Review For Waves Test

Review For Waves Test

Page 1

v = ff = 1/T

= 720. x10-9 m, v = c = 3.00 x 108 m/s

f = 4.17E+14 HzT = 1/f = 2.4E-15 s

4.17E+14 Hz, 2.4E-15 sW

What is the frequency of 720. nm (1 nm = 1x10-9 m) light? What is its period(Speed = 3.00 x 108 m/s)

n = c/v

n = 1.33, c = 3.00 x 108 m/s

2.25 x 108 m/sW

What is the speed of light in water? n = 1.33

fair = f water

n = c/vv = ffind f, then v, then f = v/ = 4.16667E+14 Hzvwater = c/n = 225563909.8 m/s = v/f = (225563909.8 m/s)/(4.16667E+14 Hz) = 541. nmor 720/1.33 = 541 nm

541 nmW

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)

Draw the red rays, be able to label the angle of incidence (θ1 in this picture) and the refracted angle (θ2)

n1 sin 1 = n2 sin 2

n1 = 1.33, c = ??, 2 = 90o, n2 = 1.00c = sin-1(1.00xsin(90o)/1.33)

48.8oW

What is the Critical angle for an air-water interface?

θc

n1 sin 1 = n2 sin 2

n1 = 2.42, c = ??, 2 = 90o, n2 = 1.33c = sin-1(1.33xsin(90o)/2.42)

33.3o in the diamondW

What is the Critical angle for an water-diamond interface? Where does the critical angle occur?

θc

More than one polarizer:I = Iocos2

Io – incident intensity of polarized lightI – transmitted intensity (W/m2) – angle twixt polarizer and incident angle of polarization

Demo two polarizers

Io½Io (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 = Iocos2After 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

.45 m = 5/4 = 4/5(.45 m) = .36 m

.36 m 36 cmW

The waveform is 45 cm long. What is the ?

.62 = 2/4 = 4/2(.62 m) = 1.24 mv = f, f = v/ = (343 m/s)/(1.24 m) = 277 Hz1.24 m, 277 Hz

W

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)

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 mv = 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 mv = 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 mv = f, f = 343/1.372 = 250. Hz

Page 3

Moving sourcehigher frequencyf’ = f{ v }

{v + us }

f = 256 hz, us = 40.0 m/s, v = 343 m/s, and -

290. HzW

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 sourcelower frequencyf’ = f{ 1 }

{1 + vs/v }

f’ = 213 Hz, f = 256 Hz, v = 343 m/s, and +

69 m/s away from youW

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 observerhigher frequencyf’ = f{v ± uo}

{ v }

f = 440.0 Hz, f’ = 463 Hz, v = 343 m/s, and +17.9 m/s

W

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)

Question D on this page is a tricky little one about wavelength and Doppler effect. What you need to know is this:

1. v = fλ2. That the wavelength gets shorter by the same amount in front of a

moving object, that it gets longer in backe.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/sW

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 HzW

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/sW

TOC

If the difference in distance from the sources is an integer number of wavelengths, you get constructive interference

A

B

TOC

If the difference in distance from the sources is an integer number of wavelengths, you get constructive interference

A

B

TOC

If the difference in distance from the sources is an integer number of wavelengths, you get constructive interference

Difference is:0 , 1 , 2 , 3 …

A

B

If the difference in distance from the sources has a remainder of a half wavelength, you get destructive interference:

A

B

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 distance3. Find out how many it is4. 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 HzW

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, destructiveW

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, constructiveW

≈ b

= Angular Spread = Wavelengthb = 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 mb = 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 mThen 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

= 1.22 b

= ?, = 550 x 10-9 m, b = 2.54 m = 2.64173E-07

2.6 x 10-7 radians W

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

= 6.00 x 10-7, = 550 x 10-9 m, b = ?b = 1.12 m

1.1 mW

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)