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)
Young’s Double Slit (like two
speakers)
Wave
crests
c
d
c
d
c
d
c
d
c
Wave
troughs
Single
frequency
source

In phase at
the slits
Dark and
bright
“fringes” on
a screen
Always true for any interference
problem
Sources
In Phase:
Constructive if PLD = m
Destructive if PLD = (m + 1/2)
PLD = “path length difference”
Sources Out Constructive if PLD = (m + 1/2)
Destructive if PLD = m
of Phase:
m = 0, 1, 2, 3,… (I used “n” the other day)
Two slit geometry (screen far away)
PLD = d sin (d = slit separation)
d

d
PLD
(close

enough)

Screen
Two slit geometry
PDL = d sin (d = slit separation)
d

Screen
d sin = m constructive interference
d sin = (m+ 1/2) destructive interference
When the sources (slits) and “in phase”
A simpler picture
Two slits very close together (d)
Screen
very

far
away
d sin = m constructive interference (L)
d sin = (m+ 1/2) destructive interference
When the sources (slits) and “in phase”
The m’s
0 “zeroth order” fringe
1 “first order” fringe
2 “second order” fringe
d sin = m
d sin = (m+ 1/2)
m=2
m=1
m=1
m=0
m=0
m=0
m=1
m=1
m=2
Distance between fringes, y
tan  = y/L

L
m=2
m=1
m=1
y m=0
m=0
m=0
m=1
m=1
m=2
Example:
m=2
m=1
m=1
m=0
5mm

m=0
2m
m=0
m=1
m=1
Light with a wavelength of 500 nm passes
m=2
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
Example:
Twin radio antennas broadcast in phase at a frequency of
93.7 MHz. Your antenna is located 150 m from one tower and
158 m from the other. How is the reception, good or bad?
vwave = c = 3108 m/s
PLD = 8 m
v=f
m
0
1
2
3
4
Does this equal some m or some (m + 1/2) ?
 = 3.2m
m
0
3.2m
6.4m
9.6m
12.8m
Make two lists
(m + 1/2)
1.6m
4.8m
8.0m
11.2m
14.4m
The condition is met for
destructive interference.
Reception at that
location is bad.
Your Homework
Two slits very close together (d)
Screen
very

far
away
d sin = m constructive interference (L)
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
10 Hz
12 Hz
0.5 s
2 Hz
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.
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
c
Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes
by.
Source 1
c
Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes
by.
Source 1
c
Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes
by.
Source 1
c
Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes
by.
Source 1
c
Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes
by.
Source 1
c
Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes
by.
Source 1
c
Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes
by.
Source 1
c
Source 2
Sources emitting different frequencies.
In this case they are alternately in and out of phase as time goes
by.
Source 1
c
Now the locations of constructive (and destructive)
interference move in time. A stationary listener
hears “Beats.”
Source 2
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.