Interference of Waves Beats Double Slit

1
Interference of Waves BeatsDouble Slit
2
Beats
Two waves of different frequencies arriving
together produce a fluctuation in power or
amplitude. Since the frequencies are different,
the two vibrations drift in and out of phase with
each other, causing the total amplitude to vary
with time.
y
time
1 beat
3
time
in phase 180o out of
phase in phase
t
4
The math
Same amplitudes, different frequencies
Trigonometry cos a cos b 2 cos
(a-b)/2 cos (ab)/2
Result
slowly-varying amplitude
SHM at average frequency
5
Note
2 beats per cycle of
beats/second
The beat frequency (number of beats per second)
is equal to the difference between the
frequencies
6
Quiz

• Two guitar strings originally vibrate at the same
  400-Hz frequency. If you hear a beat of 5Hz,
  what is/are the other possible frequencie(s) ?
• 10 Hz
• 395 Hz
• 405 Hz
• 395 Hz and 405 Hz

7
Interference
2 waves, of the same frequency arrive out of
phase. Eg y1Asin wt
y2Asin (wtf)
Then yR y1 y2 AR
sin(wtfR), and the resultant amplitude is
AR2Acos(½f).
Identical waves which travel different distances
will arrive out of phase and will interfere, so
that the resultant amplitude varies with
location.
8
Phase difference
Define
Then, at detector
(pick starting time so initial phase is zero here)
9
Example Two sources, in phase waves arrive by
different paths
P
S1
r1
detector
r2
At detector P
S2
10
8 m
detector
x

• 2 speakers, in phase f 170 Hz (so l 2.0 m
  the speed of sound is about 340 m/s)
• As you move along the x axis, where is the
  sound
• a minimum (compared to nearby points)?
• a maximum (compared to nearby points)?

11

• Solution

12
10 min rest
13
Interference of Light
Light is an electromagnetic (EM) wave. Wave
properties Diffraction bends around corners,
spreads out from narrow
slits Interference waves from two or more
coherent sources interfere
14
Electromagnetic Waves
B (magnetic field)
E
Eo
v
Usually we keep track of the electric field E
Electric field amplitude
Electromagnetic waves are transverse waves
15
Infrared Red 780 nm Yellow 600
nm Green 550 nm Blue 450 nm Violet
380 nm Ultraviolet
l
Visible-Light Spectrum
16
The Electromagnetic Spectrum
l (m) f (Hz) 300 106
3 108 3 x 10-3 1011 3 x 10-6
1014 7 x 10-7 5x1014 4 x 10-7 3 x 10-9
1017 3 x 10-12 1020
Radio TV Microwave Infrared Visible Ultraviolet
X rays g rays
17
Our galaxy (Milky Way) at viewed at different
wavelengths
18
Radio lobes (jets) from a supermassive black hole
at the center of the galaxy NGC 4261
19
Double Slit (Thomas Young, 1801)
m2 m1 m0 (center) m-1 m-2
?
incident light
double slit separation d
screen
Result Many bright fringes on screen, with
dark lines in between.
20
The slits act as two sources in phase. Due to
diffraction, the light spreads out after it
passes through each slit. When the two waves
arrive at some point P on the screen, they can be
in or out of phase, depending on the difference
in the length of the paths. The path difference
varies from place to place on the screen.
P
r1
To determine the locations of the bright fringes
(interference maxima), we need to find the points
for which the path difference Dr is equal to an
integer number of wavelengths.For dark fringes
(minima), the path difference is integer
multiples of half of a wavelength.
r2
d
?r r1-r2
21
For light, the slits will usually be very close
together compared to the distance to the screen.
So we will place the screen at infinity to
simplify the calculation.
move P to infinity
r1
r gtgt d, r1 r2 nearly parallel
P
r2
?
d
?r r1-r2
?
d
?r
?r d sin ?
22
Interference 2 coherent waves, out of phase due
to a path difference Dr
Constructive Interference (maximum intensity)
for f 0, 2p, 4p, 6p, -gt ?r
0, l, 2 l, 3 l,
Destructive Interference (minimum
intensity) for f p, 3p, 5p, -gt ?r
?/2, 3?/2, 5?/2,
23
Constructive Interference (bright) ?r m?
or d sin ? m?, m 0, 1, 2, But, if the
slit-screen distance (L) is large, then
sin?? and so sin??y/L (in radians)
d
So we have
24
Destructive Interference (no light) ?r (m
½)? or d sin ? (m ½) ?, m 0, 1, 2,
So, we have
25
Quiz
Two slits are illuminated with red light to
produce an interference patter on a distant
screen. If the red light is replaces with blue
light, how does the pattern change?

1. The bright spots move closer together
2. The bright spots move farther apart
3. The pattern does not change
4. The patter doesnt chance, but the width of the
   spots changes

26
Example
2 slits, 0.20 mm apart red light (l 667 nm)

y
0
3 m
screen
Where are a) the bright fringes? b) the dark
lines? (give values of y)
27

• Solution

28
Example

• A double slit interference patter is observed on
  a screen 1.0m behind two slits spaced 0.3mm
  apart. Ten bright fringes span a distance of
  1.65 cm.
• What is the wavelength of light used ?

29
Quiz
Which of the following would cause the separation
between the fringes to decrease?

1. Increasing the wavelength
2. Decreasing the wavelength
3. Moving the slits closer together
4. Moving the slits farther apart
5. None of the above

30
10 min rest
31
Refractive Index
The speed of light depends on the material. We
define the refractive index n as
n (speed of light in vacuum)/(speed of light in
a material)
material refractive index speed of
light vacuum 1 c ? 300,000 km/s air 1.0003
glass about 1.5 200,000 km/s water
1.333 225,000 km/s diamond 2.4
125,000 km/s
32
QuestionA beam of yellow light (wavelength 600
nm), travelling in air, passes into a pool of
water. By what factor do the following
quantities change as the beam goes from air into
water?

1. speed
2. frequency
3. wavelength

33
Reflection and Phase Change
Light waves may have a 180 phase change when
they reflect from a boundary
optically dense medium (larger refractive index)
no phase change at this reflection
180 phase change when reflecting from a denser
medium
Just remember this low to high, phase shift of
pi !
34
Example Thin film
What is the minimum thickness of a soap film (n
? 1.33) needed to produce constructive
interference for light with a 500nm wavelength ?
(air n ? 1.00).
What about destructive interference ?
35
Example Antireflection coatings
To reduce reflections from glass lenses (n ?
1.5), the glass surfaces are coated with a thin
layer of magnesium fluoride (n ? 1.38). What is
the correct thickness of the coating for green
light (550 nm vacuum wavelength)?
MgF2
glass
air
36
Example
A beam of 580 nm light passes through two closely
spaced glass plates (nglass1.6), as shown in the
figure below. For what minimum nonzero value of
the plate separation d is the transmitted light
dark?
37
Quiz
Why do we see many colours on a soap bubble?
A) because white light is made up of
different wavelengthsB) because the bubble has
different thicknessC) both A and BD)
because the bubble is round and light
reflects from the other side