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AP Physics 7: Waves Name __________________________A. Wave Motion
1. generala. oscillation (energy) spreads out without carrying
matter b. mechanical wave: individual particles move a
short distance, transmit their energy to the next particle, and then return to their original position
c. pulses are generated by a single “bump”, which is not repeated
d. periodic wave is generated by an oscillating source
e. characteristics of a periodic wave
1. amplitude A (m): maximum oscillation distance from midpoint
2. wavelength (m): distance between any two successive identical points of the wave
3. frequency f (Hz or s-1): the number of waves that pass a point per unit time
4. period T = 1/f (s): time for one wave to pass 5. velocity, vw = /T = f (m/s)
2. types of waves
a. transverse wave (a) 1. forms in solids only (guitar string)2. disturbance wave 3. light waves (next section)
b. longitudinal wave (b)1. forms in all states (sound)2. disturbance wave
3. sounda. requires a mediumb. speed of sound (in air) depends on air density
1. density = velocity 2. humidity, elevation temperature = density
c. pitch is related to frequency of sound1. human ear: 20 Hz to 20,000 Hz2. piano range: 27.5 Hz to 4186 Hz
d. loudness, , is measured in decibels (dB)dB 0 dB 60 dB 120 dB 160 dB
example threshold of hearing
normal conversation
rock concert(damage)
ruptures eardrum
loudness 20 26 212 216
power 100 106 1012 1016
e. decrease of 6 dB for each doubling of distancedistance 10 m 20 m 40 m 80 mloudness 100 dB 94 dB 88 dB 82 dB
f. high frequencies sound louder (ear is most sensitive between 2000 Hz and 4000 Hz)
4. Interferencea. amplitudes combine when two waves meet
(superposition principle)constructive interference destructive interference
b. similar frequency produce beats: fbeat = |fB – fA|
c. same frequency = standing wave1. maximum displacement (constructive
interference) is “antinode”2. minimum displacement (destructive
interference) is “node”3. vibrating string (transverse wave)
a. velocity: vw = [Ft/(m/L)]½
b. harmonics (first, second, etc.)1. all possible standing waves where
nodes form at both ends2. number of harmonic (n) = # of loops3. n = 2L/n and fn = nf1
1= 2L/1
2 = 2L/2
3 = 3L/2
d. resonance: vibrating a structure at its natural frequency will cause amplitude magnification
5. Doppler effect a. when source(vs) and/or observer(vo) are moving
with respect to each other, the perceived frequency changes
b. frequency heard, f', compared to frequency generated, f1. f’ = f(vw ± vo)/(vw ± vs) 2. f' > f when approaching (+ vo and/or – vs)3. f' < f when receding (– vo and/or + vs)
c. approximation formula f/f v/vw
1. f’ = f + f when approaching2. f’ = f – f when receding
B. Light1. general properties
a. speed in a vacuum: c = 3 x 108 m/s b. no medium is needed c. transverse waved. polarization
e. Doppler shift alters frequency (red shift for receding source, blue shift for approaching source)
2. reflectiona. law of reflection i = r
1. angles measured from perpendicular (normal)2. "r" for reflection, "R" for refraction
b. phase shift occurs when reflecting surface has greater n (index of refraction)
phase shift no phase shift
3. refractiona. light slows when entering a transparent medium
1. atom-atom interactions2. vn = c/n = 3 x 108/n3. v = f and fn = f1 n = 1/n
b. rays that enter at an angle from normal1. bend toward normal if ni < nR
2. bend away from normal if ni > nR
c. Snell's law: n1sin1 = n2sin2
d. dispersion1. n increases with frequency (violet > red)2. color separation by a prism
3. when light travels from air to glass; violet light bends toward normal more than red
4. when light travels from glass to air; violet light bends away from normal more than red
e. total reflection1. light in glass can be trapped because the
refraction angle 90o (total reflection)2. only when ni > nR
3. critical angle c (when R = 90o): sinc = (nlow/nhigh)
C. Lenses and Mirrors1. parallel light rays refract toward (converge) or away
from (diverge) a focus, F, by a parabolic lensa. distance from lens to F = focal length fb. depends on nlens/nmedium lenses work differently
under water (less refraction) 2. parallel light rays reflect off of a parabolic mirror
radius of curvature r, where f = ½rCB bisector angles equal
isosceles CBF CF = FB
BF AF CF AF
AF ½AC or f = ½r
3. shapea. convex—center is thicker than the edges
1. lens is converging in function2. mirror is diverging in function
b. concave—center is thinner than the edges1. lens is diverging in function2. mirror is converging in function
4. ray tracingsa. parallel rays go toward/away from focus mirror diverging lens
f f mirror converging lensb. rays through center are straight lens mirror
f f mirror lensc. image forms where the two rays intersect
5. Lens/mirror equation: 1/do + 1/±di = 1/±fa. + f for converging lens or mirror
– f for diverging lens or mirrorb. do = object distance, use + for single optic systemc. di = image distance
1. +, image is real—visible on screen2. –, image is virtual—visible through lens/mirror
d. ho = object height, use + for upright objecte. hi = image height
1. +, image is upright2. –, image is upside down (inverted)
f. magnification, M = hi/ho = -di/do 1. > |1|, image is larger than object2. < |1|, image is smaller than object3. generalizations
a. real image: inverted, form on opposite side as object for lens; same side for mirror
b. virtual image: upright, form on same side as object for lens, opposite side for mirror
c. converging lens or mirror image1. do > 2f: real and small -1 < M < 02. do = 2f: real and same size M = -13. 2f > do > f: real and larger M < -1 4. do = f: no image5. f > do > 0: virtual and larger M > 16. |M| increases as do approaches f
d. diverging lens or mirror image1. all images are virtual and smaller2. M increases as object approaches
lens/mirrorg. small or partially covered lens reduces amount of
light (brightness of image) but not size or nature
D. Interference1. two or more openings (slits)
a. waves of synchronized, monochromatic light traveling from different openings to a common point interfere with each other
b. light and dark bands (fringes) on a screen1. center antinode = 0 order bright2. adjacent nodes = 0 order dark3. next antinode/node = 1st order, etc4. m = order #
c. light intensity decreases as m increasesd. angular deflection () from center to band
1. tan = x/L2. x = distance from center to band 3. L = distance between slits and screen
e. path length difference1. path difference = m, then constructive
interference—bright band2. path difference = (m + ½), then
destructive interference—dark bandf. angular deflection vs. order
1. Constructive: sinC = m/d 2. Destructive: sinD = (m + ½)/d3. d = distance between slits
g. multiple slits (grating) 1. more pronounced effect 2. d = grating space = length(m)/lines
(i.e. 500 lines/mm: d = 1E-3 m/500 = 2E-6 m) 3. atomic spectra produced by a grating
a. transitions between energy levels generate multiple wavelengths
b. constructive interference at different locations x on screen
2. one opening—W 2L/d
d W spot light
L3. partial reflection
a. interference occurs when light partially reflects off of a thin film or air space at both boundaries
incident light n1 nf n2
film
interference Tb. path difference = 2(T) + phase shift
1. when nfilm (nf) is middle value: phase shift = 02. when nf is largest or smallest: phase shift = ½ 3. constructive interference when total path
difference = even-half wavelengths (2/2, 4/2, ...)4. destructive interference when total path
difference = odd-half wavelengths (1/2, 3/2, ...)c. minimum thickness of a film, T (f = 1/nf)
