Name: ________________________ Class: ___________________ Date: __________
ID: A
Quiz 5 PRACTICE--Ch12.1, 13.1, 14.1
Multiple Choice
Identify the choice that best completes the statement or answers the question.
1.
A beam of light in air is incident at an angle of 35° to the surface of a rectangular block of clear plastic (n = 1.49). What is the angle of
refraction?
a. 12° b. 42° c. 23° d. 57°
2.
Sound waves
a. are longitudinal waves. b. are transverse waves. c. do not require a medium for transmission. d. are a part of the
electromagnetic spectrum.
3.
When a light ray moves from air into glass, which has a higher index of refraction, its path is
a. bent toward the normal. b. bent away from the normal. c. parallel to the normal. d. not bent.
4.
Which portion of the electromagnetic spectrum is used in a microscope?
a. gamma rays b. infrared waves c. ultraviolet light d. visible light
5.
A wave travels through a medium. As the wave passes, the particles of the medium vibrate in a direction perpendicular to the direction of
the wave’s motion. The wave is
a. a pulse. b. electromagnetic. c. transverse. d. longitudinal.
6.
At a large distance from a sound source, spherical wave fronts are viewed as
a. plane waves. b. wavelengths. c. rays. d. troughs.
7.
In the diagram above, use the superposition principle to find the resultant wave of waves Q and R.
a. b b. a c. d d. c
8.
Carbon tetrachloride (n = 1.46) is poured into a container made of crown glass (n = 1.52). If a light ray in the glass is incident on the
glass-to-liquid boundary and makes an angle of 30.0° with the normal, what is the angle of the corresponding refracted ray with respect to
the normal?
a. 28.7° b. 25.6° c. 64.4° d. 31.4°
9.
If you are reading a book and you move twice as far away from the light source, how does the brightness at the new distance compare with
that at the old distance? It is
a. one-half as bright. b. one-fourth as bright. c. twice as bright. d. one-eighth as bright.
10.
In general, sound travels faster through
a. gases than through liquids. b. solids than through gases. c. gases than through solids. d. empty space than through matter.
1
Name: ________________________
ID: A
11.
The distance between wave fronts of plane waves corresponds to ____ of a sound wave.
a. two rarefactions b. one compression c. one wavelength d. two amplitudes
12.
The trough of the sine curve used to represent a sound wave corresponds to
a. the wavelength. b. a compression. c. a rarefaction. d. the amplitude.
13.
Refraction is the bending of a wave disturbance as it passes at an angle from one ____ into another.
a. medium b. area c. glass d. boundary
14.
Consider two identical wave pulses on a rope having a fixed end. Suppose the first pulse reaches the end of the rope, is reflected back, and
then meets the second pulse. When the two pulses overlap exactly, what will be the amplitude of the resultant pulse?
a. same as the original pulses b. half the amplitude of the original pulses c. zero d. double the amplitude of the original pulses
15.
Pitch depends on the ____ of a sound wave.
a. speed b. power c. frequency d. amplitude
16.
Atmospheric refraction of light rays is responsible for which of the following effects?
a. spherical aberration b. total internal reflection in a gemstone c. mirages d. chromatic aberration
17.
What is the wavelength of an infrared wave with a frequency of 4.2 × 10 14 Hz?
a. 7.1 × 10 5 m b. 1.4 × 10 6 m c. 1.4 × 10 −6 m d. 7.1 × 10 −7 m
18.
Part of a pencil that is placed in a glass of water appears bent in relation to the part of the pencil that extends out of the water. What is this
phenomenon called?
a. diffraction b. reflection c. refraction d. interference
19.
A train moves down the track toward an observer. The sound from the train, as heard by the observer, is ____ the sound heard by a
passenger on the train.
a. lower in pitch than b. the same as c. higher in pitch than d. a different timbre than
20.
When a mechanical wave’s amplitude is reduced by half, the energy the wave carries in a given time interval is
a. decreased to one-fourth. b. doubled. c. decreased to one-half. d. increased by a factor of 1.4.
21.
Two mechanical waves that have positive displacements from the equilibrium position meet and coincide. What kind of interference
occurs?
a. complete destructive b. destructive c. none d. constructive
22.
What is the frequency of an electromagnetic wave with a wavelength of 1.0 ´ 10 5 m?
a. 3.0 × 10 13 Hz b. 3.3 × 10 −4 Hz c. 3.0 × 10 3 Hz d. 1.0 × 10 13 Hz
23.
