Physics 1161: Lecture 26
Interference
• textbook sections 28-1 -- 28-3
Superposition
Constructive Interference
+1
t
-1
+
+1
In Phase
t
-1
+2
t
-2
Superposition
Destructive Interference
+1
t
-1
+
+1
Out of Phase
t
-1
+2
t
-2
180 degrees
Which type of interference results from the
superposition of the two waveforms shown?
1. Constructive
2. Destructive
3. Neither
1.5
1
0.5
0
-0.5
-1
-1.5
+
1.5
1
0.5
Different f
0
-0.5
-1
-1.5
0%
1
0%
2
0%
3
Which type of interference results from the
superposition of the two waveforms shown?
1. Constructive
2. Destructive
3. Neither
1.5
1
0.5
0
-0.5
-1
-1.5
+
1.5
1
0.5
Different f
0
-0.5
-1
-1.5
2.5
2
1.5
1
0.5
0
-0.5
-1
-1.5
0%
0%
0%
-2
1
2
3
Interference for Light …
• Can’t produce coherent light from separate
sources. (f  1014 Hz)
• Need two waves from single source taking
two different paths
– Two slits
– Reflection (thin films)
– Diffraction*
Two different paths
Interference possible here
Single source
Coherent & Incoherent Light
Double Slit Interference Applets
• http://www.walterfendt.de/ph14e/doubleslit.htm
• http://vsg.quasihome.com/interfer.htm
Young’s Double Slit Applet
http://www.colorado.edu/UCB/AcademicAffairs/ArtsSciences/physics/PhysicsInitiative/
Physics2000/applets/twoslitsa.html
Young’s Double Slit Layout
Interference - Wavelength
Light waves from a single source travel through
2 slits before meeting at the point shown on the
screen. The interference will be:
1. Constructive
2. Destructive
3. It depends on L
2 slits-separated by d
d
Single source of
monochromatic light 
L
Screen a distance L from slits
0%
1
0%
2
0%
3
Light waves from a single source travel through
2 slits before meeting at the point shown on the
screen. The interference will be:
1. Constructive
2. Destructive
3. It depends on L
2 slits-separated by d
d
Single source of
monochromatic light 
The rays start in phase, and
travel the same distance, so they
will arrive in phase.
L
Screen a distance L from slits
0%
1
0%
2
0%
3
Preflight 26.1
The experiment is modified so that one of the waves has its phase
shifted by ½ . Now, the interference will be:
½  shift
1)
Constructive
2)
Destructive
3)
Depends on L
d
Single source of
monochromatic light 
L
2 slits-separated by d
Screen a distance L from slits
Preflight 26.1
The experiment is modified so that one of the waves has its phase
shifted by ½ . Now, the interference will be:
½  shift
1)
Constructive
2)
Destructive
3)
Depends on L
d
Single source of
monochromatic light 
L
The rays start out of phase,
and travel the same distance,
so they will arrive out of
phase.
2 slits-separated by d
Screen a distance L from slits
Young’s Double Slit Concept
At points where the
difference in path
length is 0, ,2, …,
the screen is bright.
(constructive)
d
Single source of
monochromatic light 
L
2 slits-separated by d
At points where the
difference in path
 3 5
length is
, , 
2 2 2
the screen is dark.
(destructive)
Screen a distance L from slits
Young’s Double Slit Key Idea
L
Two rays travel almost exactly the same distance.
be very far away: L >> d)
Bottom ray travels a little further.
Key for interference is this small extra distance.
(screen must
Young’s Double Slit Quantitative
sin(θ)  tan(θ) = y/L
d
d


Path length difference =
Constructive interference
Destructive interference
(Where m = 0, 1, 2, …)
d sin 

Young’s Double Slit Quantitative
d
d



Path length difference =
Constructive interference
Destructive interference
where m = 0, or 1, or 2, ...
d sin 
dsin   m
1
d sin   (m  )
2
Need  < d
Young’s Double Slit Quantitative
L
y
d


A little geometry…
sin()  tan() = y/L
Constructive interference
Destructive interference
where m = 0, or 1, or 2, ...
dsin   m
1
d sin   (m  )
2
mL
y
d
 m  1 L