Interference nf is middle value nf is extreme valueBright T = ½f T = ¼f
Dark T = ¼f T = ½f
A. Wave MotionQuestions 1-8 Briefly explain your answer.Questions 1-3 Consider the following graph of a periodic wave.
wave direction
1. If this is a transverse wave in a steel beam, which way are the iron atoms moving? (A) (B)
2. If this is a longitudinal wave in a steel beam, which way are the iron atoms moving? (A) (B)
3. If this is a water wave, which way is the black dot moving?(A) (B) (C) (D)
4. If you wanted to make a sound by hitting a steel beam with a hammer, which way would you strike the left side of the beam?(A) (B) (C) (D) , , or
5. The figure shows two wave pulses that are approaching each other.
Which of the following best shows the shape of the resultant pulse when points P and Q, coincide? (A) (B)
(C) (D)
Questions 6-8 A child swings from A to C back to A
A B C
6. Where would the child hear the highest frequency?
7. Where would the child hear the lowest frequency?
8. Where would the child hear the same frequency as the man?
9. Complete the following chart.
Velocity Frequency Wavelength
340 m/s 510 s-1
337 m/s 3.5 m
0.067 s-1 75 m
10. 100-Hz and 85-Hz whistles are blown simultaneously. How many beats per second are heard?
11. A guitar string 0.5-m long has a mass of 0.0125 kg and is under tension of 4 x 103 N. Determine thea. wave velocity
b. wavelength of the first harmonic
c. frequency of the first harmonic
12. Determine the observed frequency of sound (f’) for 600 Hz sound (f) for each of the following situations. (vw = 340 m/s)a. The source approaches the observer at 34
m/s.
b. The observer approaches the source at 34 m/s.
c. The source recedes from the observer at 34 m/s.
d. The observer recedes from the source at 34 m/s.
13. Choice the correct word concerning sound waves in air.temperature increases, speed increase decreases
s
power increases, loudness increases decreases
distance increases, loudness increases decreases
frequency increases, sensitivity increases decreases
frequency increases, pitch increases decreases
14. Two whistles are blown simultaneously. The wavelengths of the sound emitted are 9.0 m and 9.5 m, respectively. How many beats per second are heard? The speed of sound in air is 340 m/s.
15. When the E-string of an old piano is sounded simultaneously with a tuning fork of frequency 660 Hz, a 2 Hz beat is heard. What are the two possible frequencies of the E-string?
16. A guitar string 0.5-m long has a mass of 0.010 kg and is under tension of 4500 N.a. What is the wave velocity?
b. What is the frequency of the first harmonic?
c. What tension would produce a first harmonic frequency of 600 Hz?
17. The frequency of the second harmonic is 400 Hz. What is the frequency of the fourth harmonic?
18. An ambulance, generating a 600 Hz sound, approaches a listener at 17 m/s. What frequency does the listener hear? (vw = 340 m/s)
B. Light19. Reflection and Refraction Lab
a. Reflection: Place a plane mirror with its back along the line XY below. Stick a pin through point P from the back side of the paper. Point the ray-box so that the ray passes over R1, reflects off of the mirror and shines on the pin. Make a mark on the paper where the ray strikes the mirror. Repeat with R2 and R3. Extend the marks that you made to the line XY. Draw a line from R1 to the mark on the line XY and extend it to the top of the template. Draw a second line from P to the same mark on line XY so that you make a "V" at the XY line. Repeat with R2 and R3. Draw a perpendicular line from where the two lines intersect (this is the normal). Make a mark, P', in the middle of the intersections of the lines from R1, R2 and R3 on the opposite side of the mirror.
X Y mirrori r i ri r
P• R1• R2• R3•
R1 R2 R3
Angle between R and Normal i
Angle between P and Normal r
% = 100|i – r|/½(i + r)b. Index of Refraction: Place the semi-circle lens on the template below. Make sure the lens is aligned correctly by shining
the laser light along the 0o incident angle, which must refract at 0o. Move the laser light to 1 and mark where the light exits the lens. Repeat with 2 and 3. Remove the lens. Draw lines from each mark to the origin.
refracted angle 0o
place the half-circle lens here with flat side along this line
Collect the following data and calculate n for the three values of i and the average n.
1 2 3
Incident angle (i)refracted angle (R)
n = sini/sinR
Average n
Incident angle 0o 1 2 3
Questions 20-30 Briefly explain your answer.20. What is the mirror image of the letter "d"?
(A) d (B) b (C) p (D) q
21. You look at your self in a full length mirror, where the bottom half of the mirror is covered. What would you see?(A) The top half of your body(B) The bottom half of your body(C) Your whole body full size(D)Your whole body half size
22. Where would the image form? observer A
B C
(object) X mirror D
Questions 23-25 Light travels from air to glass and bends as shown in the diagram below.
23. Which side of the vertical line is the glass?
(A) left (B) right
24. Would the diagram be different if the light went from glass to air?(A) yes (B) no
25. Would the diagram be different if the glass was under water? (A) yes (B) no
Questions 26-28 A light ray R in medium I strikes a sphere of medium II with angle of incidence . The index of refraction for medium I is n1 and medium II is n2.
