Physics 112
Exam 3
Summer 2015
41. A beam of light in air strikes a slab of glass (n  1.50) at a 30.0° angle to the vertical. What is
the angle of refraction?
A)
B)
C)
D)
E)
30.0°
25.7°
19.5°
15.0°
10.3°
Solution:
n1 sin 1  n2 sin  2 
sin  2 
1.0
1
sin 30   ;
1.5
3
sin  2 
n1
sin 1
n2
 2  19.5°
42. A diver shines a flashlight upward from beneath the water at a 60.0° angle to the vertical.
Index of refraction water is 1.33. At what angle does the light leave the water?
A)
B)
C)
D)
E)
60°
50°
40°
30°
It does not leave the water
Solution:
The critical angle for the total internal refractions is given by the equation:
 c  48.8   60.0  .
sin  c  n2 n1 
sin  c  1 1.33 
43. The speed of light in a certain substance is 89% of its value in water. What is the index of
refraction of this substance?
A)
B)
C)
D)
E)
1.00
1.21
1.49
1.60
3.00
Solution:
c
n
We find the index of refraction from the equation: v  ;
v  0.89 vwater  0.89
c
c
 
1.33 n
n  1.33 / 0.89 ,
which gives n  1.49.
Page 1 of 8
Physics 112
Exam 3
Summer 2015
44. Which statement concerning mirrors is wrong?
A)
B)
C)
D)
E)
Real image is always inverted
Inverted image is always real
Virtual image is always erect
Erected image is always virtual
Real and virtual images could be erect or inverted
Solution: In mirrors and lenses real image is always inverted; it cannot be erect.
45. Which statement concerning lenses is wrong?
A)
B)
C)
D)
E)
Real image is always inverted
Inverted image is always real
Virtual image is always erect
Erected image is always virtual
Real and virtual images could be erect or inverted
Solution: In mirrors and lenses real image is always inverted; it cannot be erect.
46. A 4.0-cm-tall object is placed 10 cm in front of a spherical mirror. It is desired to produce a
virtual image 6.0 cm tall. What is the focal length of the mirror?
A)
B)
C)
D)
E)
30 cm
25 cm
20 cm
15 cm
10 cm
Solution:
Virtual image is allays erect. We find the image distance from the magnification:
m
hi  di


ho do
di  d o
hi

ho
di   10cm 
6.0cm
 15cm .
4.0cm
We find the focal length from the equation:
 1  1 1
    ;
 d o   di  f
 1  
 1
1


  , which gives f  30cm.
 10cm     15cm   f
Note!
 To produce a larger virtual image the mirror should be concave. As expected f>0.
 Virtual image is located behind the surface of the mirror. As expected, di  0. The image is
located 15cm behind the surface.
Page 2 of 8
Physics 112
Exam 3
Summer 2015
47. Find magnification, if a real image is located at the same distance from a mirror as the
object?
A)
B)
C)
D)
E)
1
-1
2
-2
It is impossible to have this kind of image
Solution:
If d o  di , then from equation m  
di
follows that magnification m=-1.
do
48. Parallel light rays are directed from air into a double convex glass lens. These rays will
converge at:
A)
B)
C)
D)
E)
Focal point
Two focal distances from the lens
One half of the focal from the lens
At the center of the lens
On the surface of the lens
Solution:
By definition, focal point is the point where parallel rays converge.
49. Monochromatic light falls on two very narrow slits. Successive fringes on a screen 5.00 m
away are 6.5 cm apart near the center of the pattern. What is the distance between the slits if
the wavelength of the light is 650 nm?
A)
B)
C)
D)
E)
500 µm
100 µm
50 µm
25 µm
5.0 µm
Solution:
 d sin  m  m
We use equations: 
tan   y x
For small angles: dy m x  m 




d  x y1 
d  650  10 9 m 5.0m 6.5  10 2 m  5.0  10 5 m  50m
Page 3 of 8
Physics 112
Exam 3
Summer 2015
50. 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 on one side to dark
fringe on another side), 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 equation:
D sin  m  m 
  D sin 1
  2.60  10 6 msin 15.0 , which gives   6.73  10 7 m  673nm
51. The separation between adjacent maxima in a double-slit interference pattern using
monochromatic light is
A)
B)
C)
D)
E)
Greatest for red light
Greatest for green light
Greatest for blue light
The same for all colors of light
It is smallest for green light
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.
52. A 3500 lines/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 equation: d sin  m;
1

 2

 
10 m / cm sin 28.0  / 3 , which gives
 3500lines / cm 
   d sin   / m
  4.47  107 m  447nm.
Page 4 of 8
Physics 112
Exam 3
Summer 2015
53. An important reason for using a very large diameter objective in an astronomical telescope is
A)
B)
C)
D)
E)
To increase the magnification
To increase the resolution
To form a virtual image
To form a real image
To increase the width of the field of view
Solution:
Recall Rayleigh criterion:
 R  1.22   D . Increasing diameter we increase resolution.
54. A lens appears greenish yellow (   570 nm is strongest) when white light reflects from it.
What minimum thickness of coating n=1.25 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.
55. 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
tan  p 
ndiamond 2.42

 1.82, which gives  p  61.2.
nwater
1.33
Page 5 of 8
Physics 112
Exam 3
Summer 2015
56. What kinds of image observe people wearing glasses?
A)
B)
C)
D)
E)
Always real
Always virtual
Real if glasses’ power is positive, and virtual if glasses’ power is negative
Real if glasses’ power is negative, and virtual if glasses’ power is positive
It could be real and virtual for any power of glasses
Solution:
Image produced by glasses is on the same side from glasses as object. Also, erect image is
always virtual.
57. 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 
58. A nearsighted person wears contact lenses whose power is -5.0D. What is the person's far
point?
A)
B)
C)
D)
E)
5 cm
10 cm
20 cm
50 cm
1.0 m
Solution:
If d 0   , then d i  f  1 / P  1 /  5.0m 1   0.20m  20cm . So, far point is 20cm
Page 6 of 8
Physics 112
Exam 3
Summer 2015
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
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:
The magnification of the telescope is given by
f
M   o   f o P e  0.85m 35m 1  29.75
fe


Page 7 of 8
Physics 112
Exam 3
Summer 2015
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
C) 19.5°
51
A) Greatest for red light
42
E) It does not leave the
water
52
B) 447 nm
43
C) 1.49
53
B) To increase the
resolution
44
54
E) Real and virtual images E) 228 nm
could be erect or inverted
45
55
E) Real and virtual images D) 61°
could be erect or inverted
46
A) 30 cm
56
B) Always virtual
47
B) -1
57
E) -8.3
48
A) Focal point
58
C) 20 cm
49
C) 50 µm
59
D) 5.0
50
E) 673 nm
60
C) -30
Page 8 of 8