Physics 112
Exam 3
Summer 2014
41. How far are you from your image when you stand 0.75 m in front of a vertical plane mirror?
A)
B)
C)
D)
E)
0.75 m
1.5 m
3.0 m
3.75m
4.5m
Solution:
Distance to the image in the plane mirror is equal to the distance to the object ( d i  d 0 ). So, the
distance between the image and object is twice the distance to the object:
d i  d 0  d i  d 0  2d 0  2  0.75m  1.5m
42. The image formed by a plane mirror of a real object is always
A)
B)
C)
D)
E)
Real, erect, and smaller than the object
Real, inverted, and the same size as the object
Real, erect, and the same size as the object
Virtual, inverted, and larger than the object
Virtual, erect and the same size as the object
Solution:
The image formed by a plane mirror of a real object is always: virtual (behind the mirror), erect
(upright) and the same size as the object.
43. The critical angle for a substance is measured at 53.7°. Light enters from air at the angle
45.0° to the normal. At what angle will it continue?
A)
B)
C)
D)
E)
34.7°
45.0°
53.7°
58.2°
It will not continue, but be totally reflected.
Solution:
For air index of refraction is approximately 1.
n
1
1) sin  C  2  sin  C 
n1
n
1
sin 1  sin  c sin 1 
n
 2  arcsin(sin  c sin 1 )  arcsin (sin 53.7  ) sin( 45  )    2  34.7 
2) n1 sin 1  n2 sin  2  sin 1  n sin  2  sin  2 
Page 1 of 8
Physics 112
Exam 3
Summer 2014
44. A concave mirror with a radius of 20 cm creates a real image 30 cm from the mirror. What is
the object distance?
A) 30cm
B) 20 cm
C) 15 cm
D) 7.5 cm
E) 5.0 cm
Solution:
f  R 2  f  20cm / 2  10cm
1
1
1
1
1 1
1
1
2

 
 



 d 0  15cm
d0 di
f
d0
f d i 10cm 30cm 30cm
45. When a person stands 40 cm in front of a cosmetic mirror (concave mirror), the erect image
is twice the size of the object. What is the focal length of the mirror?
A)
B)
C)
D)
E)
27 cm
40 cm
80 cm
100 cm
160 cm
Solution:
f  R 2  f  40cm / 2  20cm
h
d
m  i   i  2  d i  2d o
ho
do
1
1
1
1
1
1
1
1

 

 
 
d0 di
f
d 0 2d o
f
2d o
f
f  2d o  2 * 40cm  80cm
46. An object is placed 40 cm in front of a 20 cm focal length converging lens. How far is the
image of this object from the lens?
A)
B)
C)
D)
E)
40 cm
30 cm
20 cm
13 cm
8 cm
Solution:
1
1
1
1
1 1
1
1
1

 
 




d0 di
f
di
f d o 20cm 40cm 40cm
d 0  40cm
Page 2 of 8
Physics 112
Exam 3
Summer 2014
47. Radio waves are diffracted by large objects such as buildings, whereas light is not noticeably
diffracted. Why is this?
A)
B)
C)
D)
E)
Radio waves are unpolarized, whereas light is plane polarized
Radio waves are polarized, whereas light is plane unpolarized
Radio waves are coherent and light is usually not coherent
The wavelength of light is much greater than the wavelength of radio waves
The wavelength of light is much smaller than the wavelength of radio waves
Solution:
Radio waves are diffracted by large objects such as buildings because the wavelength of radio
waves is substantially larger than buildings.
Light is not noticeably diffracted by large objects such as buildings because the wavelength of
light is much smaller than these objects.
48. A beam of light (f = 5.0 × 1014 Hz) enters a piece of glass (n = 1.5). What is the frequency
of the light while it is in the glass?
A)
B)
C)
D)
E)
5.0 × 1014 Hz
7.5 × 1014 Hz
3.3 × 1014 Hz
1.0 × 1014 Hz
2.5 × 1014 Hz
Solution:
When light is entering glass its frequency is not changing.
49. 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)
B)
C)
D)
E)
1.5 cm
3.0 cm
4.5 cm
6.0 cm
7.5 cm
Solution:
For Young's double slit experiment: d sin  m  m . For small angles: y x  tan   sin 
y m  x tan  m  x sin  m  x



600  10 9 m
m
 y 2  y1  x  2.5m 
 3.0  10  2 m  3.0cm
3
d
d
0.050  10 m


Page 3 of 8
Physics 112
Exam 3
Summer 2014
50. In a single slit diffraction experiment, if the width of the slit increases, what happens to the
width of the central maximum on a screen?
A)
B)
C)
D)
E)
It increases.
It decreases.
It remains the same.
It depends from the distance to the screen.
There is not enough information to determine.
Solution:
D sin  m  m .
If the width, D of the slit increases, than 1 decreases, so the width of the central maximum on a
screen decreases.
51. In order to obtain a good single slit diffraction pattern, the slit width could be:
A)
B)
C)
D)
E)
/100
/10