A
B
C D E
26. Which path is possible if nI < nII?
27. Which path is possible if nI > nII?
28. Which path is possible if nI = nII?
Questions 29-30 You are on shore looking at a fish in a pond.29. You want to spear the fish, where would you aim?
(A) directly at the fish(B) slightly behind the fish(C)slightly ahead of the fish
30. Where would you aim a laser light on a fish under water?(A) directly at the fish(B) slightly below the fish(C)slightly above the fish
31. 600-nm light (in air) enters glass (n = 1.50). Determine thea. frequency in glass.
b. wavelength in glass.
c. speed of light in glass.
32. The rays of the sun strike the surface of a lake at an angle of 65o with the vertical. At what angle, measured from the vertical, is the refracted ray in the water (n = 1.33)?
33. Determine the critical angle for light passing from diamond (n = 2.42) into air (n = 1.00).
34. What is the index of refraction of glass for which the critical angle at the glass-air interface is 37o?
35. Light of wavelength 450 nm passes from a vacuum into water (n = 1.33). Determine the following in water
velocity wavelength
36. Light strikes a flat piece of glass (n = 1.50) at an incident angle of 70o. Some of the light is reflected and some light passes out of the glass on the opposite side. a. What is the angle of reflection?
b. What is the angle of refraction inside the glass?
c. With what angle does light exit the glass on the opposite side?
37. Determine the critical angle for light passing from glass
(n = 1.50) into water (1.33).
C. Lenses and Mirrors38. What is the focal length of a mirror
that has a 10-cm radius of curvature?
39. State whether the lens or mirror is converge or diverge.Lens Mirror
ConcaveConvex
40. What is the + and – case for each of the following?positive negative
fdi
h41. What does it mean to have a magnification that is
greater than 1negative
42. Which type of mirror is the followingmake-up mirrorpassenger side mirror
43. A 1.5-cm-high diamond ring is placed 20 cm from a concave mirror (r = 30 cm). Determinea. the position of the image di.
b. the height of the image hi.
44. Complete the following ray tracings for a lens.Focus Straight Combined
Con
verg
ing
F F | | | | | |
Div
ergi
ng
| | | | | |
45. Convex-converging lens: Draw the two ray tracings.
F F
Calculated di Calculated hi
F F
F F
Calculated di Calculated hi
46. Concave-diverging lens: Draw the ray tracings.
F F
Calculated di Calculated hi
47. Complete the following ray tracings for a mirror.Focus Straight Combined
Con
verg
ing
F F | | | | | |
Div
ergi
ng
| | | | | |
48. Concave-converging mirror: Draw the two ray tracings.
F F
Calculated di Calculated hi
F F
F F
Calculated di Calculated hi
49. Convex-diverging mirror: Draw the ray tracings.
F F
Calculated di Calculated hi
50. The image formed by a converging lens is produced along a line from the top of the image through the center of the lens. Indicate the range of values for do in terms of f.
region I region II region III
51. An object 2.0 cm high is placed 20 cm from a convex lens of focal length 10 cm.a. Draw a ray diagram and locate the position
of the image formed. Draw in the image.
F F
b. Mathematically determine the following.image distance, di magnification, M image height, hi
c. Circle the correct descriptions of the image.real virtual inverte
d upright larger same size smaller
52. An object 2.0 cm high is placed 10 cm from a concave lens of focal length 10 cm.a. Draw a ray diagram and locate the position
of the image formed. Draw in the image.
F F
b. Mathematically determine the following.image distance, di magnification, M image height, hi
c. Circle the correct descriptions of the image.real virtual inverte upright larger same size smaller
d
53. An object 1.0 cm high is placed 30 cm from a concave mirror of focal length 20 cm.a. Draw a ray diagram and locate the position
of the image formed. Draw in the image.
F F
b. Mathematically determine the following.image distance, di magnification, M image height, hi
c. Circle the correct descriptions of the image.real virtual inverte
d upright larger same size smaller
54. A mirror is convex on one side (left) and concave on the other side (right). The center of curvature, C, is at 5.
C 1 2 3 4 5 6
a. Answer the following questions.Which side is converging in function?Which side could be used as a make-up mirror?Which side could be used as a side mirror in a car?Which point is closest to focus?b. Where would an object be placed to produce
an upright magnified image?an inverted unmagnified image?an inverted image that is smaller than the object?no image at allan upright image that is smaller than the object?an inverted image that is larger than the object?
55. A converging lens (f = 10 cm) is used to examine object A that is held 6 cm from the lens. a. Draw rays that show the position and size of the image.
b. Is the image real or virtual. Explain your reasoning.
c. Calculate the distance of the image from the lens.
d. Calculate the ratio of the image size to the object size.
e. The object A is moved to 20 cm. Describe the image position, size, and orientation.
56. Lenses and Mirrors Laba. Testing f = ½r for a converging Mirror: Shine the rays
from the ray-box at the converging mirror. Make sure the ray-box is aligned correctly by shining the center ray along the major axis and having it reflect straight back. Mark the intersection of the outside rays from the ray-box with the mirror. Also mark where the two reflected rays intersect the major axis; this is the focal point F. Measure the distance from the mirror to the intersection, which is the focal length f.
Exchange the 3-slit plate with the 1-slit plate. Shine the single ray at the mirror at one of the outside marks so that it reflects on itself. Mark where the ray crosses the major axis. Repeat with the other outside mark. Measure the distance from the mirror to the intersection of the two rays; this is the radius of curvature r.
f (cm) 2f (cm) r (cm) % between 2f and r
b. Nature of converging and diverging lenses: Light the candle. Hold C-1 at arms length from your eyes. Have your lab partner slowly move the candle from a few centimeters from the lens to 50 cm from the lens. Describe how the image of the candle flame changes as the candle is moved away. Repeat with C-2. Compare the change in image for C-1 and C-2.
Lens Description of image as candle is moved away.C-1C-2
Comparison between C-1 and C-2
Repeat the procedure from (b) with lenses D-1 and D-2.Lens Description of image as candle is moved away.D-1D-2
Comparison between D-1 and D-2
c. Determining f for converging lenses (method 1): Insert the C-1 lens in the lens holder and place it on the 50 cm line of the meter stick. Place the screen on the 80 cm line. Point the 0 cm side of the meter stick at a distant object. Move the screen toward the lens until a sharp, inverted, image forms on the screen. Measure the distance between the lens and screen, which is the focal length f. Repeat with the C-2 lens.
