Physics 2C Summer Session II
Quiz #4
Multiple Choice. Choose the answer that best completes the
statement or answers the question.
1. A plane wave of linearly polarized light is incident on a stack of 2 linear
polarizers. The polarization angle of the incident light is rotated by an angle
3 =8 with respect to the …rst polarizer whose polarization axis is rotated by
an angle
=8 with respect to the second polarizer. The intensity of the light
relative to the incident intensity, I=I0 ; is
A) :073 B) 1=16 C) 1=8 D) 1=4 E) :354
From the Law of Malus we have
I
= cos2 (3 =8) cos2 (
I0
=8) = :125 ! C
2. Red light emitted by hydrogen atoms at rest in the laboratory has a
wavelength of 656:3nm. The same spectral line emitted by hydrogen atoms
from a distant galaxy has a frequency of 4:16 1014 Hz. The relative velocity
of the galaxy relative to Earth is (speed of light is c = 3 108 m= sec)
A) 1:41 107 m= sec B) 2:82 107 m= sec C) 2:96 107 m= sec
D) 2:70 107 m= sec E) 1:48 107 m= sec
The observed wavelength of the light as it reaches Earth is
obs
=3
108 = 4:16
1014 = 721:15nm
From the expression for the Doppler shift of light we …nd
s
2
721:15
1 + v=c
1 + v=c
=
!
=
obs
em
1 v=c
656:3
1 v=c
1 + v=c =
v
=
1:2074 (1
2:82
v=c) ! v=c =
:2074
= :09396
2:2074
107 m=s ! B
3. In the …gure for problem 3 a light ray is propagating in the plane of the
page parallel to one of the mirrors (problem 4 ray in the …gure). Through what
angle will it be turned (a normal re‡ection is equivalent to a turn of 180 )?
1
Problem 3
A) 60 B) 120 C) 180 D) 240 E) 300
From geometry the angle between the incident light ray and the lower mirror
is 60 : Hence the incident and re‡ected angles are both 30 ; and the angle
between the incident and re‡ected rays is 60 : Thus the light ray is rotated
through 180
60 = 120 after the …rst re‡ection..After the re‡ection from
the lower mirror the light ray and the two mirrors form an equilateral triangle.
Hence the above scenario is repeated for the upper mirror and the light is rotated
through 2 120 = 240 ! D:
4. Light traveling in water (n = 4=3) is incident upon an water-glass interface. The refracted light makes a 51 angle with respect to the interface. If the
glass has a refractive index of n = 1:62; the incident angle of the light is
A) 54:8 B) 39:7 C) 35:6 D) 49:9 E) 70:8
The refracted and incident angles are measured from the normal to the
interface, hence the refracted angle is 39 : From Snell’s law we …nd
n1 sin
1
= n2 sin
1
=
2
! sin
1
=
3
1:62
sin 39 = :76462
4
49:9 ! D
5. What is the wavelength of the light inside the glass in problem 4, if it
has a frequency of 4:82 1014 Hz in water?
A) 467nm B) 512nm C) 622nm D) 756nm E) 384nm
Since the frequency is the same in all media, the wavelength in vacuum is
108 = 4:82
= c=f = 3
1014 = 622nm
The wavelength in the glass is found from
=n
n
!
n
= 622=1:62 = 383:95nm ! E
2
6. The prism shown in the …gure is an isosceles right triangle and has an
index of refraction of n = 1:76: If the prism is immersed in a liquid, what is the
minimum value for the index of refraction of the liquid for the light to cease
undergoing total internal re‡ection?
Problem 6
A) 1:245 B) 1:250 C) 2:150 D) 1:150 E) 1:075
The incident angle is 45 which in this case is also the critical angle. Hence
the index of the liquid is found from
p
n1 sin c = n2 ! 1:76= 2 = nliq = 1:245 ! A
If the index of the liquid is larger than this, then there is some refraction (no
internal re‡ection).
7. An object at the bottom of a wading pool appears to be 52cm below the
surface. How deep is the pool if the refractive index of water is n = 4=3?
A) 39cm B) 45cm C) 52cm D) 60cm E) 69cm
The apparent depth is given by `0 = `=n: Hence the actual depth is
d=`=
4
52 = 69:33 ! E
3
8. An image is located 36cm behind a spherical concave mirror which has a
radius of curvature of 18cm. How much is the image magni…ed (include sign to
determine if the image is inverted or not)?
A) M = 3:0 B) M = 2:0 C) M = 5:0 D) M = 3:0 E) M = 5:0
The focal length is given by f = R=2: Since the image is behind the mirror
its location is `0 = 36cm: From the mirror equation
1
f
1
`
=
=
1
1
+ 0 !
`
`
5
!`=
36
3
1
1
1
=
9
`
36
36
= 7:2
5
Hence the magni…cation is
M=
`0
=
`
36
= 5:0 ! C
36=5
9. Determine the focal length of a spherical mirror if the image magni…cation
is given by M = 2 when the object is 5cm beyond the focal point (measured
in cm) of the mirror.
A) M = 15cm B) 5cm C) 2cm D) 10=3cm E) 10cm
From the mirror equation and magni…cation equations
1
`0
=
M
=
1
f
1
5
=
f +5
f (f + 5)
`0
f
=
= 2 ! f = 10cm ! E
`
5
10. Find the Brewster angle, ; for the interface between air (n = 1) and a
material for which the critical angle in air is given by c = 36:87 :
A) 37 B) 59 C) 31 D) 53 E) 47
From the critical angle, the index of refraction for this material is found from
n sin
c
= 1 ! n = 5=3:
The Brewster angle is found from
tan
= n=1 = 5=3 !
= 59 ! B
u=v) ; f 0 = f (1
1
Some useful formulas
f0
n1 sin
1
= f (1
= n2 sin
2;
u=v)
1
1
1
= + 0; M =
f
`
`
4
; f0 = f
`0 =`; tan
s
1
1
u=c
u=c
= n2 =n1