Physics 112
Exam 3
Summer 2011
41. What is the angle of refraction when a light ray meets the boundary between two
materials perpendicularly?
A)
B)
C)
D)
E)
0
45
 45
90 
It depends on index of refraction
Solution
When a light ray meets the boundary between two materials perpendicularly, both the
angle of incidence and the angle of refraction are 0o. Snell’s law shows this to be true for
any combination of the indices of refraction of the two materials.
42. Light is passing through water at angle of incidence 55.0° to the surface. At what
angle will it leave into the air? (Index of refraction of water is 1.33)
A)
B)
C)
D)
E)
63.7°
53.2°
45.0°
38.9°
It will not leave
Solution:
The critical angle for the total internal refractions is given by the equation:
 c  48.8   55.0  .
sin  c  n2 n1 
sin  c  1 1.33 
43. The critical angle for a substance is measured at 53.7°. Light enters from air at 45.0°.
At what angle will it continue?
A) 28.2°
B) 34.7°
C) 45.0°
D) 53.7°
E) It will not continue, but be totally reflected.
Solution
Snell’s law: n1 sin 1  n2 sin  2
1) For light entering from the substance into air at critical angle 1   c  53.7  ,
n2  1 ,  2  90 ,
n1 sin  c  1  sin 90   n  n1  1 / sin  c
2) For light entering from air at 45.0°:
1  sin 45  n sin  2  sin  2  sin 45 / n  sin 45 sin 53.7  
 2  34.7 
Page 1 of 9
Physics 112
Exam 3
Summer 2011
44. An object is located 2.6 m in front of a plane mirror. The image formed by the mirror
appears to be
A)
B)
C)
D)
E)
1.3 m in front of the mirror
on the mirror's surface
1.3 m behind the mirror's surface
2.6 m behind the mirror's surface
2.6 m in front of the mirror
Solution
For plane mirror: d i d o
45. A single convex spherical mirror produces an image which is
A)
B)
C)
D)
E)
Always virtual and upright
Always virtual and inverted
Always real and upright
Always real and inverted
Real only if the object distance is less than focal distance.
Solution
From the ray diagram follows: “Always virtual and upright”.
46. A mirror at an amusement park shows an upright image of any person who stands
1.4m 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)
2.2 m
3.2 m
4.2 m
5.2 m
6.2 m
Solution
We find the image distance from the magnification:
 di
h d
3 
,
m i  i ;
ho
do
1.4 m 
which gives di   4.2 m.
We find the focal length from
 1  
 1
 1  1 1
1


 ,
   ;
 d o   di  f
 1.4 m      4.2 m   f
which gives f  2.1m.
The radius of the concave mirror is
r  2 f  2  2.1m  4.2m.
Page 2 of 9
Physics 112
Exam 3
Summer 2011
47. A negative magnification for a mirror means
A) The image is inverted, and the mirror is concave
B) The image is inverted, and the mirror is convex.
C) The image is inverted, and the mirror may be concave or convex
D) The image is upright, and the mirror is convex
E) The image is upright, and the mirror may be concave or convex
Solution
h
m  i  Negative m means that hi and h0 have opposite signs – image is inverted.
ho
For convex mirror image is always upright.
48. A negative magnification for a lens means
A) The image is inverted, virtual and the lens is converging
B) The image is inverted, real and the lens is converging
C) The image is, inverted, virtual and the lens is diverging
D) The image is inverted, real and the lens is diverging
E) The image is upright, real and the lens is diverging
Solution
h
m  i  Negative m means that hi and h0 have opposite signs– image is inverted.
ho
For diverging lens image is always upright.
49. When a light wave enters into a medium of different optical density,
A) Its speed and frequency change
B) Its speed and wavelength change
C) Its frequency and wavelength change
D) Its speed, frequency, and wavelength change
E) Its speed change, but frequency and wavelength remain unchanged
Solution
Wave frequency, f is determent by wave source. Speed, v is a characteristic of the
medium, and wavelength   v / f .
Page 3 of 9
Physics 112
Exam 3
Summer 2011
50. 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 of the light.
A) 4.2  10 7 m
B) 5.2  10 7 m
C) 6.2  10 7 m
D) 7.2  10 7 m
E) 8.2  10 7 m
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
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
L m
y 
;
d
 5.00 m   1
0.065m 
, which gives   6.2  107 m.
3
 0.048  10 m 
51. Monochromatic light falls on a slit that is 2.60  10 3 mm wide. If the angle between
the first dark fringes on either side of the central maximum is 30.0° (dark fringe to dark
fringe), what is the wavelength of the light used?
A)
B)
C)
D)
E)
450 nm
511 nm
585 nm
627 nm
673 nm
Solution
The angle from the central maximum to the first minimum is 1 = 30.0° /2 =15.0°
We find the wavelength from
D sin  m  m 
  D sin 1
  2.60  10 6 msin 15.0 , which gives   6.73  10 7 m  673nm
Page 4 of 9
Physics 112
Exam 3
Summer 2011
52. 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.
53. A 3500 -line cm grating produces a third-order fringe at a 28.0° angle. What
wavelength of light is being used?
A)
B)
C)
D)
E)
421 nm
447 nm
502 nm
631 nm
680 nm
Solution
We find the wavelength from
d sin  m;    d sin   / m