Lens PositionScreen (S) Lens (L) f = S - L
C-1: f (cm) 50 cmC-2: f (cm) 50 cmd. Determining f for converging lenses (method 2): Place
the candle, lens and screen at the positions listed in the table below. Slowly move the screen toward the lens until a sharp, inverted image of the candle forms on the screen.(1) Record the final location of the screen.
Lens
PositionCandle
(C)Lens(L)
Screeninitial final (S)
C-110 cm 60 cm 95 cm10 cm 50 cm 95 cm
C-210 cm 60 cm 95 cm10 cm 50 cm 95 cm
(2) Calculate the following from the experiment data.
variable di do f fav
formula di = S – L do = L – C 1/f = 1/do + 1/di fav =f/2
Lens
C-1
C-2
(3) Calculate the percent difference between the f from the two techniques (c and d).
C-1 C-2
D. Interference57. Hydrogen Spectrum Lab: observe the hydrogen
spectrum through a diffraction grating and measure the distance x from the light source to the color band, and then calculate the wavelength of the two color bands.a. Determine the distance d between lines on the
gratings.300 lines/mm
600 lines/mm
b. Collect the following data.Diffraction Grating 300 lines/mm 600 lines/mm
xAqua Red Aqua Red
c. Calculate the following from the data.Formula Calculation
300 lines/mm Aqua
tan = x/L
= dsin
300 lines/mm Red
tan = x/L
= dsin
600 lines/mm Aqua
tan = x/L
= dsin
600 lines/mm Red
tan = x/L
= dsin
d. Calculate the average value for for the 300 lines/mm and 600 lines/mm diffraction gratings.
Aqua
Red
e. Calculate the percent difference between the expected wavelengths (in parentheses) and the average wavelengths.
True Value %
Aqua 4.9 x 10-7 m
Red 6.6 x 10-7 m
58. Consider the diagram of a light wave.
D
A B E
C
a. Which position(s) are in phase with the dot?
b. Which position(s) are ½ out of phase with the dot?
c. Which position(s) are ¼ out of phase with the dot?
59. Triangles S1RS2 and QPoP1 are nearly similar. Therefore, the angles, , are equal.
a. Complete the trig function: _______ = x/L.
b. Complete the trig function: _______ = /d.
c. Determine the sin and tan of the following angles.
o 1 2 3 4 5 6 7 8 9 10sintand. What is the largest angle where the 3 place values
for sin and tanare equal? __________e. Assuming that is less than 7o and tan = sin,
combine the formulas into one and solve for x.tan = x/L and sin = m/d x = tan = x/L and sin = (m + ½)/d x =
60. Consider the Young's double slit experiment below, which demonstrated the wave nature of light.a. Fill in the order for each bright and dark band.b. Fill in the path difference between the rays.
Order
____ ____
____ ____
____ ____
____ ____
____ ____
____ ____
____ ____61. The diagram shows the relative intensity for each band.
What generalization can you say about the intensity as n increases?
62. A 600 nm (600 x 10-9 m) laser beam is incident on a pair of slits 0.500 mm (0.50 x 10-3 m) apart. Determine the angular deflection, , for the following.a. Second-order bright band?
b. Zero-order dark band?
c. Determine the distance from the center bright band to the zero order dark band (x) on a screen 2.0 m away.(1) Use the angular deflection from part b.
(2) Use the formula derived in 58e(2).
d. What is the width of the center bright band? (Distance between the zero order dark bands.)
e. One slit is covered, so that the light only passes through one slit which is 0.500 mm wide and 1.00 mm tall.(1) What is the width of the light spot?
(2) What is the length of the light spot?
(3) The slit is longer than it is wide. How would you describe the shape of the spot light?
63. A 500 lines/mm diffraction grating is used to observe mercury light. The first-order green band is 28.4 cm away from the light when the grating is 1.00-meter from the light.a. What is the angle of deflection for the green
band?
b. What is the spacing between each groove on the diffraction grating (d)?
c. Determine the wavelength of the green band.
d. Use the formula derived in 58e(1) to determine .
e. Which answer, part c or d, is correct? Explain.
64. 500 nm light does not reflect off of a film (n = 1.25) that is on a piece of glass (n = 1.5) viewed in air (n = 1). a. What is the wavelength of light in the film?
b. What is the minimum thickness of the film?
c. Would this coating work under water? Explain.
65. A soap bubble appears blue ( = 450 nm) in sunlight.a. What is the wavelength in the soap film (n =
1.35)?
b. What is the minimum thickness for a soap bubble?
66. A laser beam is incident on two slits separated by 0.5 mm. The interference pattern is formed on a screen 1 m from the slits and the first bright fringe is found to be 0.12 cm to the right of the central maximum. Determine the a. angular deflection of the first bright fringe.
b. wavelength of light in nanometers.
c. distance from the center to the zero order dark fringe.
d. width of the zero order bright fringe.
e. Would the width of the zero order bright fringe increase or decrease if 700 nm light is used?
67. The hydrogen spectrum is analyzed using a 600 lines/mm diffraction grating, which is 1.00 m away from the light source. An aqua line appears at 30.5 cm and a red line appears at 42.8 cm from the light source. Determine thea. wavelength of aqua light.
b. wavelength of red light.
68. What is the minimum thickness for a soap bubble that appears red ( = 750 nm, n = 1.35)?
69. What is the thickness of a coating (n = 1.22) on glass (n = 1.50) that will eliminate 550 nm light?
Practice Multiple Choice (No calculator)Briefly explain why the answer is correct in the space provided.Questions 1-3 The diagram represents a transverse wave traveling in a string.
6 m1. Which pair points is half a wavelength apart?
(A) A and D (B) D and F (C) B and F (D) D and H
2. What is the wavelength?(A) 1 m (B) 6 m (C) 2 m (D) 3 m
3. What is the speed of the wave if its frequency is 9 Hz?(A) 0.3 m/s (B) 1 m/s (C) 3 m/s (D) 27 m/s
4. If the amplitude of a transverse wave traveling in a rope is doubled, the speed of the wave in the rope will(A) decrease (B) increase (C) remain the
same
Question 5-6 Use the graph wavelength versus frequency of waves to answer the questions.