10
100
Solution:
D sin  m  m
If D  10 then sin  m  m / D  m / 10 , and one can observe several fringes on the screen.
52. Two stars 15 light-years away are barely resolved by a 55-cm (mirror diameter) telescope.
How far apart are the stars? Assume   550 nm and that the resolution is limited by
diffraction. (1 light-year = 9.46x1015 m)
A)
B)
C)
D)
E)
1.3x107 m
1.4x108 m
1.5x109 m
1.6x1010 m
1.7x1011 m
Solution:
 R  1.22   D  y / x  y  1.22  x  D
y  1.22  15 9.46  1015 m 550  10 -9 m 0.55m  1.7  1011 m



Page 4 of 8
Physics 112
Exam 3
Summer 2014
53. A diffraction grating has 5000 lines per cm. The angle between the central maximum and the
fourth order maximum is 60°. What is the wavelength of the light?
A)
B)
C)
D)
E)
138 nm
183 nm
367 nm
433 nm
637 nm
Solution:
d  1 5000cm  2.000  10 4 cm  2.000  10 6 m
d sin  m  m    d sin  m m   2.000  10 6 m sin 60  4  0.433  10 6 m  433nm



54. What is the minimum thickness of a nonreflecting film coating (n = 1.30) on a glass lens (n =
1.50) for wavelength 500 nm?
A)
B)
C)
D)
E)
250 nm
192 nm
167 nm
96.2 nm
57.3 nm
Solution:
nair  1  n film  1.3  nlens  1.50 . This means that there is phase shift on both sides of the film,
and for destructive interference (nonreflecting film) we have the following equation:

500nm
2t  m  12  film  m  12  / n 
t min 
t min 
 96.2nm
;
4n
4  1.3
55. An ideal polarizer is placed in a beam of unpolarized light and the intensity of the transmitted
light is I. A second ideal polarizer is placed in the beam with its preferred direction rotated
40° to that of the first polarizer. What is the intensity of the beam after it has passed through
both polarizers?
A)
B)
C)
D)
E)
0.77I
0.64I
0.59I
0.41I
0.36I
Solution:
I 2  I cos 2   I cos 2 40   0.59 I
Page 5 of 8
Physics 112
Exam 3
Summer 2014
56. A beam of unpolarized light in air strikes a flat piece of glass at an angle of 57.3°. If the
reflected beam is completely polarized, what is the index of refraction of the glass?
A)
B)
C)
D)
E)
1.50
1.52
1.54
1.56
1.58
Solution:
tan  p  n2 n1  n glas nair  n glas  n air tan  p ;


nglas  tan 57.3  1.56
57. A near-sighted person has a far point of 20 cm. What lens (in diopters) will allow this person
to see distant objects clearly? Assume that the lens is 2.0 cm from the eye (typical for
eyeglasses)
A)
B)
C)
D)
E)
+ 5.6 D
- 5.6 D
+ 0.056 D
- 0.056 D
-5.0 D
Solution:
d i  20cm  2cm  18cm  0.18m ; d o  
1
1
1
1

 5.6 D

P P
d i  0.18m
d0 di
58. A magnifying glass with a focal length of 8.5 cm is used to read print placed at a distance 7.5
cm. Calculate the angular magnification.
A)
B)
C)
D)
E)
2.3x
3.3x
4.3x
5.3x
6.3x
Solution:
25cm 25cm
M 

 3.3
d0
7.5cm
Page 6 of 8
Physics 112
Exam 3
Summer 2014
59. A person is designing a 10X telescope. If the telescope is limited to a length of 20 cm, what
is the approximate focal length of the objective?
A)
B)
C)
D)
E)
16 cm
17 cm
18 cm
19 cm
20 cm
Solution:
f
f
M  o  o 
fe
l  fo
M l  f o    f o  f o 
Ml
;
M 1
fo 
 1020cm 
 18cm
 10  1
60. What power lens is needed to correct for farsightedness where the uncorrected near point is
75 cm?
A)
B)
C)
D)
E)
+ 2.7 D
- 2.7 D
+ 5.3 D
- 5.3 D
+6.0 D
Solution:
d i  75cm  0.75m ; d o  25cm
1
1
1
1
2


 2.7 D

P P
0.25m  0.75m 0.75m
d0 di
Page 7 of 8
Physics 112
Exam 3
Summer 2014
Answer Sheet
41
B) 1.5 m
51
D) 10
42
E) Virtual, erect and the
same size as the object
52
E) 1.7x1011 m
43
A) 34.7°
53
D) 433 nm
44
C) 15 cm
54
D) 96.2 nm
45
C) 80 cm
55
C) 0.59I
46
A) 40 cm
56
D) 1.56
47
E) The wavelength of
light is much smaller than
the wavelength of radio
waves.
48
A) 5.0 × 1014 Hz
57
B) - 5.6 D
49
B) 3.0 cm
59
C) 18 cm
50
B) It decreases.
60
A) + 2.7 D
58
B) 3.3x
Page 8 of 8