Review For Waves Test
Page 1
What is the frequency of 720. nm (1 nm = 1x10-9
m) light? What is its period(Speed = 3.00 x 108
m/s)
v = f
f = 1/T
 = 720. x10-9 m, v = c = 3.00 x 108 m/s
f = 4.17E+14 Hz
T = 1/f = 2.4E-15 s
4.17E+14 Hz, 2.4E-15 s
W
What is the speed of light in water? n = 1.33
c
n = /v
n = 1.33, c = 3.00 x 108 m/s
2.25 x 108 m/s
W
What is the wavelength of 720. nm light in water?
n = 1.33 (720. nm is its wavelength in a vacuum,
the frequency remains the same) (4)
fair = f water
c
n = /v
v = f
find f, then v, then 
f = v/ = 4.16667E+14 Hz
vwater = c/n = 225563909.8 m/s
 = v/f = (225563909.8 m/s)/(4.16667E+14 Hz) = 541. nm
or 720/1.33 = 541 nm
541 nm
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Draw the red rays, be able to label the angle of incidence (θ1 in
this picture) and the refracted angle (θ2)
What is the Critical angle for an air-water interface?
θc
n1 sin 1 = n2 sin 2
n1 = 1.33, c = ??, 2 = 90o, n2 = 1.00
c = sin-1(1.00xsin(90o)/1.33)
48.8o
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What is the Critical angle for an water-diamond
interface? Where does the critical angle occur?
θc
n1 sin 1 = n2 sin 2
n1 = 2.42, c = ??, 2 = 90o, n2 = 1.33
c = sin-1(1.33xsin(90o)/2.42)
33.3o in the diamond
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More than one polarizer:
I = Iocos2
Io – incident intensity of polarized light
I – transmitted intensity (W/m2)
 – angle twixt polarizer and incident angle of polarization
Io
½Io
Demo two polarizers
(½ Io)cos2
Two polarizers are at an angle of 37o with each other.
If there is a 235 W/m2 beam of light incident on the first
filter, what is the intensity between the filters, and after
the second?
I = Iocos2
After the first polarizer, we have half the intensity:
I = 235/2 = 117.5 W/m2
and then that polarized light hits the second filter at
an angle of 37o:
I = (117.5 W/m2) cos2(37o) = 74.94 = 75 W/m2
Page 2
The waveform is 45 cm long. What is the ?
.45 m = 5/4 
 = 4/5(.45 m) = .36 m
.36 m 36 cm
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The waveform is 62 cm long. What is the ?
If it is a sound wave (v = 343 m/s),
what is its frequency (v = f)
.62 = 2/4 
 = 4/2(.62 m) = 1.24 m
v = f, f = v/ = (343 m/s)/(1.24 m) = 277 Hz
1.24 m, 277 Hz
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A string is 32.0 cm long, and has a wave speed of
281.6 m/s. Draw the first three modes of resonance.
Find for each mode 1. The wavelength, 2. The
frequency. Hint v = f
.32 m = 2/4,  = .64 m,
v = f, f = 281.6/.64 = 440 Hz
.32 m = 4/4,  = .32 m,
v = f, f = 281.6/.32 = 880 Hz
.32 m = 6/4,  = .2133 m
v = f, f = 281.6/ .2133 m = 1320 Hz
A pipe with both ends open is 1.715 m long, sound
travels at 343 m/s along the pipe. Draw the first three
modes of resonance. Find for each mode 1. The
wavelength, 2. The frequency. Hint v = f
1.715 m = 2/4,  = 3.43 m,
v = f, f = 343/3.43 = 100. Hz
1.715 m = 4/4,  = 1.715 m,
v = f, f = 343/1.715 = 200. Hz
1.715 m = 6/4,  = 1.14333 m
v = f, f = 343/ 1.14333 = 300. Hz
A pipe with one end closed, one end open is also 1.715 m
long, sound travels at 343 m/s along the pipe. Draw the
first three modes of resonance. Find for each mode 1. The
wavelength, 2. The frequency. Hint v = f
1.715 m = 1/4,  = 6.86 m,
v = f, f = 343/6.86 = 50. Hz
1.715 m = 3/4,  = 2.28666 m,
v = f, f = 343/2.28666 = 150. Hz
1.715 m = 5/4,  = 1.372 m
v = f, f = 343/1.372 = 250. Hz
Page 3
A car with a 256 Hz horn approaches you at 40.0
m/s. What frequency do you hear? (3)
(use v sound = 343 m/s)
Moving source
higher frequency
f’ = f{ v
}
{v + us }
f = 256 hz, us = 40.0 m/s, v = 343 m/s, and -
290. Hz
W
What speed in what direction is the same car (f =
256 Hz) moving if you hear 213 Hz (use v sound =
343 m/s)
Moving source
lower frequency
f’ = f{ 1
}
{1 + vs/v }
f’ = 213 Hz, f = 256 Hz, v = 343 m/s, and +
69 m/s away from you
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A running person who is late for a concert hears
the concertmaster who is playing an A 440. Hz.
How fast and in what direction are they running if
they hear a frequency of 463 Hz.
(use v sound = 343 m/s)
Moving observer
higher frequency
f’ = f{v ± uo}
{ v }
f = 440.0 Hz, f’ = 463 Hz, v = 343 m/s, and +
17.9 m/s
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Question D on this page is a tricky little one about wavelength and
Doppler effect. What you need to know is this:
1.
v = fλ
2.
That the wavelength gets shorter by the same amount in front of a
moving object, that it gets longer in back
e.g. – suppose the wavelength of a car horn is 2.0 m when the car is sitting
still, if it moves so that the wavelength is 1.8 m in front of the car, it
will be 2.2 m long behind the car.
The problem can be solved without knowing this through the use of some
fairly difficult algebra.
What is the velocity of a 1.12 m wave
with a frequency of 32 Hz?
v = f
= (32 Hz)(1.12 m) = 35.84 m/s = 36 m/s
36 m/s
W
What is the frequency of a sound wave
that has a wavelength of 45 cm, where
the speed of sound is 335 m/s
v = f
f = v/ = (335 m/s)/(.45 m) = 744.444 = 740 Hz
740 Hz
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Be able to draw the circles, and know where the approaching
wavelength is, and the receding wavelength is
Page 4
What is the velocity of a 1.12 m wave
with a frequency of 32 Hz?
v = f
= (32 Hz)(1.12 m) = 35.84 m/s = 36 m/s
36 m/s
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
A
B
If the difference in distance from the sources is
an integer number of wavelengths, you get
constructive interference
TOC