5. What is the wavelength when the frequency is 2.0 Hz?(A) 5.0 m (B) 2.5 m (C) 2.0 m (D) 1.0 m
6. What is the speed of the waves generated in the spring?(A) 2 m/s (B) 5 m/s (C) 7 m/s (D) 10 m/s
7. Sound in air can best be described as which of the following types of waves? (A) Longitudinal (B) Transverse (C) Gravitational (D) Electromagnetic
8. Maximum destructive interference will occur when two waves having the same amplitude and frequency (A) meet crest to crest (B) meet crest to trough
9. The motion of the individual particles in the medium compared to the direction of the transverse wave, is(A) perpendicular (B) parallel
Questions 10-12 A standing wave of frequency 5 Hz is set up on a string 2 m long with nodes at both ends and in the center.
10. What is the harmonic of this standing wave?(A) first (B) second (C) third (D) fourth
11. The speed at which waves propagate on the string is (A) 0.4 m/s (B) 2.5 m/s (C) 5 m/s (D) 10 m/s
12. The first harmonic of vibration of the string is(A) 1 Hz (B) 2.5 Hz (C) 5 Hz (D) 7.5 Hz
13. A ringing bell is located in an airless chamber. The bell be seen vibrating but not be heard because (A) Light can travel through a vacuum, but sound cannot. (B) Sound waves have greater amplitude than light waves. (C) Light waves travel slower than sound waves.(D) Sound waves have higher frequencies than
light waves.
14. Resonance occurs when a vibrating object transfers energy to another object causing it to vibrate. The energy transfer is most efficient when the two objects have the same (A) frequency (B) amplitude(C) loudness (D) speed
15. The frequencies of the third harmonics of a vibrating string is f. What is the fundamental frequency of this string? (A) f/3 (B) f/2 (C) f (D) 2f
16. In the Doppler effect for sound waves, factors that affect the frequency that the observer hears include which of the following?
I. The speed of the source II. The speed of the observerIII. The loudness of the sound
(A) I only (B) II only (C) III only (D) I and II
17. A train whistle has a frequency of 100 Hz as heard by the engineer on the train. Assume that the velocity of sound in air is 330 m/s. If the train is approaching the station at a velocity of 30 m/s, the whistle frequency that a stationary listener at the station hears is most nearly (A) 90 Hz (B) 110 Hz (C) 120 Hz (D) 240 Hz
18. The product of a wave's frequency and its period is(A) one (B) its velocity(C) its wavelength (D) Plank's constant
19. The figure shows two waves that are approaching each other.
Which of the following best shows the shape of the resultant pulse when points P and Q, coincide? (A) (B)
(C) (D)
20. A sound wave has a wavelength of 5.5 m as it travels through air, where the velocity of sound is 330 m/s. What is the wavelength of this sound in a medium where its speed is 1320 m/s?(A) 1.4 m (B) 2.2 m (C) 14 m (D) 22 m
21. A cord of fixed length and uniform density, when held between two fixed points under tension T, vibrates with a fundamental frequency f. If the tension is doubled, the fundamental frequency is (A) 2f (B) √2f (C) f (D) f/√2
Questions 22-23 An object, slanted at an angle of 45°, is placed in front of a vertical plane mirror.
mirror• A
• B• C
• D 22. Which of the labeled points is the position of the
image?
23. Which of shows the orientation of the object's image? (A) (B) (C) (D)
Questions 24-26 A ray of light ( = 6 x 10-7 m) in air (n = 1) is incident on quartz glass (n = 2).
Air
Quartz glass24. What is the angle of reflection measured from normal?
(A) 35o (B) 55o (C) 22o (D) 33o
25. The angle of refraction measured from normal is closest to(A) 25o (B) 35o (C) 55o (D) 75o
26. Which is correct about the light in quartz glass is correct?(A) v = 3 x 108 m/s (B) = 6 x 10-7 m(C) v = 1.5 x 108 m/s (D) = 1.2 x 10-6 m
Question 27-31 A curved surface, with a 10-cm focal length is mirrored on both sides.
A B C D| | |
10 cm 10 cm 20 cm27. Which location would you place an object to form a virtual
image that is smaller than the object?
28. Which location would you place light source in order to produce parallel rays of reflected light?
29. Which location would you place an object to form a real image?
30. Which location would you place an object to form a virtual image that is larger than the object?
31. Which location is the radius of curvature for this mirror?
32. Light leaves a source at X and travels to Y along the path.
Which of the following statements is correct? (A) n1 = n2 (B) n1 > n2 (C) v1 > v2 (D) f1 > f2
33. The critical angle for a transparent material in air is 30°. The index of refraction is most nearly (A) 0.33 (B) 0.50 (C) 1.0 (D) 2.0
34. A light ray passes through substances 1, 2, and 3. The indices of refraction for these three substances are n1, n2, and n3, respectively. Rays 1 and 3 are parallel.
1 2 3
n1 n2 n3
Which medium has the largest n?(A) 1 only (B) 2 only (C) 3 only (D) 1 and 3
35. Which statement is correct about lenses and mirrors?(A) Converging lenses are thinnest in the middle.(B) Convex mirrors are curved inward.(C) A real image formed by a convex lens is dimmed
when half of the lens is covered.(D) A diverging lens or mirror can make a real or
virtual image depending on where the object is placed.
36. 600-nm light passes through two slits. The first-order interference maximum appears at 6°. What is the separation of the slits? (sin6o = 0.10)(A) 1500 nm (B) 4500 nm(C) 3000 nm (D) 6000 nm
37. A thin film with index of refraction nf separates two materials, each of which has an index of refraction less than nf. A monochromatic beam of light is incident normally on the film.
If the light has wavelength in air, maximum constructive interference between the incident beam and the reflected beam occurs for which of the following film thicknesses? (A) 2/nf (B) /nf (C) /2nf (D) /4nf
38. How are electromagnetic waves that are produced by oscillating charges and sound waves in air that are produced by oscillating tuning fork similar? (A) Both have the same frequency as their respective
sources. (B) Both require a matter medium for propagation. (C) Both are longitudinal waves. (D) Both are transverse waves.
39. When observed from Earth, the wavelengths of light emitted by a star are shifted toward the red end of the spectrum. This red shift occurs because the star is (A) at rest relative to earth (B) moving away from earth (C) moving toward earth at decreasing speed(D) moving toward earth at increasing speed
40. Parallel wave fronts incident on an opening in a barrier are diffracted. For which combination of wavelength and size of opening will diffraction effects be greatest? (A) short wavelength and narrow opening(B) short wavelength and wide opening (C) long wavelength and narrow opening(D) long wavelength and wide opening
Question 41-44 Consider a converging lens with focal length, f.
