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Lecture 5: Interference and diffraction of light (II)
Interference in Thin Films (Y&F 35.4)
 Thin films on a substrate undergo the
phenomenon of multiple wave interference.
 Examples are the bright bands of colour from
a thin layer of oil floating on water, the patterns
in soap bubbles, antireflective coatings in
glasses.
Thin film of oil on water
n1
n2
t
d
n3
n1< n2 < n3
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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Geometric description of thin film interference
 Interference between bc and ef:
• At b there is a reflection: change of phase = p (D r = l2 /2) if n1<n2
• At d there is a reflection: change of phase = p (D r = l2 /2) if n2<n3
So constructive interference when:
bc + l2 /2 - (bd + de +ef +l2 /2) = ml2
a
c
n1
f
l2 = wavelength in medium n2
b
e
 If light perpendicular:
bd + de = 2t
bc=ef
n2
t
d
n3
n1< n2 < n3
Therefore:
2t=ml2
for constructive interference m=0,1,2,…
2t=(m+1/2)l2 for destructive interference m=0,1,2,...
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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 Note: When light is reflected off a medium that has higher refractive
index than the initial medium, there is a phase shift of p rads
(equivalent to l/2). If the second medium has a smaller refractive
index, there is no phase shift. If the two refractive indices are equal
there is no reflection.
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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Examples of thin film interference
 Non-reflective coatings:
• We want to create a thin film on a substrate that is non-reflective,
so we want destructive interference.
• If we have a “quarter-wave” plate: t=l2 / 4 then:
Path difference = 2t = l2 /2 (destructive interference)
v1=c/n1 so: fl1 = c/n1
l = c/ (f n2) = l1 (n1/n2)
v2=c/n2 so: fl2 = c/n2  2
If n1 = 1 (air) then:
l2 = l1/n2
therefore: 2t= l2/2 = l1/(2n2)
 The optical path-length between two
interfering waves is:
2t n2 = l1/2
physical path-length
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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 Example 35-8 (Y&F):
Non reflective coatings for lenses are normally designed for 550 nm
(central yellow-green part of the visible spectrum). Magnesium fluoride
(MgF2) with n=1.38 is used as a lens coating material for a glass of
n=1.52. What thickness should the coating be?
The wavelength of light in the coating:
l2 = l1/n2 = 550 nm/1.38 = 400 nm
Air n1=1
Therefore:
t = l2 /4 = 400/4 = 100 nm.
MgF2 n2=1.38
t
d
Glass n3=1.52
n1< n2 < n3
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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 Interference in an air wedge between two glass plates:
• Ray 1: reflected at A (no phase change)
• Ray 2: travels path ABC (phase change of p at B)

Difference is ABC = 2t - l/2 (it q is small).
1
2
 Constructive interference:
2t - l/2 = ml
n1
C
n2
n3
A
q
xm
t
h

 Destructive interference:
2t + l/2 = (m+1/2)l
B
2t = (m+1/2)l

2t =ml
L
 Distance xm from line of contact to mth dark fringe:
Since 2tm =ml and xm = tm(L/h) then:
xm = (ml/2)(L/h)
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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 Newton’s rings: interference between a curved lens (radius R)
and a flat glass plate inside a medium of refractive index n
(e.g. n=1 for air and n= 1.33 for water).
Observation of concentric dark and light fringes with centre in
the centre of the lens.
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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 Phase change of p (equivalent to l/2)
at the bottom surface reflection:
r1-r2= 2t therefore:
Optical path length = n(r1-r2) = 2tn + l/2
• Dark rings (destructive interference):
dark centre ring
2tn + l /2 = (m+1/2)l Therefore:
mlR
2
2nt  ml  xm 
n
•Bright rings (constructive interference):
2tn + l /2 = ml Therefore:
1 

2nt   m  l
2


2
xm   m 

R
n
t
xm
xm2 = R2 - (R-t)2 =
R2 - (R2-2Rt+t2) = 2Rt -t2
When R>>t then xm2 ~ 2Rt
1  lR

2 n
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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