What is the frequency of infrared light of 1.0 × 10 −4 m wavelength?
a. 3.0 × 10 −13 Hz b. 3.0 × 10 12 Hz c. 3.0 × 10 4 Hz d. 3.0 × 10 2 Hz
24.
The Doppler effect occurs with
a. only transverse waves. b. only water waves. c. all waves. d. only sound waves.
25.
Which is an example of refraction?
a. Light is bent slightly around corners. b. In a mirror, when you lift your right arm, the left arm of your image is raised. c. A
parabolic mirror in a headlight focuses light into a beam. d. A fish appears closer to the surface of the water than it really is when
observed from a riverbank.
26.
A ray of light in air is incident on an air-to-glass boundary at an angle of exactly 30.0° with the normal. If the index of refraction of the
glass is 1.65, what is the angle of the refracted ray within the glass with respect to the normal?
a. 34.4° b. 18.0° c. 37.3° d. 58.3°
27.
If you know the wavelength of any form of electromagnetic radiation, you can determine its frequency because
a. the speed of light increases as wavelength increases. b. the speed of light varies for each form. c. all wavelengths travel at the
same speed. d. wavelength and frequency are equal.
2
Name: ________________________
ID: A
28.
In the diagram above, use the superposition principle to find the resultant wave of waves W and Z.
a. c b. b c. d d. a
29.
Two mechanical waves can occupy the same space at the same time because waves
a. are matter. b. cannot pass through one another. c. are displacements of matter. d. do not cause interference patterns.
30.
The relationship between frequency, wavelength, and speed holds for light waves because
a. all forms of electromagnetic radiation travel at a single speed in a vacuum. b. light travels slower in a vacuum than in air.
c. different forms of electromagnetic radiation travel at different speeds. d. light travels in straight lines.
31.
Which of the following wavelengths would produce standing waves on a string approximately 3.5 m long?
a. 4.55 m b. 2.33 m c. 2.85 m d. 3.75 m
32.
When two mechanical waves coincide, the amplitude of the resultant wave is always ____ the amplitudes of each wave alone.
a. the sum of b. the same as c. less than d. greater than
33.
Which is not correct when describing the formation of rainbows?
a. Sunlight is internally reflected on the back side of a raindrop. b. All wavelengths refract at the same angle. c. Sunlight is spread
into a spectrum when it enters a spherical raindrop. d. A rainbow is really spherical in nature.
34.
The ____ of light can change when light is refracted because the medium changes.
a. transparency b. medium c. wavelength d. frequency
35.
The highness or lowness of a sound is perceived as
a. ultrasound. b. wavelength. c. compression. d. pitch.
36.
When a light ray passes from water (n = 1.333) into diamond (n = 2.419) at an angle of 45°, its path is
a. parallel to the normal. b. bent toward the normal. c. bent away from the normal. d. not bent.
37.
What is the wavelength of microwaves of 3.0 × 10 9 Hz frequency?
a. 0.20 m b. 0.050 m c. 0.10 m d. 0.060 m
38.
Which of the following is not an example of approximate simple harmonic motion?
a. a car’s radio antenna waving back and forth b. a piano wire that has been struck c. a child swinging on a swing d. a ball
bouncing on the floor
39.
When red light is compared with violet light,
a. both have the same wavelength. b. both travel at the same speed. c. both have the same frequency. d. red light travels faster
than violet light.
3
Name: ________________________
40.
ID: A
The ____ of light can change when light is refracted because the velocity changes.
a. medium b. transparency c. frequency d. wavelength
Problem
41.
The objective lens of a compound microscope has a focal length of 1.00 cm. A specimen is 1.25 cm from the objective lens. The image
formed by the objective lens is 0.180 cm inside the focal point of the eyepiece whose focal length is 1.50 cm. What is the distance from the
eyepiece to the image formed by the eyepiece lens?
42.
Yellow-green light has a wavelength of 560 nm. What is its frequency?
43.
A student wishes to construct a mass-spring system that will oscillate with the same frequency as a swinging pendulum with a period of
3.45 s. The student has a spring with a spring constant of 72.0 N/m. What mass should the student use to construct the mass-spring
system?
44.
A certain radio wave has a frequency of 2.0 × 10 6 Hz. What is its wavelength?
45.
The frequency of an X ray is 6.50 × 10 18 Hz. What is the X ray’s wavelength?
46.
A fiber-optic cable (n = 1.53) is submerged in water (n = 1.33). Predict whether light will be refracted or whether it will undergo total
internal reflection if the angle of incidence is between 65° and 70°.
47.
An object is placed along the principal axis of a thin converging lens that has a focal length of 28 cm. If the distance from the image in
front of the lens is 24 cm, what is the distance from the object to the lens?