41. Where is the image formed for an object that is placed on the left at 3/2f?(A) On the right at 3f. (B) On the left at 3f.(C) On the right at ⅓f. (D) On the left at f.
42. Where is the image formed for an object that is placed on the left at ½f?(A) On the right at 3f. (B) On the left at 3f.(C) On the right at ⅓f. (D) On the left at f.
43. Where would the object be placed in order to produce an upright image that is larger than the object?(A) at 2f (B) between 2f and f(C) at f (D) between f and the lens
44. Where would the object be placed in order to produce an inverted image that has a magnification of 2?(A) 2f (B) 3/2f (C) 4/3f (D) 5/4f
45. Which is true of a single-slit diffraction pattern? (A) It has equally spaced fringes of equal intensity. (B) It has a relatively strong central maximum. (C) It can be produced only if the slit width is less than one
wavelength. (D) It can be produced only if the slit width is
exactly one wavelength.
46. Two plane mirrors are positioned perpendicular to each other as shown. A ray of light is incident on mirror 1 at an angle of 55°. This ray is reflected from mirror 1 and then strikes mirror 2.
Which is correct?I. The incident ray for mirror 2 is 55o measured from
normal.II. The reflected ray for mirror 2 is 35o measured from
normal.III. The reflected ray for mirror 2 is parallel to incident ray
for mirror 1.(A) I only (B) II only (C) I and III (D) II and III
Questions 47-49 A light ray R in medium I strikes a sphere of medium II with angle of incidence . The index of refraction for medium I is n1 and medium 2 is n2.
47. Which path is possible if n1 < n2?
48. Which path is possible if n1 > n2?
49. Which path is possible if n1 = n2?(A) A or B (B) C or D (C) All (D) None
50. A beam of white light is incident on a triangular glass prism (n = 1.5) for visible light, producing a spectrum. Initially, the prism is in an aquarium filled with air.
Which is true if the aquarium is filled with water (n = 1.3)? (A) No spectrum is produced. (B) The positions of red and violet are reversed. (C) The spectrum produced has greater separation
between red and violet than that produced in air. (D) The spectrum produced has less separation between
red and violet than that produced in air.
Practice Free Response1. The second harmonic standing wave is generated on a 0.1
kg string and has a frequency of 20 s-1.
a. What is the wave velocity?
b. What is the tension in the 0.1 kg string?
c. Determine the following for the fourth harmonic.(1) Wavelength
(2) Frequency
2. The figure shows a converging mirror, focal point F, center of curvature C, and an object represented by the arrow.a. Draw a ray diagram showing two rays and the image.
b. Is the image real or virtual? Justify your answer.
c. The focal length of this mirror is 6.0 cm, and the object is located 8.0 cm away from the mirror. Calculate the position of the image formed by the mirror.
Suppose that the converging mirror is replaced by a diverging mirror with the same radius of curvature that is the same distance from the object, as shown below.d. Draw a ray diagram showing two rays and the image.
e. For this mirror, determine
(1) the image distance from the mirror.
(2) the magnification.
3. A diffraction grating with 600 lines/mm is used to study the line spectrum of the light produced by a hydrogen discharge tube. The grating is 1.0 m from the source (a hole at the center of the meter stick). An observer sees the first-order red line at a distance yr = 428 mm (0.428 m) from the hole.
Determine a. the angular deflection.
b. the wavelength of the red light.
The 600 line/mm grating is replaced by a 800 lines/mm grating. Determinec. the angular deflection.
d. the distance, yr, where the observer sees the first-order red line.
4. A beam of red light of wavelength 6.65 x 10-7 m in air is incident on a glass prism at an angle 1. The glass has index of refraction n = 1.65 for the red light. When 1 = 40o, the beam emerges on the other side at 4 = 84o.
a. Calculate the angle of refraction 2.
b. What minimum angle 3 would result in total internal reflection.
c. Would the incident angle 1 be greater than or less than 40o?
d. The glass is coated with a thin film that has an index of refraction nf = 1.38 to reduce the partial reflection. (1) Determine the wavelength of the red light in the film.
(2) Determine the minimum thickness of the film.
Answers (Don't look until after you have tried the problem)1 B—Transverse wave particle move perpendicular to wave2 A—Longitudinal wave particles move parallel to wave3 D—Wave is moving right, trough is heading toward dot
