University of the Fraser Valley
Physics 100
Waves, Optics and Relativity Practice
Name: __________________________
Directions: Fill in the scantron form with the following information:
1. ID number (student number)
2. Name at top of form
3. Name bubbled into the columns labelled "C1C2C3C4"
4. Test Version Number (shown below); bubble in spaced labelled "version"on the form
Version # 0
1. A ticker tape timer produces 100 dots in 5.0 seconds. What is the period of the timer?
a) 20.0 seconds
b) 0.05 seconds
c) 0.5 seconds
d) 2.0 seconds
T = seconds/cycle
T = 5.0 s ÷ 100 dots
T = 0.05 seconds
2. A tape from a laboratory experiment has the appearance shown below. The dots were made by a
vibrator with a steady frequency of 20.0 Hz.
The total elapsed time between dot W and dot M is:
a) 6.0 s
b) 0.30 s
c) 0.35 s
d) 3.0 s
e) 3.5 s
frequency = 1÷T, So T = 1÷f
T = 1÷20 = 0.05 sec. The time between dots is the period which is 0.05 s.
Since there is a total of 6 spacings, then the total time = 6 x T.
total time = 6 x 0.05 = 0.30 s.
3. A ticker tape timer produces 200 dots in 2.0 seconds. What is the period of the timer?
(2.00 marks)
T = sec/cycle
T = 2 s ÷ 200 dots
T = 0.01 sec
(2.00 marks)
4. Using the following diagram calculate the frequency of the camera taking the pictures of the
ball as it progresses across the screen:
(3.00 marks)
total time = d ÷ v
t = 6.4 ÷ 4.0
t = 1.6 sec
(1 mark)
total number of time intervals = 5
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time per interval = 1.6 ÷ 5
time per interval = 0.32 sec
(1 mark)
f = 1/T
f = 1 ÷ 0.32
f = 3.1 sec
(1 mark)
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5. The number of wave crests passing a point per second is called
a) velocity
b) period
c) frequency
d) amplitude
e) wavelength
Frequency (f) is defined as: the number of cycles (wave crests) per second.
6. What is the amplitude of the wave in the diagram below?
a) 2.0 cm
b) 3.0 cm
c) 4.0 cm
d) 6.0 cm
Solution:
amplitude is distance from zero position to the maximum vertical position = 2 cm.
7. Height is to man as ______________ is to a wave.
a) wavelength
b) frequency
c) period
d) amplitude
e) beat
Height is to man as "amplitude" is to a wave. These both refer to the maximum vertical distance
that the objects display.
8. A wave has a frequency of 10 Hz. The period of the wave is
a) 1 s
b) 0.1 s
c) 0.01 s
Solution:
1
1
T= =
= 0.1 s
f 10
d) 0.001 s
9. The volume or intensity of sound waves is an indication of the wave's
a) velocity
b) period
c) frequency
d) amplitude
e) wavelength
10. Which waves require a material medium for transmission?
a) sound waves
b) radio waves
c) x rays
d) visible light
11. A beam of white light passes obliquely from air into glass. Which color experiences the greatest
change in direction ?
a) red
b) yellow
c) green
d) blue
Red light travels fastest, it has the smallest index of refraction. Blue light travels slower, it has
the largest index of refraction. As a result blue light is bent the most.
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12. A waves travels along a spring at a speed of 1.7 m/s. If the frequency of the source is 7.5 Hz,
then what is the wavelength?
a) 0.23 m
b) 4.4 m
c) 13 m
d) 44 m
Step 1. Recognize that this is a Universal wave equation problem. Therefore, to solve for the
wavelength, we
must use the equation v = fλ
Step 2. λ = ?
v = 1.7 m/s
f = 7.5 Hz
Step 3. λ = v ÷ f
λ = (1.7 m/s)÷(7.5 Hz)
λ = 0.23 m
13. A boat at anchor is rocked by waves with a speed of 10 m/s. These waves are 40 m apart, and
would reach the boat once every
a) 400 s
b) 4 s
c) 30 s
d) 0.25 s
Step 1. Recognize that this is a Universal wave equation problem. Therefore, to solve for
period, we
must use the equation v = fλ = λ÷T;
Step 2. T = ?