48.
A 0.20 kg mass suspended from a spring moves with simple harmonic motion. At the instant the mass is displaced from equilibrium by
–0.050 m, what is its acceleration? (The spring constant is 10.0 N/m.)
49.
A mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. If the mass of the object is 0.20 kg and the spring
constant is 130 N/m, what is the frequency?
50.
A portion of infrared light in the electromagnetic spectrum has a wavelength of 725 µm. What is the frequency of this portion of infrared
light?
4
ID: A
Quiz 5 PRACTICE--Ch12.1, 13.1, 14.1
Answer Section
MULTIPLE CHOICE
1.
ANS: C
Given
θ i = 35°
n i = 1.00
n r = 1.49
Solution
Rearrange Snell’s law, n i sinθ i = n r sinθ r , and solve for θ r .
ÍÈÍ
˙˘˙
ÈÍ
˘˙
Í 1.00
˙
Í ni Ê
˙
ÁË sinθ i ˜ˆ¯ ˙˙ = sin −1 ÍÍÍ
θ r = sin −1 ÍÍ
( sin 35° ) ˙˙˙ = 23°
ÍÍ n r
˙˙
1.49
Í
˙˚
Î
Î
˚
2.
3.
4.
5.
6.
7.
8.
PTS: 1
ANS: A
ANS: A
ANS: D
ANS: C
ANS: A
ANS: A
ANS: D
Given
θ i = 30.0°
DIF:
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
IIIA
1
1
1
1
1
1
OBJ:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
14-1.3
I
II
I
I
I
II
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
12-1.1
14-1.2
13-1.1
11-3.1
12-1.4
11-4.1
14-1.3
II
I
II
I
I
IIIA
I
I
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
13-1.4
12-1.3
12-1.4
12-1.1
14-1.1
11-4.3
12-1.2
14-3.2
n i = 1.52
n r = 1.46
Solution
Rearrange Snell’s law, n i sinθ i = n r sinθ r , and solve for θ r .
ÍÈÍ
˙˘˙
ÍÈÍ
˙˘˙
n
1.52
˙
−1 ÍÍ i ÁÊ
ˆ
˜
θ r = sin Í
sinθ i ¯ ˙˙ = sin −1 ÍÍÍ
( sin 30.0° ) ˙˙˙ = 31.4°
Ë
ÍÍ n r
˙˙
ÍÎ 1.46
˙˚
Î
˚
9.
10.
11.
12.
13.
14.
15.
16.
PTS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
1
B
B
C
C
A
C
C
C
DIF:
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
IIIB
1
1
1
1
1
1
1
1
OBJ:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
1
ID: A
17.
ANS:
Given
D
f = 4.2 × 10 14 Hz = 4.2 × 10 14 s −1
c = 3.00 × 10 8 m/s
Solution
Rearrange the wave speed equation, c = f λ, to isolate λ, and calculate.
λ=
18.
19.
20.
21.
22.
ÊÁÁ
ˆ˜
ÁÁ 3.00 × 10 8 m/s ˜˜˜
Ë
¯
c
= Ê
= 7.1 × 10 −7 m
f
ÁÁ
14 −1 ˆ˜˜
Á 4.2 × 10 s ˜
Ë
¯
PTS:
ANS:
ANS:
ANS:
ANS:
ANS:
Given
1
C
C
A
D
C
DIF:
PTS:
PTS:
PTS:
PTS:
IIIA
1
1
1
1
OBJ:
DIF:
DIF:
DIF:
DIF:
13-1.2
I
I
II
I
OBJ:
OBJ:
OBJ:
OBJ:
14-1.1
12-1.5
11-3.5
11-4.2
OBJ:
OBJ:
12-1.5
14-1.1
λ = 1.0 × 10 5 m
c = 3.00 × 10 8 m/s
Solution
Rearrange the wave speed equation, c = f λ, to isolate f, and calculate.
ÁÊÁ
˜ˆ
ÁÁ 3.00 × 10 8 m/s ˜˜˜
Ë
¯
c
f=
= Ê
= 3.0 × 10 3 s −1 = 3.0 × 103 Hz
λ
ÁÁ 1.0 × 10 5 m ˆ˜˜
Á
˜
Ë
¯
23.
PTS:
ANS:
Given
1
B
DIF:
IIIA
OBJ:
13-1.2
λ = 1.0 × 10 −4 m
c = 3.00 × 10 8 m/s
Solution
Rearrange the wave speed equation, c = f λ, to isolate f, and calculate.
f=
24.
