PES 2130 Fall 2014, Spendier
Lecture 23/Page 1
Lecture today: Chapter 35 Interference
1) Intensity in Double-Slit Interference
2) Thin Film Interference
Announcements:
- Shortened office hours this Thursday (10-10:30am). Please e-mail me to find
another time in case you planned on stopping by my office between 10:3011am
Last lecture: started diffraction & interference of light
Waves can interfere: that is, combine together to form complex wave patterns
Interference is best demonstrated using coherent waves.
Coherence means that waves have a phase relationship that is maintained for many
cycles
- Laser light is coherent
- light from slits S1 and S2 in the double slit experiment is said to be completely coherent
- direct sunlight/light bulb is partially coherent
- Young’s two slit Interference Experiment: example of the interference of light waves
Constructive interference, bright fringes (two slits)
d sinθ = mλ (m = 0, ±1, ±2, ±3,…)
Destructive interference, dark fringes (two slits)
d sinθ = (m+1/2) λ (m = 0, ±1, ±2, ±3,…)
PES 2130 Fall 2014, Spendier
Lecture 23/Page 2
1) Intensity in Double-Slit Interference
Goal is to find the intensity at any point in the fringe pattern of double slit experiment
A) Superposition of two waves: combine the two sinusoidally varying fields from two
sources at a point P, taking proper account of the phase difference ϕ of the two waves at
point P
Etotal(x,t) = E1(x,t) + E2(x,t) = E0 cos(kx-ωt) + E0cos(kx-ωt+ ϕ)
Etotal (x,t) = 2 E0cos (ϕ/2)cos(kx-ωt+ ϕ/2)
B) The intensity is then proportional to the square of the resultant electric-field
amplitude
E 2 max
I
2 0 c
amplitude: Emax =| 2 E0cos (ϕ/2)|
Intensity of interfering waves as a function of phase constant ϕ
E 2 max  2E0 cos ( /2) 
E02
I

4
cos2 ( /2) = 4 I 0 cos2 ( /2)
20 c
20 c
20 c
2
I  4 I 0 cos2  / 2 
E 20
Here we used Intensity for original wave: I 0 
2 0 c
Now recall that recall that a phase difference of 2π rad corresponds to one wavelength
 L

2

L
2 d

sin 
 2  


Fully constructive interference (bright fringes) occurs when  / 2  m , for m = 0,1,2,....
Fully destructive interference (dark fringes) occurs when  / 2  ( m  1/ 2) 
Therefore at a bright fringe
I  4 I 0 cos2  / 2   4 I 0 cos2  m 
since m is an integer
I bright  4 I 0 at   0 , 2 , 4 ,....
and dark fringe:
I  4 I 0 cos2  / 2   4 I 0 cos2  m  1/ 2  
I dark  0
at   1 , 3 , 5 ,......
PES 2130 Fall 2014, Spendier
Lecture 23/Page 3
Since we used any integer in this proof, it does not matter which intensity peak we
choose, the maximum intensity is the same for each bright fringe.
This is true only if d << λ (We will see in chapter 36 that is d << λ, then the intensity
profile of light hitting the screen will have an envelope and fall off.
Also, in reality, the waves get less intense as they propagate. So the further away you are,
the less intense the pattern. On the screen, the intensity maxima wil be dimmer the wider
out you go.
2) Thin Film Interference
The colors we see when sunlight illuminates a soap bubble or an oil slick are caused by
the interference of light waves reflected from the front and back surfaces of a thin
transparent film.
To understand, why soap bubbles show vibrant color patterns, even though soapy water is
colorless, we need to reconsider the change of phase of an EM wave due to reflection.
PES 2130 Fall 2014, Spendier
Lecture 23/Page 4
Thin and Thick Films:
We emphasized thin films, since for two waves to cause a steady interference pattern, the
waves must be coherent!
Sun light and a light bulb emits light in a stream of short bursts, each of which is only a
few micrometers long.
- if light reflects from two surfaces of a thin film, the two reflected waves are part of the
same burst and they are coherent ==> steady interference pattern
- if the film is two thick, the two reflected waves will belong to different bursts and they
are not coherent ==> no steady interference pattern
Phase shifts during reflection:
Depending on the differences in index of refraction, an EM wave can undergo or not
undergo a phase change upon reflection:
Recall wave on string:
a)Free boundary
c)fixed boundary
Hence: An electromagnetic wave undergoes a phase change of 180o upon reflection
from a medium that has a higher index of refraction than the one in which the wave
is traveling.
PES 2130 Fall 2014, Spendier
Lecture 23/Page 5
For simplicity, we assume that the incident light ray is almost perpendicular to the film
(θ ≈ 0)
Case 1: Reflection with no phase shift:
Consider a thin film in a medium that has a larger
index of refraction than the film. There are reflections
at surfaces A and B
Now recall that a phase difference of 2π rad
corresponds to one wavelength
 L

2 b
λb: Light wavelength in the film. The path difference
between waves 1 and 2 is 2L for normal incidence.
Fully constructive interference occurs when
 / 2  m , for m = 0,1,2,....
∆∅ 2𝐿
=
𝜋 = 𝑚𝜋
2
𝜆𝑏
or
2𝐿 = 𝑚𝜆𝑏
Fully destructive interference occurs when  / 2  ( m  1/ 2) 
 2 L

   m  1/ 2  
2
b
or
2 L   m  1/ 2  b
NOTE: there relations also hold both waves have a π phase shift.
Case 2: Reflection with phase shift:
Consider a thin film in a air.
Since nair < nfilm, phase shift of π, or halfwavelength occurs upon reflection.
Hence in this case:
Fully constructive interference occurs when
2 L   m  1/ 2   film
Fully destructive interference occurs when
2 L  m film
since λfilm = λair/n
PES 2130 Fall 2014, Spendier
Lecture 23/Page 6
Fully constructive interference (maxima - bright film in air)

2 L   m  1/ 2  air
n film
Fully destructive interference (minima - dark film in air)

2 L  m air
n film
SUMMARY:
Normal incidence
2L =mλfilm
2L = (m+1/2) λfilm
m=0,1,2,3...
No phase shift or both have
π-shift
Constructive reflection
Destructive reflection
One of the two wave has πshift
Destructive reflection
Constructive reflection
PES 2130 Fall 2014, Spendier
Lecture 23/Page 7
Example: White light in air shines on an
oil film (n = 1.50) that floats on water (n
= 1.33). When looking straight down at
the film, the reflected light is red, with a
wavelength of 636 nm. What is the
minimum possible thickness of the film?
If 106 nm gives constructive interference for red light, what about the other colors? They
are not completely cancelled out, because 106 nm is not the right thickness to give
completely destructive interference for any wavelength in the visible spectrum. The other
colors do not reflect as intensely as red light, so the film looks red.