λ = 40 m
v = 10 m/s
Step 3. T = λ ÷ v
T = (40 m)÷(10 m/s)
T=4s
14. Radio astronomers detect radio waves at a frequency of 660 Hz. Given that radio waves travel
at the speed of light (3.0 x 108 m/s), the wavelength of these radio waves would be:
a) 2.2 x 10-6 m
b) 4.55 x 103 m
c) 4.55 x 105 m
d) 1.98 x 1014 m
Step 1. Recognize that this is a Universal wave equation problem. Therefore, to solve for
wavelength, we
must use the equation v = fλ
Step 2. λ = ?
v = 3.0 x 108 m/s
f = 660 Hz
Step 3. λ = v ÷ f
λ = (3.0 x 108 m/s)÷(660 Hz)
λ = 4.55 x 105 m
15. A periodic wave has a frequency of 10 Hz. The period of the wave is
a) 1 s
b) 0.1 s
c) 0.01 s
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d) 0.001 s
Step 1. Recognize that this is a simple period frequency relation problem. Therefore, to solve
for the period, we
must use the equation T = 1÷f
Step 2. Τ = ?
f = 10 Hz
Τ = (1)÷(10 Hz)
Τ = 0.1 s
Step 3. Τ = 1÷ f
16. An image formed in a plane mirror is
a) vertically inverted
b) laterally inverted
c) further behind the mirror
d) closer to the mirrors surface
All images are laterally inverted in plane mirrors.
17. Draw a diagram which represents the reflection of water waves in a ripple tank?
Include wave ray and wave front.
(1.00 mark)
(1.00 mark)
18. The image of the letter "P" in a plane mirror looks like
a)
b)
c)
All images are laterally inverted in plane mirrors. This
becom
d)
es
19. When straight waves are incident on a parabolic reflector, after they are reflected they
become…
a) circular waves
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b) standing waves
c) parabolic waves
d) point waves
20. A wave moves obliquely from shallow to deep water. What will you notice about the wave's
new direction, speed, and wavelength in this order?
a) Bends away from the normal, increase and increase
b) Bends towards the normal, increase and increase
c) Bends away from the normal, decrease and increase
d) Bends towards the normal, increase and decrease
A wave moves obliquely from shallow to deep water bends away from the normal, the speed and
wavelength increase.
21. Draw the refracted waves as they pass into the 'Fast Medium' in the diagram below. Marks are
given for correct wavelength, path, and pattern of distorted waves.
(3.00 marks)
(3.00 marks)
22. As a wave moves towards a shore line, what happens to the wavelength, frequency, speed and
amplitude?
a) increase, increase, increase, increase
b) decrease, same, increase, decrease
c) decrease, same, decrease, increase
d) decrease, decrease, decrease, decrease
As a wave moves towards a shore line the wavelength and speed decrease, the amplitude
decreases, and the frequency stays the same. This is due to the fact that kinetic energy is being
converted to potential energy.
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23. Sketch the waves as they might appear after they undergo the interaction shown. (show
the direction and the wavefronts)
(2.00 marks)
24. A wave moves obliquely from deep to shallow water. What will you notice about the wave's
new direction, speed, and wavelength in this order?
a) Bends away from the normal, decrease and decrease
b) Bends towards the normal, decrease and decrease
c) Bends away from the normal, decrease and increase
d) Bends towards the normal, increase and decrease
A wave moves obliquely from deep to shallow water bends towards the normal, the speed and
wavelength decrease.
25. Draw the reflected waves in the diagram below. Marks are given for correct wavelength, path,
and pattern of distorted waves.
(2.00 marks)
(2.00 marks)
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26. Which of the following statements is not true?
a) Diffraction is the dending of waves around a corner
b) Minimum diffraction is produced by short wavelenghts and large openings
c) Diffraction is caused when waves pass through an opening in a barrier
d) Diffraction is the change in direction when waves pass from one medium to another
A change in direction when waves pass from one medium to another is referring to refraction.
27. The bending of waves around an obstacle is called
a) refraction.
b) diffraction.
c) reflection.
Diffraction is defined as the bending of waves around an obstacle
d) interference.
Use the following information to answer the next 4 question(s).
Diagrams below, labelled A-D, represent four interference patterns produced by monochromatic light
passing through a diffraction grating and onto a screen. The dark bars indicate areas of maximum
intensity, or antinodes as they appear on the screen.
28. In pattern C, the distance between the central maximum and the first bright line is
2.0 x 10-2 m. The separation of the slits in the grating is 1.0 x 10-4 m and the distance to the
screen is 4.0 m. Calculate the wavelength of the source.
a) 2.2 x 10-3 m
b) 8.0 x 102 m
c) 5.0 x 10-7 m
d) 1.5 x 10-6 m
Step 1. Recognize that this is an interference problem. Therefore, to solve for wavelength, we
must use the equation λ = xd÷L
Step 2.