25.
c
λ
PTS:
ANS:
ANS:
ÊÁÁ
ˆ˜
ÁÁ 3.00 × 10 8 m/s ˜˜˜
Ë
¯
12 −1
12
= Ê
ˆ˜ = 3.0 × 10 s = 3.0 × 10 Hz
ÁÁ
−4
Á 1.0 × 10 m ˜˜
Ë
¯
1
C
D
DIF:
PTS:
PTS:
IIIA
1
1
OBJ:
DIF:
DIF:
13-1.2
I
II
2
ID: A
26.
ANS: B
Given
θ i = 30.0°
n i = 1.00
n r = 1.65
Solution
Rearrange Snell’s law, n i sinθ i = n r sinθ r , and solve for θ r .
ÈÍ
˘
ÍÈÍ n
˙˘˙
ˆ ˙˙
Í 1.00 ÊÁ
Í i Ê
˙
ÁÁ sin 3.0 × 10 1 ° ˜˜˜ ˙˙ = 18.0°
θ r = sin −1 ÍÍ
ÁË sinθ i ˆ˜¯ ˙˙ = sin −1 ÍÍÍ
ÍÍ n r
˙˙
¯ ˙˙˚
ÍÎ 1.65 Ë
Î
˚
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
PTS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
Given
1
C
B
C
A
B
A
B
C
D
B
C
DIF:
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
IIIA
1
1
1
1
1
1
1
1
1
1
OBJ:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
DIF:
14-1.3
I
II
I
II
IIIC
I
I
I
I
II
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
OBJ:
13-1.3
11-4.1
11-4.1
13-1.3
11-4.4
11-4.2
14-3.3
14-1.1
12-1.2
14-1.2
OBJ:
OBJ:
OBJ:
11-1.1
13-1.3
14-1.1
f = 3.0 × 10 9 Hz = 3.0 × 109 s −1
c = 3.00 × 10 8 m/s
Solution
Rearrange the wave speed equation, c = f λ, to isolate λ, and calculate.
ÁÊÁ
˜ˆ
ÁÁ 3.00 × 10 8 m/s ˜˜˜
Ë
¯
c
λ=
= Ê
= 0.10 m
f
ÁÁ 3.0 × 10 9 s −1 ˆ˜˜
Á
˜
Ë
¯
38.
39.
40.
PTS:
ANS:
ANS:
ANS:
1
D
B
D
DIF:
PTS:
PTS:
PTS:
IIIA
1
1
1
OBJ:
DIF:
DIF:
DIF:
13-1.2
I
II
II
3
ID: A
PROBLEM
41.
ANS:
−11 cm
Given
F o = 1.00 cm
p o = 1.25 cm
F e = 1.50 cm
p e = 1.50 cm − 0.180 cm = 1.32 cm
Solution
The focal length and object distance of the objective lens do not enter into the calculation.
The image of the objective lens is the object of the eyepiece lens.
Rearrange the thin-lens equation,
1
1
1
+
=
, and solve for q.
p q
f
1
1
1
1
1
0.667 0.758
0.091
=
−
=
−
=
−
=−
qe
fe
pe
1.50 cm 1.32 cm
1 cm
1 cm
1 cm
q e = −11 cm (since q is negative, the image is virtual and in front of the lens)
42.
PTS:
ANS:
1
DIF:
IIIC
OBJ:
14-2.2
5.4 × 10 14 Hz
Given
λ = 560 nm = 560 × 10 − 9 m = 5.6 × 10 − 7 m
c = 3.00 × 10 8 m/s
Solution
Rearrange the wave speed equation, c = f λ, to isolate f , and calculate.
f=
c
λ
PTS:
ÊÁÁ
ˆ˜
ÁÁ 3.00 × 10 8 m/s ˜˜˜
Ë
¯
14 − 1
14
= Ê
ˆ˜ = 5.4 × 10 s = 5.4 × 10 Hz
ÁÁ
−7
Á 5.6 × 10 m ˜˜
Ë
¯
1
DIF:
IIIA
OBJ:
13-1.2
4
ID: A
43.
ANS:
21.7 kg
Given
T pendulum = 3.45 s
k = 72.0 N/m
Solution
If both systems have the same frequency, they will also have the same period.
Therefore, the given period may be substituted into the equation for a mass-spring system.
m
k
T = 2π
ÁÊÁ m ˜ˆ˜
T 2 = 4π 2 ÁÁ ˜˜
Á k ˜
Ë ¯
m=
44.