2
y
d
Preflight 26.3
L
y

d


When this Young’s double slit experiment is placed under
water. The separation y between minima and maxima
1) increases
2) same
3) decreases
Preflight 26.3
L
y

d


When this Young’s double slit experiment is placed under
water. The separation y between minima and maxima
1) increases
2) same
…wavelength is shorter under water.
3) decreases
Preflight 26.2
In the Young’s double slit experiment, is it possible to
see interference maxima when the distance between
slits is smaller than the wavelength of light?
1) Yes
2) No
Preflight 26.2
In the Young double slit experiment, is it possible to see
interference maxima when the distance between slits is
smaller than the wavelength of light?
1) Yes
Need: d sin  = m 
If
>d
2) No
=> sin  = m  / d
then
 /d>1
so sin  > 1
Not possible!
Reflections at Boundaries
Slow Medium
to
Fast Medium
Free End Reflection
No phase change
Fast Medium
to
Slow Medium
Fixed End Reflection
180o phase change
Newton’s Rings
Iridescence
Iridescence
Soap Film Interference
• This soap film varies in
thickness and produces a
rainbow of colors.
• The top part is so thin it
looks black.
• All colors destructively
interfere there.
Thin Film Interference
1
t
2
n0=1.0 (air)
n1 (thin film)
n2
Get two waves by reflection off of two different
interfaces.
Ray 2 travels approximately 2t further than ray 1.
Reflection + Phase Shifts
Incident wave
Reflected wave
n1
n2
Upon reflection from a boundary between two transparent
materials, the phase of the reflected light may change.
• If n1 > n2
• If n1 < n2
Reflection + Phase Shifts
Incident wave
Reflected wave
n1
n2
Upon reflection from a boundary between two transparent
materials, the phase of the reflected light may change.
• If n1 > n2 - no phase change upon reflection.
• If n1 < n2 - phase change of 180º upon reflection.
(equivalent to the wave shifting by /2.)
Thin Film Summary
Determine d, number of extra wavelengths for each ray.
1
2
n = 1.0 (air)
n1 (thin film)
t
This is
important!
n2
Reflection
Distance
Ray 1: d1 = 0 or ½
Ray 2: d2 = 0 or ½
+ 2 t/ film
If |(d2 – d1)| = 0, 1, 2, 3 ….
(m)
If |(d2 – d1)| = ½ , 1 ½, 2 ½ ….
Note: this is
wavelength in film!
(film= o/n1)
constructive
(m + ½) destructive
Thin Film Practice
1
t
2
n = 1.0 (air)
nglass = 1.5
nwater= 1.3
Blue light (o = 500 nm) incident on a glass (nglass = 1.5) cover slip (t = 167
nm) floating on top of water (nwater = 1.3).
Is the interference constructive or destructive or neither?
d1 =
d2 =
Phase shift = d2 – d1 =
Thin Film Practice
1
2
n = 1.0 (air)
nglass = 1.5
t
nwater= 1.3
Blue light (o = 500 nm) incident on a glass (nglass = 1.5) cover slip (t = 167
nm) floating on top of water (nwater = 1.3).
Is the interference constructive or destructive or neither?
d1 = ½
Reflection at air-film interface only
d2 = 0 + 2t / glass = 2t nglass/ 0= 1
Phase shift = d2 – d1 = ½ wavelength
Blue light  = 500 nm incident on a thin film (t =
167 nm) of glass on top of plastic. The
interference is:
1
2
t
n=1 (air)
nglass =1.5
44%
44%
nplastic=1.8
1. Constructive
2. Destructive
3. Neither
11%
1
2
3
Blue light  = 500 nm incident on a thin film (t =
167 nm) of glass on top of plastic. The
interference is:
1. Constructive
1
t
2
n=1 (air)
2. Destructive
3. Neither
nglass =1.5
56%
nplastic=1.8
d1 = ½
d2 = ½ + 2t / glass = ½ + 2t nglass/ 0= ½ + 1
33%
11%
Phase shift = d2 – d1 = 1 wavelength
1
2
3
Preflights 26.4, 26.5
A thin film of gasoline (ngas=1.20)
and a thin film of oil (noil=1.45)
are floating on water
(nwater=1.33). When the thickness t = 
of the two films is exactly one
wavelength…
The gas looks:
• bright 67 %
• dark 33 %
nair=1.0
ngas=1.20
noil=1.45
nwater=1.3
The oil looks:
• bright 35 %
• dark 65 %
Preflights 26.4, 26.5
A thin film of gasoline (ngas=1.20)
and a thin film of oil (noil=1.45)
are floating on water
(nwater=1.33). When the thickness t = 
of the two films is exactly one
wavelength…
The gas looks:
• bright
• dark
d1,gas = ½
d2,gas = ½ + 2
| d2,gas – d1,gas | = 2
constructive
nair=1.0
ngas=1.20
noil=1.45
nwater=1.3
The oil looks:
• bright
• dark
d1,oil = ½
d2,oil = 2
| d2,oil – d1,oil | = 3/2
destructive