λ=?
x = 1.0 x 10-4 m
L = 4.0 m
d = 2.0 x 10-6 m
Step 3. λ = xd÷L = ((2.0 x 103 m ) (1.0 x 10-4 m)) ÷ 4.0 m = 5.0 x 10-7 m
29. If pattern B is produced by using monochromatic green light and the source is changed to
monochromatic red light (longer wavelength), then the pattern would become like pattern
a) A
b) B
c) C
d) D
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Step 1. Recognize that this is an interference problem. Therefore, to solve for the distance
between antinodes, we
must use the equation x = λL÷d
Step 2. Since L÷d is a constant for this problem, an increase in λ will produce an increase
between antinodes.
30. Which phenomenon is primarily responsible for producing all four interference patterns?
a) polarization
b) dispersion
c) refraction
d) diffraction
As light passes through two or more openings in a barrier it is diffracted. It is this diffracted
light that causes the interference patterns.
31. If the distance from the diffraction grating to the screen upon which pattern B is displayed is
increased, then the most likely pattern to appear would be
a) A
b) B
c) C
d) D
Step 1. Recognize that this is an interference problem. Therefore, to solve for the distance
between antinodes, we
must use the equation x = λL/d
Step 2. Since λ/d is a constant for this problem, an increase in L will produce an increase
between antinodes.
32. Nodes are regions of
a) complete constructive interference.
b) complete destructive interference.
c) points of resonance in a wave.
d) standing waves.
Nodes are the result of positive and negative pulses of equal magnitude interfering, thus
complete destuctive interference.
33. Interference patterns could be produced by
a) water waves
b) your voice
c) X-rays
d) all of these could produce interference patterns
e) only two of the above could produce interference patterns
Interference patterns applies to all types of waves
34. Which of the following sketches best illustrates the sound waves produced by a jet travelling at
super-sonic speeds? (The jet is represented by the black dot)
a)
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b)
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c)
d)
One diagram is moving at the speed of sound. One diagram is not moving. One diagram is
moving slower than the speed of sound. Super-sonic speeds are greater than the speed of sound,
so B is correct.
35. The pitch of the siren from a rapidly approaching police car would be
a) lower than the pitch heard by the driver
b) the same as the pitch heard by the driver
c) lower than the pitch heard by the observer after the car passes
d) higher than the pitch heard by the driver
Since the frequency of the sound is increasing the pitch also increases.
36. The image formed when an object is placed inside the focus of a concave lens is
a) virtual and smaller
b) virtual and larger
c) no image will be formed
d) real and smaller
e) real and larger
37. An object is located 12 cm from a converging lens of focal length 10 cm. The image distance is
a) 5.5 cm
b) – 5.5 cm
c) 60 cm
d) – 60 cm
do = 12 cm, f = 10 cm
1/f = 1/do + 1/ di
1/f – 1/do = 1/ di
1/10cm – 1/12 cm = 1/ di
6/ (60 cm ) – 5/ (60) cm = 1/ di
1/(60 cm )=1/ di
then 60 cm = di
Use the following diagrams to answer the next 1 questions (38-38)
The following diagrams show light rays interacting with various objects.
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38. Which of the diagrams shows a convex lens.
a) A
b) B
c) C
d) D
39. Select all that apply. Marks are subtracted for incorrect answers...
During an experiment you will be working with a virtual image when:
a) an object is placed inside the focus of a converging lens
b) an object is placed outside the focus of a converging lens
c) the image can only be viewed on a screen
d) the image can only be seen by placing your eye in the path of the light
e) the object is placed outside the focus of a diverging lens
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A virtual image is an image that has no light passing through it. The image appears to be in a
certain location because light appears to be coming from there, but it really is not. An image in a
plane mirror is a common example of a virtual image. It appears that the image is behind the
wall, but this is not true.
The blue lines represent where your eye
would think the light is coming from.
40. How can a drop of water act as a lens?
(2.00 marks)
Place a drop of water on an oily but clear piece of glass (such as a microscope slide). Look
through the water at some printing. The drop of water has a shape that is somewhat spherical
(due to surface tension in the water). Water will refract light, and so this drop of water will act
in a manner similar to a convex lens.
(2.00 marks)
41. Draw a diagram showing how five parallel rays of light would change direction entering and
leaving (a) a convex lens and (b) a concave lens.
