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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 10 8 m/s). v = f f = 1/T = 720. x10 -9 m , v = c = 3.00 x 10 8 m/s f = 4.17E+14 Hz T = 1/f = 2.4E-15 s. W. 4.17E+14 Hz, 2.4E-15 s.

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slide3

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

W

4.17E+14 Hz, 2.4E-15 s

slide4

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

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

W

2.25 x 108 m/s

slide5

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

n = c/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

W

541 nm

slide6

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

slide7

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

θc

n1sin 1 = n2sin 2

n1 = 1.33, c = ??, 2 = 90o, n2 = 1.00

c = sin-1(1.00xsin(90o)/1.33)

W

48.8o

slide8

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

θc

n1sin 1 = n2sin 2

n1 = 2.42, c = ??, 2 = 90o, n2 = 1.33

c = sin-1(1.33xsin(90o)/2.42)

W

33.3o in the diamond

slide9

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

(½Io)cos2

½Io

Io

Demo two polarizers

slide10

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

slide12

The waveform is 45 cm long. What is the ?

.45 m = 5/4

 = 4/5(.45 m) = .36 m

W

.36 m 36 cm

slide13

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

W

1.24 m, 277 Hz

slide14

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

slide15

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

slide16

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

slide18

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 -

W

290. Hz

slide19

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 +

W

69 m/s away from you

slide20

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 +

W

17.9 m/s

slide21
Question D on this page is a tricky little one about wavelength and Doppler effect. What you need to know is this:
  • v = fλ
  • 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.

slide22

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

W

36 m/s

slide23

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

W

740 Hz

slide24

Be able to draw the circles, and know where the approaching wavelength is, and the receding wavelength is

slide26

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

W

36 m/s

slide27

A

B

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

TOC

slide28

A

B

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

TOC

slide29

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

slide30

A

B

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

slide31

A

B

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

slide32

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 …

slide33

To figure out two source problems:

  • Calculate the 
  • Find the difference in distance
  • Find out how many  it is
  • Decide:
    • __.0 = constructive
    • __.5 = destructive
    • __.1 = mostly constructive
    • __.25 = ???
slide34

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

W

715 Hz

slide35

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

W

3.5 wavelengths, destructive

slide36

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

W

6.0 wavelengths, constructive

slide37

 ≈

  • 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

slide38

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

slide39

Central maximum of one is over minimum of the other

  • 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.
slide40

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

W

2.6 x 10-7 radians

slide41

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

W

1.1 m