, you should use blue/violet light. The Rayleigh


criterion   
1.22 
 shows that the shorter wavelengths of light have higher resolution. (Of
D 
course, you would need very special lenses, since most glass absorbs blue/violet light very
strongly.)
52. The first-order line of 589-nm light falling on a diffraction grating is observed at a
15.5° angle. At what angle will the third order be observed?
A)
B)
C)
D)
E)
31.0°
46.5°
53.3°
62.1°
71.4°
Solution:
sin  m 
sin  3 3
m

 
d
sin 1 1
sin  3  3 sin 1  3 sin 15.5   0.8017 
 3  53.3
53. 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, and why?
A)
B)
C)
D)
E)
152 nm
228 nm
320 nm
440 nm
570 nm
Solution:
The total phase shift is zero. For constructive interference: 2t  m film  m n .
The minimum non-zero thickness occurs for m  1:
 570nm   228nm.

tmin 

2nfilm
2 1.25
Page 5 of 10
Physics 112
Exam 3
Summer 2012
54. What is Brewster’s angle for an air–glass (n  1.52) surface?
A)
B)
C)
D)
E)
31.2°
41.0°
45.0°
48.1°
56.7°
Solution:
Because the light is coming from air to glass, we find the angle from the vertical from
tan p  nglass  1.52, which gives  p  56.7.
1
55. A light meter reports that a camera setting of 200
s at f 8 will give a correct
exposure. But the photographer wishes to use f 16 to increase the depth of field.
What should the shutter speed be (time shutter is open)?
A)
B)
C)
D)
E)
(1/50)s
(1/100)s
(1/200)s
(1/400)s
(1/800)s
Solution:
f D8
f D '  16
t  1 / 200 s
2
2
2
 f ' D'
D
 D' 
1
 1  16 
  
t    t '    t '  t 
s   
s
50
 200  8 
 f 
 f '
 f D 
2
56. What statement is wrong?
A)
B)
C)
D)
E)
Far point is farthest distance at which object can be seen clearly.
Normal far point is about 25 cm.
Nearsightedness means that far point is too close.
Near point is closest distance at which eye can focus clearly.
Farsightedness means near point is too far away.
Solution:
Normal FP is at infinity.
Normal NP is about 25 cm.
Page 6 of 10
Physics 112
Exam 3
Summer 2012
57. Reading glasses of what power are needed for a person whose near point is 115 cm,
so that he can read a computer screen at 55 cm? Assume a lens–eye distance of 1.8cm.
A)
B)
C)
D)
E)
-1D
-2D
+1D
+2D
+3D
Solution:
With the lens, the screen placed 55 cm from the eye, or 53.2 cm from the lens, is to produce a
virtual image 115 cm from the eye, or 113.2 cm from the lens. We find the power of the lens from
 1  1 1
       P, when distances are in m;
 d o   di  f
 1  

1


  P  1.0D.
 0.532m   1.132m 
58. A small insect is placed 5.00 cm from a 6.00 -cm-focal-length lens. Calculate the
angular magnification.
A)
B)
C)
D)
E)
2
3
4
5
6
Solution:
 ' 25cm
25cm
M  

 5.00

d0
5.00cm
59. An astronomical telescope has an objective with focal length 85 cm and a 35-D
eyepiece. What is the total magnification?
A)
B)
C)
D)
E)
-17
+17
-30
+30
-33
Solution:
f
M   0   f 0 Pe ;
fe


M  0.85m 35m 1  30 .
Page 7 of 10
Physics 112
Exam 3
Summer 2012
60. A microscope has a 1.8-cm-focal-length eyepiece and a 0.85-cm objective lens. The
distance between the lenses is 16.0 cm. Calculate the total magnification (ignore
minus sign).
A)
B)
C)
D)
E)
55
83
110
150
232
Solution
 25cm  l  f e

M  
 f e  d 0
  25cm  16cm  1.8cm 
  

  232
  1.8cm  0.85cm 
41-23.28
42-23.38
43-23.10
44-L18
45-L19
46-L19
47-23.21 (hw10)
48-24.4
49-L20
50-24-24
51-Q25.17
52-24.31
53-24.42
54-24.54
55-25.3 (modified)
56-L23 (modified)
57-25.12
58-25.26
59-25.26b
60-25.43b (hw13)
Page 8 of 10
Physics 112
Exam 3
Summer 2012
#39. 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?
Solution:
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 
1
nliquid
, or nliquid  1.5.
Page 9 of 10
Physics 112
Exam 3
Summer 2012
Record Sheet
41
B) 43◦
51
D) Blue
42
A) R>71 cm
52
C) 53.3°
43
D) 4.2 m
53
B) 228 nm
44
B) always virtual and
smaller than the actual
object
54
E) 56.7°
45
A) virtual and erect
55
A) (1/50)s
46
D) at a distance greater
than 2f from the lens
56
B) Normal far point is
about 25 cm
47
C) -98 cm
57
C) +1D
48
E) 3.9 cm
58
D) 5
49
A) It increases to 2.0 cm.
59
C) -30
50
E) 23m
60
E) 232
Page 10 of 10