T2 k
4π 2
PTS:
ANS:
=
( 3.45 s ) 2 ( 72.0 N/m )
1
4π 2
= 21.7 kg
DIF:
IIIC
OBJ:
11-2.3
1.5 × 10 2 m or 150 m
Given
f = 2.0 × 10 6 Hz = 2.0 × 106 s −1
c = 3.00 × 10 8 m/s
Solution
Rearrange the wave speed equation, c = f λ, to isolate λ, and calculate.
ÊÁÁ
ˆ˜
ÁÁ 3.00 × 10 8 m/s ˜˜˜
Ë
¯
c
λ=
= Ê
= 1.5 × 10 2 m or 150 m
ˆ
f
ÁÁ
˜
6 −1 ˜
Á 2.0 × 10 s ˜
Ë
¯
45.
PTS:
ANS:
1
DIF:
IIIA
OBJ:
13-1.2
4.62 × 10 −11 m
Given
f = 6.50 × 10 18 Hz = 6.50 × 10 18 s −1
c = 3.00 × 10 8 m/s
Solution
Rearrange the wave speed equation, c = f λ, to isolate λ, and calculate.
λ=
ÊÁÁ
ˆ˜
ÁÁ 3.00 × 10 8 m/s ˜˜˜
Ë
¯
c
= Ê
= 4.62 × 10 −11 m
f
ÁÁ
18 −1 ˆ˜˜
Á 6.50 × 10 s ˜
Ë
¯
PTS:
1
DIF:
IIIA
OBJ:
13-1.2
5
ID: A
46.
ANS:
Since the angle of incidence is greater than the critical angle, 60.4°, the light ray will undergo total internal refraction.
Given
n optic cable = 1.53
n water = 1.33
θ i = 65° − 70°
Solution
Rearrange the critical angle equation, sin θ c =
nr
, to find θ c .
ni
ÁÊÁ n ˜ˆ˜
ÁÊÁ n
˜ˆ˜
ÊÁÁ 1.33 ˆ˜˜
water
Á r ˜˜
Á
˜˜
˜ = 60.4°
= sin −1 ÁÁ
= sin −1 ÁÁ
˜
˜
Á 1.53 ˜˜
ÁÁ n i ˜˜
ÁÁ n optic cable ˜˜
Ë
¯
Ë
¯
Ë
¯
Since the angle of incidence is greater than the critical angle, the light ray will undergo total internal refraction.
θ c = sin −1 ÁÁÁ
47.
PTS:
ANS:
13 cm
1
DIF:
IIIB
OBJ:
14-3.1
Given
q = −24 cm (q is negative, since the image is in front of the lens)
f = 28 cm (f is positive, since this is a converging lens)
Solution
Rearrange the thin-lens equation,
1
1
1
+
=
, and solve for p.
p
q
f
1
1
1
1
1
0.036 0.042
0.078
=
−
=
−
=
+
=
p
f
q
28 cm −24 cm
1 cm
1 cm
1 cm
p = 13 cm
48.
PTS:
ANS:
1
DIF:
IIIB
OBJ:
14-2.2
IIIA
OBJ:
11-1.3
2.5 m/s 2
Given
m = 0.20 kg
k = 10.0 N/m
x = –0.050 m
Solution
F = −kx and F = ma
ma = −kx
−(10.0 N/m)(–0.050 m)
−kx
a=
=
m
0.20 kg
a = 2.5 N/kg = 2.5 m/s 2
PTS:
1
DIF:
6
ID: A
49.
ANS:
4.0 Hz
Given
x = 8.0 cm (This value is not relevant to the problem.)
m = 0.20 kg
k = 130 N/m
Solution
m
1
and f =
, so
k
T
T = 2π
1
f=
2π
50.
PTS:
ANS:
=
m
k
1
2π
k
1
=
m
2π
1
DIF:
130 N/m
= 4.1 Hz
0.20 kg
IIIB
OBJ:
11-2.3
4.14 × 10 11 Hz
Given
λ = 725 µm = 725 × 10 −6 m = 7.25 × 10−4 m
c = 3.00 × 10 8 m/s
Solution
Rearrange the wave speed equation, c = f λ, to isolate f, and calculate.
f=
c
λ
PTS:
ÊÁ
ˆ
ÁÁ 3.00 × 10 8 m/s ˜˜˜
Ë
¯
= Ê
= 4.14 × 10 11 s −1 = 4.14 × 10 11 Hz
ÁÁ
˜ˆ
ÁÁ 7.25 × 10 −4 m ˜˜˜
Ë
¯
1
DIF:
IIIA
OBJ:
13-1.2
7