People often use interference between the reflected light from front and back surface to measure the thickness of a film. However, there is this a very important formula:

d=nu/(2*n)*), where nu is the frequency difference (difference of 1/lambda ), n is the refractive index, and d is the film thickness.

The caveat is that, we should watch out for the absorption (or the extinction coefficient k).

The formula comes from interference between reflection of from and back surface of a material. However, it does not consider the phase shift of the reflectance, and besides, it does not consider the phaseshift of the transmission of the back-surface reflection.

The problem only goes away when k is very small.

Because the phaseshift dependce on

tan(theta)=epsilon2/epsilon1=2nk/(n^2-k^2),

where theta is the phase shift.

If k<<n, theta~0. Then there is no problem, fortunately, it is the case most of the time.

Interesting color

If the material is has a single band (band gap Eg), the color depends on that only.

Eg	Tranmission/ color	reflection/ color	color of powder
>3.2 eV	large transmission / no color	small reflection / no color	white!
1.6eV< <3.2 eV	depends on thickness	depends on thickness	change color with powder size
<1.6 eV	none	metallic lusters may have different color (e.g. copper: red)	never change color with size

The color is very interesting and confusing. Actually, all the color we see is the color of transmittance color!!

Some may say, there are a lot of non-transparent colorful stuff. That's illusion of color. Let's use paint as an example. The paint contains many small transparent small particles (which is also called pigments). It is the reflection between interfaces makes the light not able to go all the way through, and in turn make the paint not transparent. The real color we normally see is the double transmitted light from the back surface.

Relation between optical functions

We measure reflection R in the expreiment.

Using R we can calculation the phase shift:

θ(ω) = ω/π ∫ log(√(R/R₀))/(ω₀²-ω²)dω₀

Given R and θ as functions of ω, one can calculate all the optical functions.

1) n and k (refractive index and extinction factor):

because R=((n-1)²+k²)/((n+1)²+k²)

k²=4n/(1-R)-(n+1)²

therefore:

n=(1-R)/(1+R+2*√(R)cos(θ))

hence:

k=√(R(n+1)²-(n-1)²)/√(1-R)

2) other functions:

ε₁=n²-k²;

ε₂=2nk

σ₁=ε₂*ω*ε₀

α=k*ω/c, where c is the speed of light