1

 2

 10 m / cm sin 28.0 / 3 , which gives   4.47  107 m  447nm.
 3500lines / cm 
 
54. 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.
Page 5 of 9
Physics 112
Exam 3
Summer 2011
55. 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)
B)
C)
D)
E)
570 nm
512 nm
450 nm
321 nm
228 nm
Solution
There are phase shifts on both surfaces: air-film and film-glass. For constructive
interference we have: 2t  m film  m / n film , and
tmin 

2nfilm

 570nm  
2 1.25
228nm.
56. 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.
Page 6 of 9
Physics 112
Exam 3
Summer 2011
57. What is Brewster’s angle for a diamond submerged in water if the light is hitting the
diamond (n  2.42 ) while traveling in the water (n =1.33)?
A)
B)
C)
D)
E)
33°
41°
52°
61°
72°
Solution
Because the light is coming from water to diamond, we find the angle from the vertical from
n
2.42
tan  p  diamond 
 1.82, which gives  p  61.2.
nwater
1.33
58. A person has a far point of 14 cm. What power glasses would correct this vision if
the glasses were placed 2.0 cm from the eye?
A)
B)
C)
D)
E)
+2.0
-2.0
-4.6
-6.5
-8.3
Solution
With the glasses, an object at infinity would have its image 14 cm from the eye or
14cm  2cm  12cm from the lens; di  12cm.
P

1  1  1 1 
1
       
   8.3D
f  do   di       0.12m 
59. A small insect is placed 5.0 cm from a 6.00 -cm-focal-length lens. Calculate the angular
magnification.
A)
B)
C)
D)
E)
1.2
2.4
4.1
5.0
6.0
Solution
Magnification is
M 
 ' 25cm 25cm


 5.0

d0
5.0cm
Page 7 of 9
Physics 112
Exam 3
Summer 2011
60. 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)
-15
-20
-30
-35
-41
Solution
We find the focal length of the eyepiece from the power:
fe 
1
1
1
1



m
1
P 35D 35m
35
The magnification of the telescope is given by
M 


fo
  f o P  0.85m 35m 1  29.75
fe
Page 8 of 9
Physics 112
Exam 3
Summer 2011
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
A) 0
51
E) 683 nm
42
E) It will not leave
52
D) 10
43
B) 34.7°
53
B) 447 nm
44
D) 2.6 m behind the
mirror's surface
54
A) greatest for red light
45
A) Always virtual and
upright
55
E) 228 nm
46
C) 4.2 m
56
B) n1  n3  n2
47
A) The image is inverted,
and the mirror is concave
57
D) 61°
48
B) The image is inverted,
real and the lens is
converging
49
B) Its speed and
wavelength change
58
E) -8.3
50
C) 6.2  10 7 m
60
C) -29
59
D) 5.0
Page 9 of 9