(4.00 mark)
a)
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b)
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(2 marks each)
42. Complete the ray diagram below, identifying the location and height of the image.
Marks will be assigned for accuracy!
(4.00 marks)
(4.00 marks)
43. Which of the following rays correctly shows the path of light through the lens?
a) a
b) b
c) c
A light ray that approaches a diverging lens in a direction parallel to the lens' principal axis is
refracted so that it diverges in such a way that it appears to go through the focus of the lens.
44. In all four diagrams, the object is 5 cm high. Which set ups will produce a real, inverted image
that is greater than or equal to 5 cm in height.
a)
b)
c)
d)
e)
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Diagram i
Diagram ii
Diagram iii
Diagram iv
Diagrams ii and iv
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Converging lenses produce real, inverted images as long as the object is beyond the lens' focus.
The size of the real, inverted image will be smaller than the object if the object is beyond the
centre of curvature. The image will be the same size as the object if the object is at the centre of
curvature (diagram iv). The image will be larger than the object if the object is between the
focus and the centre of curvature (diagram ii).
45. Complete the ray diagram below, identifying the location and height of the image.
Marks will be assigned for accuracy!
(4.00 marks)
the image is virtual, larger than, and behind the object,
and it is upright (not inverted)
(4.00 marks)
46. A pencil 10 cm long is placed 30 cm in front of a mirror of focal length 50 cm. The image is
a) 2.5 cm long and upright
b) 0.25 cm long and inverted
c) 25 cm long and upright
d) 25 cm long and inverted
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do = 30 cm, f = 50 cm, ho = 10 cm
1/f = 1/do + 1/ di
1/f - 1/do = 1/ di
1/50cm - 1/30 cm = 1/ di
-0.013cm -1 =1/ di
then -75 cm = di
so the image is virtual. this indicates that we are using a diverging mirror, so we
Magnification = -di/do = hi/ho
Then (-di/do)* ho= hi
(- (-75cm) / (30cm))*10 cm = hi
25 cm= hi so the image is upright and 25 cm long
47. A real image is formed when
a) light rays converge and pass through the image
b) light rays seem to diverge from behind the mirror
c) the image cannot be projected onto the screen
d) rays farthest from the principle axis meet at the mirror's surface
The definition of a real image requires that light rays actually converge at the location of the
image. The rays will not just stop, but continue past the image.
48. Which type of mirror may be used to produce an inverted image?
a) flat or plane
b) convex
c) concave
d) convex or concave
A concave or converging mirror will produce an image that is real and inverted if the object is
beyond the mirror's focus.
Use the following diagrams to answer the next 1 questions (49-49)
The following diagrams show light rays interacting with various objects.
49. Which of the diagrams shows a concave mirror.
a) A
b) B
c) C
d) D
A concave mirror will reflect light back towards a focal point. Diagram C shows such a mirror.
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50. A student places a candle 30 degrees to the normal of a flat upright piece of glass. The candle
is located exactly 10 cm from the glass. The student observes the image of the candle apparently
behind the glass. How far behind the glass does the image appear to be?
a) 10 cm
b) 3 cm
c) 30 cm
d) 6.7 cm
The glass acts like a plain mirror, and so the image is virtual and the same distance behind the
glass as the object is in front of the glass (that is 10 cm).
51. The following diagram of an extended radius and a tangent line of a circle demonstrate how the
law of reflection is used to show the following principle of spherical mirrors:
a) The reason that a light ray through the focus of a mirror is reflected back parallel to the
principal axis.
b) The reason that diverging mirrors have virtual images.
c) A light ray that hits the mirror along a path through the mirror's centre of curvature will
bounce straight back.
d) The reason that a light ray parallel to the principal axis is reflected back through the mirror's
focus.
e) The reason that light rays that are through the centre of curvature are reflected through the
mirror's focus.
A light ray that goes through the centre of curvature (or appears to) hits the mirror normal to the
mirror's surface at that point. This is evident from the diagram which shows a radius that is
perpendicular to the tangent line (the "flat surface" of the mirror at the point of contact). Since
the law of reflection holds for any flat surface, and the tangent represents a flat surface, the
angle of incidence (zero degrees) equals the angle of reflection. Thus, the light ray reflects back
in the exact same direction it came from.
52. All of the diagrams have at least one missing light ray that has not been drawn in. Which
diagram correctly show the image that would form?
