# Thin film interference

1. Apr 6, 2005

### stunner5000pt

A physicist wishes to deposit a thin film on both sides of a glass (with index of refraction 1.50) window to reduce reflected light of wavelength 532nm to 0%

In order to achieve 100% destructive interference for light reflectedo ff of the air /thing film inteference and the light reflected off the thin film/glass interface what must hte index of refraction of the thin film be??

so its is purely destructive
$$2d + \frac{1}{2} \lambda = (m+ \frac{1}{2}) \frac{\lambda}{n}$$
what is the value of m though?? So would i assume m=0?? But then what is the value of the thickness?? Sincei want to find the index of refraction ,n.

Perhaps the subsequent sub questions have somethin to do with this answer?

Sketch a graph of the reflected intensity for lambda = 532nm as a function of thickness T 0<T<2000 nm
I could do this if i knew the index of refraction...

What thickness of a film will give this 100% AR effect at 532nm?
What fraction of light at twice the laser wavelength lambda = 1264nm will be reflected from this AR coated window?

P.S. Dont ask me what AR is, i dont know either!

Last edited: Apr 6, 2005
2. Apr 6, 2005

### stunner5000pt

can anyone help??

3. Apr 6, 2005

### whozum

Just wondering what level physics is this? course title i mean

4. Apr 6, 2005

### Staff: Mentor

AR = Anti-Reflection

There are several things to consider:

(1) Interference: You want the light reflected off the first and second surfaces to destructively interfere. This happens when the thickness (d) meets this criteria: $2dn/\lambda = k + 1/2$, where k = 0, 1, 2... The thinnest coating meeting this criteria will be when k = 0. Of course, you need to know what n is for the material of the coating. (In this case you know it will be between 1 and 1.5.)

(2) Reflectance: To have perfect anti-reflection you must choose the index of refraction of the coating so that the intensity of both reflections is equal. Thus you want the reflectances at each surface to be equal. (Reflectance depends on the indices of refraction of the surfaces involved: look it up!)

Last edited: Apr 7, 2005
5. Apr 6, 2005

### stunner5000pt

This is second year level course named Optics and Spectra
It is supposed to be part two of the EM course incorporating its applications to optics (such as maxwell's equations)

6. Apr 6, 2005

### stunner5000pt

I cna certainly find the reflction coefficient for this surface using $$r = \frac{n_{2}-n_{1}}{n_{1}+n_{2}}$$ does that have something to do with this?
I do not know the significance of the reflection coefficinet does it give the fracton of intensity reflected??
in $2dn/\lambda = k$ we known lambda and we know k we dont know n and d but we cant find n as a function of d. So any thickness would give a reasonable value for n?? I dont think guessing is what im supposed ot be doing here!

Last edited: Apr 6, 2005
7. Apr 7, 2005

### Staff: Mentor

Yes!
For normal incidence, the intensity of the reflected light will be $I_0 r^2$. This will allow you to find the optimal index of refraction for your coating material.
First off, I made an error in that formula (I left out the 1/2). So it should be: $2dn/\lambda = k + 1/2$. You will choose k = 0 to get the thinnest coating.
You'll figure out the optimal n for the coating material by matching the intensity of the reflections.
Right. No guessing!