Today 1/23 Today: Young’s Double Slit experiment, Text 27.2 Beats, Text 17.4 HW: 1/23 Handout “Young’s Double Slit” due Monday 1/24 Friday: Phase Shifts on Reflection, Text 27.3 (See fig 27.11) Peer Guidance Center 1-4 Mon-Thur Wit 211 (Not 209)
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Today 1/23 Today:
Young’s Double Slit experiment, Text 27.2Beats, Text 17.4
HW: 1/23 Handout “Young’s Double Slit” due Monday 1/24
Friday: Phase Shifts on Reflection, Text 27.3 (See fig 27.11)
Peer Guidance Center 1-4 Mon-Thur Wit 211 (Not 209)
Young’s Double Slit (like two speakers)
Single frequency source
Wave crests
Wave troughs
Dark and bright “fringes” on a screen
In phase at the slits
c
c
c
c
c
d
d
d
d
Always true for any interference problem
Sources In Phase:
Constructive if PLD = mDestructive if PLD = (m + 1/2)
Sources Out of Phase:
Constructive if PLD = (m + 1/2)Destructive if PLD = m
PLD = “path length difference”
m = 0, 1, 2, 3,… (I used “n” the other day)
Two slit geometry (screen far away)
Screen
PLD = d sin (d = slit separation)
d
PLD
d
(close enough)
Two slit geometry
Screen
PDL = d sin (d = slit separation)
d sin = m constructive interference
d sin = (m+ 1/2) destructive interference
When the sources (slits) and “in phase”
d
A simpler picture
Screen very far away (L)
Two slits very close together (d)
d sin = m constructive interference
d sin = (m+ 1/2) destructive interference
When the sources (slits) and “in phase”
The m’s
m = 0
m = 1
m = 2
m = 1
m = 2m = 1
m = 0
m = 0
m = 1d sin = md sin = (m+ 1/2)
0 “zeroth order” fringe1 “first order” fringe2 “second order” fringe
Distance between fringes, y
m = 0
m = 1
m = 2
m = 1
m = 2m = 1
m = 0
m = 0
m = 1
Ly
tan = y/L
Example:
m = 0
m = 1
m = 2
m = 1
m = 2m = 1
m = 0
m = 0
m = 1
Light with a wavelength of 500 nm passes through two closely spaced slits and forms an interference pattern on a screen 2m away. The distance between the central maximum and the first order bright fringe is 5 mm. What is the slit spacing? The light is in phase at the slits.
tan = 5mm / 2m
= 0.14°
d sin = m = 1(500 nm)
d = 0.2 mm
5mm
2m
Example:Twin radio antennas broadcast in phase at a frequency of 93.7MHz. Your antenna is located 150 m from one tower and 158 m from the other. How is the reception, good or bad? vwave=c=3108 m/s
PLD = 8 m Does this equal some m or some (m + 1/2) ?
v = f = 3.2m Make two lists
03.2m6.4m9.6m12.8m
1.6m4.8m8.0m11.2m14.4m
m (m + 1/2)m01234
The condition is met for destructive interference. Reception at that location is bad.
Your Homework
Screen very far away (L)
Two slits very close together (d)
d sin = m constructive interference
d sin = (m+ 1/2) destructive interference
When the sources (slits) and “in phase”
Beats
Occur when the frequencies of the sources are not the same
Frequencies must be close Locations for constructive interference move
over time Causes sound to get loud and soft fb “beat frequency” depends on source
frequency difference
0.5 s
10 Hz
12 Hz
2 Hz
Sources emitting different frequencies.
Source 1 Source 2
In this case they are alternately in and out of phase as time goes by.
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
c
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
c
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
c
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
c
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
c
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
c
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
c
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
c
Source 1 Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes by.
c
Now the locations of constructive (and destructive) interference move in time. A stationary listener hears “Beats.”
Beats
fb = f1 - f2
The beat frequency tells you the difference between the two source frequencies.
You want to know the frequency of a tuning fork. You test it by playing it at the same time as a tuning fork with a known frequency of 342 Hz and you hear beats at a rate of 5 per second. You then play it at the same time as one with a known frequency of 349 Hz and the beats are heard at a rate of 12 per second. What is the frequency of the tuning fork?
a. 347 Hz b. 361 Hz c. 345.5 Hz
d. 337 Hz e. 354 Hz.
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