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a)
b)
c)
d)
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From the object, draw a horizonal ray towards the mirror
and then reflecting through the focal point. From the
object draw a ray through the focal point to the mirror
and then reflect horizontally. The intersection of the two
reflected rays will be where the image is.
53. Draw a ray diagram to locate the image.
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(3.00 marks)
(3.00 marks)
54. Complete the ray diagram for the mirror shown below, identifying the location and height of the
image. Note that the object is placed beyond the center of curvature for the mirror
Marks will be assigned for accuracy!
(4.00 marks)
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(4.00 marks)
55. Using the measurements from your ray diagram, calculate...
a) the magnification of the image
b) the location of the image
a) M = hi÷ho
M = 1.3 ÷ 0.56
M = approximately 2.3
(2 marks)
(2 marks)
(2 marks)
b) approximately 9 cm
(2 marks)
56. If you were to travel at a speed close to light, you would observe that your:
a) mass increased.
b) pulse decreased
c) shape changed
d) all of the above.
e) none of the above.
Relativistic effects are measured when there is a difference in relative velocities. Measurements
made within a moving frame would be the same as measurements made within a stationary
frame.
57. Compared to clocks in a stationary reference frame, clocks in a moving reference
frame run ...
a) slower
b) faster
c) at the same speed
In different reference frames, measured times can be different. According to the theory of
special relativity, moving clocks appear to run shorter than clocks at rest.
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58. Relativity equations for time, length, and mass hold true for…
a) relativistic speeds
b) every day low speeds
c) both of the above
d) none of the above
Solution:
Relativistic equations hold true for low speeds, since the ratio v2÷c2 becomes very small and so
the dilation factor becomes 1.
59. A truck is travelling toward a flashlight at 40.0 m/s. The flashlight is approaching at
12.0 m/s. At what speed does light strike the windshield of the truck?
a) 52 m/s
b) c + 40m/s
c) c + 52 m/s
d) c + 12 m/s
e) c
Solution:
One of the postulates of special relativity is that the speed of light (c) is constant regardless of
the motion of the source or observer.
60. What will be the mass of an electron travelling 0.80 c if its rest mass is 9.0 x 10-31 kg?
a) 5.4 x 10-31 kg
b) 1.5 x 10-30 kg
c) 2.5 x 10-30 kg
d) 1.2 x 10-15 kg
Solution:
mo
-31
Use the mass increase formula: m =
v 2 where v = 0.80 c, and mo = 9.0 x 10 kg
1− 2
c
m = 9.0 x 10-31 ÷ (0.36)1/2
m = 1.5 x 10-30 kg
61. When an object is pushed to relativistic speeds, its mass is measured to be…
a) greater than at rest
b) smaller than at rest
c) the same as at rest
Solution:
The mass of an object is measured to increase as its speed increases.
62. We are actually looking into the past when we look at ...
a) starlight
b) our physics book
c) actually both of the above
d) none of the above
Solution:
It would take light 1.7 x 10-9 seconds to travel 50 cm (distance from book to your eyes).
Technically, we are looking into the past when we see anything.
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63. The mass of a rocket is calculated as 1250.0 kg while stationary on the Earth. It is then
accelerated to a speed of 0.80 c relative to the Earth. At this speed, what would an
astronaut on board measure the mass of the rocket to be?
a) 750 kg
b) 1000 kg
c) 1250 kg
d) 2080 kg
Solution:
An observer on board would be at rest with respect to the rocket, so measure the mass to be the
same as when at rest on Earth.
64. A star emits light at c. Our rocket travels toward the star at 0.5 c. What speed would
the rocket measure for the beam of light?
a) 0 c
b) 0.5 c
c) 1.0 c
d) 1.5 c
Solution:
One of the postulates of special relativity is that the speed of light (c) is constant regardless of
the speed of the source or observer.
65. Calculate the dilation factor for an object travelling at 240 000 000 m/s
a) 240 000 000
b) 0.8
c) 0.6
d) 0.36
e) 300 000 000
v2
1− 2
c
Solution:
Expressing 2.4 x 108 m/s as a fraction of c
The dilation factor is:
(.8c) 2
γ = 1− 2
c
gives v = 0.8 c
γ = 0.6
66. What is the energy released when there is a 25 gram mass 'loss' in a nuclear explosion?
a) 7.5 x 106 J
b) 2.25 x 1015 J
E = mc2
E = 0.025 kg x (3.0 x 108 m/s)2
E = 2.25 x 1015 J
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c) 7.5 x 10-3 J
d) 1.8 x 1019 J