Phys 123
Thin film constr.,destr. interference
A thin film floats between two
liquids. In this problem how
many 180o reflection phase
shifts are there?
(Normal incidence)
A. from top
B. from bottom
C. from both
D. from neither
n=2.2
n=1.5
n=1.8
Phys 123
Thin film constr.,destr. interference
Normal incidence
Thickness for greatest reflection if l = 300 nm?
Principles
n=2.2
Constructive int.
Reflection phase shifts
n=1.5
When I say “l = 300 nm” it
means l in vacuum!
Pathlength phase shifts
  k l 
n 2
lvac
n=1.8
l
Find the smallest thickness for greatest reflection if l = 300
nm (vacuum), at normal incidence.
a)
b)
c)
d)
e)
50nm
100 nm
150 nm
300 nm
400 nm
n=2.2
n=1.5
n=1.8
Why does top of soap film become black in R?
So thin at top that only
reflection phase shift matters
A common cheap anti-reflection coating on glass is to put a
(lower index) layer thickness of
a)
b)
c)
d)
l /4
l /(4n1 )
l /2
l /(2n1 )
n=1
n1 =1.5
n2 =1.8
An anti-reflection coated
window, shown at a 45° and a
0° angle of incidence.
More complicated
Effect of significant incident angle:
To get high reflection, will I need a thicker or thinner film (vs
normal incidence)?
Non-intuitive principle: only the
part of k perpendicular to the
interface causes phase shifts.
Why: Snell’s law already
enforces phase matching in
parallel direction.
Need thicker d at higher angles.
Comprehensive theory will give
us all this.
Double boundary r,t
Double boundary r,t
Key ideas:
• How do we handle the multiple reflections in the thin film?
lumped into one left and one right-going waves
• How do we join the fields across each interface?
Fresnel coefficients
A
• How do we introduce phase shifts due to thickness of thin film?
Study y = 0 case, and use only the phase shifts in the z direction
exp[ik z d ]
A
Where do n’s enter into
the above?
Where do phase shifts
due to reflections enter?
http://www.youtube.com/watch?v=tu3T8f-8Fgw Start at 3:26 .
Soap film vs time
above: http://media.efluids.com/galleries/interfacial?medium=518
video of turbulence