Interference Effects
6.2 Two slit interference
Coherence
Two-Slit Interference
Thin film Interference
Interference is a general property of waves.
A condition for interference is that the wave
source is coherent.
Interference between two waves gives
characteristic interference patterns due to
constructive and destructive interference.
Coherence
Coherence
For two waves to show interference they
must have coherence.
Light from
two separate
light bulbs is
Incoherent
Two waves are coherent if one wave has a constant
phase relation to the other
incoherent
coherent
Light from a single
light bulb passing
through two slits is
coherent
φ = 2π
∆x
λ
phase
shift
Interference
δ=0
Constructive Interference
Interference
λ
δ= 2
Destructive Interference
Sum
λ
1
Interference
path difference
δ=λ Constructive Interference
path difference =δ =r2 – r1
r1
A
Superposition of waves
at A shows interference
due to path differences
r2
In phase
Condition for constructive interference δ
Condition for destructive interference δ
Order number m
= mλ
1
= (m + )λ
2
m = 0 + 1, + 2,….
Interference pattern due to
two spherical wavesAmplitude on screen
Destructive
δ=λ
Constructive
m=1
δ=0
δ=-λ
Interference pattern of water waves
Young’s two slit experiment
Coherent
waves
m=0
m=-1
Interference
pattern
Path difference from the two slits
Does light show wave properties?
In the limit L>>d, the
rays are nearly parallel
Thomas Young
path difference
δ = dsin θ
2
Interference pattern
maxima
m=2
d
θ
m=1
m=0 Central maximum
m= -1
Question
Light from a laser is passed through two slits a distance of
0.10 mm apart and is hits a screen 5 m away. The
separation between the central maximum and the first
bright interference fringe is 2.6 cm. Find the wavelength
lf the light.
m= -2
Bright
constructive interference
Dark
destructive interference
L
dsin θbright = mλ
dsinθ=mλ
dsin θdark = (m + 1/ 2)λ
solve for λ
m = 0, + 1, + 2, ............
Thin film interference
λ=
y
θ
d
sinθ ~ y
L
for small
angles
d
y
= mλ
L
for m= 1
yd (2.6x10 −2 )(0.1x10−3 )
=
= 5.2x10−7 = 520nm
mL
(1)(5)
Phase shift due to reflection
Reflection with inversion
phase shift = 180o
n1 < n2
phase shift=180o
•
•
In thin film interference the phase difference is due to
reflection at either side of a thin film of transparent
material.
The phase difference is due to two factors:
– Path difference through the film (corrected for the
change in speed of light in the material)
– Phase shift due to reflection at the interface
Phase shift due to reflection
n1 > n2
Phase shift = zero
Reflection without inversion
Phase shift = zero
Thin film Interference
For a thin film in air the
phase difference due to
reflection is 180o
Condition for constructive
interference
δ=2t= (m +
1
1 λ
)λ film = (m + )
2
2 n
The wavelength in the film is longer than in air.
3
Thin film Interference
Soap film interference pattern
For film in air the phase
difference due to
reflection is 180o
Condition for destructive
interference
δ=2t=
m λ film = m
λ
n
The wavelength in the film is longer than in air.
Soap film
Question
A vertical soap film displays a series of
colored band due to reflected light. Find
the thickness of the film at the position of
the 5th green band (λ=550 nm, n =1.33)
Constructive Interference
The 5th band has m=4 (the first is m=0)
1 λ
2t = (m + )
2 n
1 λ
t = (m + )
2 2n
Anti-reflective Coating
no coating
Anti-reflective coatings are used
to reduce reflections at the air-glass
interface.
1 550x10 −9
= (4 + )
= 9.3x10−7 m = 930nm
2 2(1.33)
Anti-reflective Coating
n1
n2
n3
Anti-reflective coatings consists of
a thin-layer of material with a
refractive index in between that of
air and glass. Destructive
interference between light reflected
at the two surfaces reduces the
intensity of reflected light.
t
anti-reflective
coating
n1=1.00 < n2 < n3
phase shift of 180o at
both interfaces.
Total phase difference due
to reflection is zero
Condition for destructive interference.
1 λ
2t = ( m + )
2 n2
4
Question
Optical compact disc
An anti-reflective coating of MgF2 (n=1.38) is used
on a glass surface to reduce reflections. Find
the minimum thickness of the coating that can
be used for green light (λ=550 nm).
For destructive interference
1 λ
2t = (m + )
2 n2
minimum
at m=0
Solve for t
t=
λ
4n
=
2t =
1λ
2n
550
= 100nm
4(1.38)
Quarter wavelength (in coating) thickness
Cd store information in a series
of pits and bumps in the plastic.
The information is read by a reflected
laser beam.
The intensity of the beam is changed by
constructive or destructive interference
t=
λ
4n
destructive
interference
5