4 D—Any hit will generate a wave, but how you hit the beam determines if it is transverse or longitudinal
5 A—The left sides will interfere constructively and the right sides will interfere destructively
6 B—when traveling toward the whistle7 B—when traveling away from the whistle8 A and C—when the child is momentarily stationary
9340 m/s/510 s-1 = 0.67 m337 m/s/3.5 m = 96 s-1
(75 m)(0.067 s-1) = 5.0 m/s10 fbeats = fB – fA = 100 Hz – 85 Hz = 15 Hz
11v = [FT/(m/L)]½ = [(4 x 103 N)/(0.0125 kg/0.5 m)]½ = 400 m/s1 = 2L/n = 2(0.5 m)/1 = 1 mv = f f = v/ = 400 m/s/1.0 m = 400 s-1
12
f’ = f(vw + vo)/(vw
– vs) = 600 s-1(340 + 0)/(340 – 34) = 667 s-1
f’ = f(vw + vo)/(vw
– vs) = 600 s-1(340 + 34)/(340 – 0) = 660 s-1
f’ = f(vw + vo)/(vw
– vs) = 600 s-1(340 – 0)/(340 + 34) = 545 s-1
f’ = f(vw + vo)/(vw
– vs) = 600 s-1(340 – 34)/(340 + 0) = 540 s-1
13 increases increases decreases increases increases14 fbeats
= fB – fA
= vw/A – vw/B = 340 m/s/9.00 m – 340 m/s/9.5 m = 2 s-1
15 Fbeats = fB – FA 2 = 660 – Fstring or 2 = Fstring – 660 Fstring = 658 s-1 or 662 s-1
16
v = [Ft/(m/L)]½ = [(4500 N)/(0.010 kg/0.5 m)]½ = 474 m/s1 = 2L = 2(0.5 m) = 1 mv = f f = v/ = 474 m/s/1.0 m = 474 s-1
v = f = (600 s-1)(1.0 m) = 600 m/sv = [Ft/(m/L)]½ 600 m/s = [Ft/(0.010/0.5)]½Ft = 7200 N
17 F2 = nF1 F1 = 400 s-1/2 = 200 s-1
F4 = nF1 = 4(200 s-1) = 800 s-1
18 f’ = f(vw + vo)/(vw
– vs) = 600 s-1(340 + 0)/(340 – 17) = 632 s-1
20 B—Mirrors transpose left and right
21 C—As long as the mirror above the half way point between your eyes and feet is uncovered
22 D—The image is formed where the line from the reflected light is extended to the other side of the mirror
23 A—The light in glass will bend toward normal (ng > na)
24 B—The light would follow the same path if reversed (n1sin1 = n2sin2) since "1" and "2" are interchangeable
25 A—The index for water is different than air the light would not bend in the same way (less diffraction)
26 E—Bend toward normal at left interface and away from normal at right interface
27 A— Bend away from normal at left interface and toward normal at right interface
28 C—There is no refraction at either interface
29 B—The light from the fish bends away from normal into your eye you see it farther away than it is
30 A—Directly at the fish because the laser light will follow the same path as visible light
31fn = f = c/ = 3 x 108 m/s/600 x 10-9 m = 5 x 1014 s-1
n = /n = 600 nm/1.50 = 400 nmvn = c/n = 3 x 108 m/s/1.50 = 2 x 108 m/s
32 n1sin1 = n2sin2 (1.00)(sin65) = (1.33)(sin2) 2 = 43o
33 sinc = nlow/nhigh = 1.00/2.42 C = 24.4o
34 sinc = nlow/nhigh sin37 = 1.00/nhigh nhigh = 1.66
35 v = c/n = 3E8m/s/1.33 =
2.26E8m/sn = /n = 450 nm/1.33 = 338 nm
36i = r = 70o
nRsinR = nisini 1.50sinR = 1.00sin70o R = 38.8o
nRsinR = nisini1.50sin38.8 = 1.00sini i = 70o
37 sinC = n2/n1 = 1.33/1.50 C = 62o
38 f = ½r = ½(10 cm) = 5 cm
39 diverge convergeconverge diverge
40converging diverging
real virtualupright inverted
41 The image is larger than the object.The image is upside down.
42 Concave (converging) Convex (diverging)
43 1/do + 1/di = 1/f 1/20 + 1/di = 1/15 di = 60 cmhi/ho = -di/do hi = -hodi/do = (-1.5 cm)(60/20) = -4.5 cm
45 1/3 + 1/di = 1/2 di = 6 hi/1 = -6/3 hi = -21/1 + 1/di = 1/2 di = -2 hi/1 = -(-2)/1 hi = 2
46 1/2 + 1/di = 1/-2 di = -1 hi/2 = -(-1)/2 hi = 1
48 1/3 + 1/di = 1/2 di = 6 hi/1 = -6/3 hi = -21/1 + 1/di = 1/2 di = -2 hi/1 = -(-2)/1 hi = 2
49 1/2 + 1/di = 1/-2 di = -1 hi/2 = -(-1)/2 hi = 150 0 < do < f no value do > 2f 2f > do > f
511/20 + 1/di = 1/10di = 20 cm
M = -di/do
M = -20/20 = -1-1 = hi/2.0hi = -2.0 cm
real virtual inverted upright larger same smaller
521/10 + 1/di = 1/-10di = -5 cm
M = -di/do
M = -(-5)/10 = ½½ = hi/2.0hi = 1.0 cm
real virtual inverted upright larger same smaller
531/30 + 1/di = 1/20di = 60 cm
M = -di/do
M = -60/30 = -2-2 = hi/1.0hi = -2.0 cm
real virtual inverted upright larger same smaller54 right right left 3 2 5 6 3 1 4
55
Virtual, it forms on the same side as the object.1/do + 1/di = 1/f 1/6 + 1/di = 1/10 di = -15 cmhi/ho = -di/do = -(-15)/6 = 2.51/do + 1/di = 1/f 1/20 + 1/di = 1/10 di = 20 cm(M = -di/do = -20/20 = -1) Left side at 20 cm; inverted; same size
58 A B and E C and D61 The intensity decrease as n increases
62
sin = m/d = 2(6 x 10-7 m)/(0.50 x 10-3 m) = 2.4 x 10-3 = 0.14o
sin = (m+½)/d = ½(6 x 10-7 m)/(0.50 x 10-3 m) = 6.0x10-4 = .034o
tan = x/L x = Ltan = (2.0 m)tan0.034 = 1.2 x 10-3 mx = (m+½)L/d = ½(600x10-9 m)(2.0 m)/(0.500x10-3 m) = 1.2x10-3 mW = 2x = 2(1.2 x 10-3 m) = 2.4 x 10-3 m W 2L/d 2(600 x 10-9 m)(2.0 m)/(0.0005 m) = .0048 mW 2L/d 2(600 x 10-9 m)(2.0 m)/(0.001 m) = 0.0024 mThe spot light is wider than it is long.