A
B
If the difference in distance from the sources is
an integer number of wavelengths, you get
constructive interference
TOC

A
B
If the difference in distance from the sources is
an integer number of wavelengths, you get
constructive interference
Difference is:
0 , 1 , 2 , 3 …
TOC

A
B
If the difference in distance from the sources
has a remainder of a half wavelength, you get
destructive interference:

A
B
If the difference in distance from the sources
has a remainder of a half wavelength, you get
destructive interference:

A
B
If the difference in distance from the sources
has a remainder of a half wavelength, you get
destructive interference:
Difference
.5 , 1.5 , 2.5 , 3.5 …
To figure out two source problems:
1. Calculate the 
2. Find the difference in distance
3. Find out how many  it is
4. Decide:
__.0 = constructive
__.5 = destructive
__.1 = mostly constructive
__.25 = ???
Two speakers 3.0 m apart are making
sound with a wavelength of 48.0 cm.
A. What is the frequency of this sound if
v = 343 m/s?
v = f ,
343 m/s = f (.48 m)
f = 714.5833333
715 Hz
W
Two speakers 3.0 m apart are making
sound with a wavelength of 48.0 cm.
If I am 2.12 m from one speaker, and
3.80 m from the other, is it loud, or quiet,
and how many wavelengths difference in
distance is there?
3.80 m - 2.12 m = 1.68 m
(1.68 m)/(.48 m) = 3.5  = destructive interference
3.5 wavelengths, destructive
W
Two speakers 3.0 m apart are making
sound with a wavelength of 48.0 cm.
If I am 5.17 m from one speaker, and
8.05 m from the other, is it loud, or quiet,
and how many wavelengths difference in
distance is there?
8.05 m - 5.17 m = 2.88 m
(2.88 m)/(.48 m) = 6.0  = constructive interference
6.0 wavelengths, constructive
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≈
b
 = Angular Spread
 = Wavelength
b = Size of opening
b
656 nm light is incident on a single slit with a width of 0.12
mm. What is the approximate width of spread behind the slit?
 = 656E-9 m
b = 0.12E-3 m
 = (656E-9 m)/(0.12E-3 m) = 0.0055 radians or about 0.31o
Try this problem: Sound waves with a frequency of
256 Hz come through a doorway that is 0.92 m wide.
What is the approximate angle of diffraction into the
room? Use 343 m/s as the speed of sound.
Use v = f, so  = 1.340 m
Then use
≈
b
 ≈ 1.5 rad
What if the frequency were lower?
Sub Woofers
Rayleigh Criterion
 = 1.22
b
 = Angle of resolution (Rad)
 = Wavelength (m)
b = Diameter of circular opening (m)
(Telescope aperture)
the bigger the aperture, the smaller the
angle you can resolve.
Central maximum of one is over
minimum of the other
What is the angular resolution of the 100 inch
(2.54 m) diameter telescope on the top of Mt
Wilson? (use 550 nm as the wavelength)
(uh 550 nm = 550 x 10-9 m)
 = 1.22
b
 = ?,  = 550 x 10-9 m, b = 2.54 m
 = 2.64173E-07
2.6 x 10-7 radians
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What diameter telescope do you need to resolve
two stars that are separated by 1.8 x 1011 m, but
are 3.0 × 1017 m from us? (use 550 nm as the
wavelength)
(AU, 32 LY)
hint  = s/r = (1.8 x 1011 m)/(3.0 × 1017 m)
 = 1.22
b
 = 6.00 x 10-7,  = 550 x 10-9 m, b = ?
b = 1.12 m
1.1 m
W