63
tan = x/L = 0.284 m/1.00 m = 15.9o
d = Length(m)/lines = 0.001 m/500 grooves = 2 x 10-6 msin = m/d = dsin/m = (2 x 10-6 m)sin15.9o/1 = 5.46 x 10-7 mm/d = x/L = dx/mL = (2 x 10-6)(0.284)/(1)(1.00) = 5.68 x 10-7 mPart c is correct because the > 6o
64n = /n = 500 nm/1.25 = 400 nmnfilm is middle and dark T = ¼ = ¼(400 nm) = 100 nmNo, under water, nfilm is extreme ¼ would be bight
65 n = /n = 450 nm/1.35 = 333 nmnfilm is extreme and bright T = ¼ = ¼(333 nm) = 83 nm
66
tan = x/L = 0.12 x 10-2 m/1 m = 1.2 x 10-3 = 0.069o
x/L = m/d=xd/mL=(0.12x10-2 m)(0.5x10-3 m)/(1)(1m) = 600 nmx/L = (m+½)/dx = (1m)(0.5)(600x10-9m)/(0.5x10-3m) = 6x10-4m Width = 2x = 2(6 x 10-4 m) = 1.2 x 10-3 mx is proportional to wavelength increase
67
tan = (30.5 x 10-2 m)/(1.00 m) = 17.0o
sin = m/d = /d = dsin = (1 x 10-3/600)sin17.0o = 4.86 x 10-7 mtan = (42.8 x 10-2 m)/(1.00 m) = 23.2o
sin = m/d = /d = dsin = (1 x 10-3/600)sin23.2o = 6.56 x 10-7
m
68 film is extreme and brightT = ¼film = 750 nm/(4)(1.35) = 140 nm
69 film is middle and dark T = ¼film = 550 nm/(4)(1.22) = 113 nmPractice Multiple Choice
1 B—Half wavelength is half as way to next corresponding
point
2 D—Diagram is two complete waves (6 m) 1 wavelength = 3 m
3 D—v = f = (3 m)(9 s-1) = 27 m/s4 C—vw = [FT/(m/L)]½, no amplitude5 B—start at 2 Hz on x-axis, do up to line and left to y-axis = 6 B—v = f = (2.5 m)(2.0 s-1) = 5 m/s
7 A—Only longitudinal waves in air (no bonding between particles)
8 B—Amplitudes combine, which is minimized with crest + trough
9 A—Transverse: particle move at right angles to the wave motion
10 B—This is the second harmonic (number of loops)11 D—v = f = (2 m)(5 s-1) = 10 m/s12 B—fn = nf1 5 s-1 = (2)f1 f1 = 2.5 s-1
13 A—Sound (mechanical wave) requires a medium; not light14 A—Same frequency produces reinforcing impulses15 A—fn = nf1 f = 3f1 f1 = f/316 D—f’ = f(vw ± vo)/(vw ± vs)f' depends on f, vo, vs and vw
17 B—f/f v/vw f/100 30/330 f 10 Hz, approaching 110
18 A—T = 1/f fT = 119 A—Left sides add together while the right sides cancel20 D—f1
= f2: v1/1 = v2/2 2 = (v2/v1)1 = (1320/330)5.5 m = 22 m21 B—v = [Ft/(m/L)]½ = f and f (Ft)½ f' = 2f
22 D—Image is located where reflected ray extends from the object
23 A—Same top-bottom, front-back orientation, reverses left-right
24 B—i = r, which is 90o – 35o = 55o (measured from normal)25 A—n1sin1 = n2sin2 sin(55) = 2sin2 2 < ½(55o)
26 C—n = /n = 6x10-7/2 = 3x10-7m,vn
= c/n = 3x108 m/s/2 = 1.5x108 m/s
27 A—Convex: smaller virtual (concave: large virtual when do < f)
28 C—Light from focus produces parallel rays rather than an image
29 D—Real with concave where the object is outside the focus30 B—Large virtual on concave side when do < f31 D—Radius of curvature is twice the focal length (f = ½r)32 C—Bends toward normal in 2 n2 > n1 and 1 > 2 and v1 > v2
33 D—sinc = nlow/nhigh sin(30) = 1/nhigh ½ = 1/nhigh nhigh = 2
34 B—n1sin1 = n2sin2 n2 > n1 because 2 < 1 and n1 = n3 because 1 = 3, n2 > n1 = n3
35 C—Converging lens are thick in the middle (convex), partially covering dims image, diverging make virtual
36 D—sinc = m/d 0.10 = (1)(600 nm)/d d = 6000 nm37 D—nf is extreme and bright: T = ¼f
38 A—Sound requires matter, not light; sound is longitudinal, light is transverse; source f = wave f
39 B—When source and observer separate, the wavelength increases (stretches out) = red shift
40 C—sinC = m/d: Angular deflection C is greatest when /d is largest (long wavelength and narrow opening)
41 A—1/do + 1/di = 1/f 2/3f + 1/di = 1/f di = 3f (on the right side)
42 D—1/do + 1/di = 1/f 2/f + 1/di = 1/f di = f (on the left side)
43 D—1/do + 1/di = 1/f: an upright image is virtual di < 0 and do < f
44 B—A 2 x inverted image is real di = 2 do
1/do + 1/di = 1/f 1/do + 1/2do = 1/f do = 3/2f
45 B—Most energy goes to the central maximum (sharp shadow edge), width > wavelength (sin = /d < 1)
46 B—The normal for mirror 2 forms a right triangle with the normal from mirror 1 and the reflected ray
47 C—If n1 < n2, then ray bends toward normal (C-D path) and away from normal on the right (path C)
48 A—If n1 > n2, then ray bends away from normal (A-B path) and toward normal on right (path A)
49 D—If n1 = n2, then there is no refraction at either surface and the light passes straight through.
50 D—R decreases when n decreases sinR = ni(sini)/nR the smaller angle produces less separation
Practice Free Responsevw = f = (2 m)(20 s-1) = 40 m/s
vw = [FT/(m/L)]½
FT = vw2m/L = (40 m/s)2(0.1 kg)/(2 m) = 80 N
n = 2L/n 4 = 2(2 m)/4 = 1 mfn = nf1 f1 = 20 s-1/2 = 10 s-1
f4 = (4)(10 s-1) = 40 s-1
The image is real because the rays converge on the same side of the mirror as the object.1/do + 1/di = 1/f1/di = 1/f – 1/do = 1/6.0 cm – 1/8.0 cm di = 24 cm1/di = 1/f – 1/do = 1/-6.0 cm – 1/8.0 cm di = -3.4 cmM = -di/do = -(-3.40)/8 = 0.43tan = x/L = 0.428 m/1.0 m = 0.428 = 23.2o
sin = m/d = dsin = (0.001 m/600)sin(23.2o) = 6.55 x 10-7 m (655 nm)sin = m/d = 6.55 x 10-7 m/(0.001 m/800) = 0.524 = 31.6o
tan = x/L x = Ltan = (1.0 m)tan(31.6o) = 0.615 mn1sin1 = n2sin2 (1.00)sin40 = (1.65)sin2 2 = 23o
n3sin3 = n4sin4 (1.65)sin3 = (1.00)sin90 3 = 37o
For 4 to be 90o, 3 has to be greater, which makes 2 smaller, which in turn makes 1 smallerf = /n = 6.65 x 10-7 m/1.38 = 4.83 x 10-7 mMinimum reflection and middle nf T = ¼f
T = ¼(4.83 x 10-7 m) = 1.21 x 10-7 m
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