# Thin-Film Interference and Antireflective coating

1. Jul 21, 2013

### aChordate

1. The problem statement, all variables and given/known data

A pair of eyeglasses has an index of refraction n-glass = 1.52. It is covered with an
antireflection layer that has a thickness of 120 nm and an index of refraction that is less
than n-glass. If the antireflection layer is designed to minimize reflection in the middle of
the visible spectrum (l = 565 nm), what is its index of refraction?

2. Relevant equations

index of refraction
n=c/v
c=3.00*108m/s

thin film interference
n=λvacuumfilm

3. The attempt at a solution

n-glass = 1.52

n-layer=?

λvacuum=565nm

λfilm=120nm ???

ncoating= 565nm/120nm = 4.71

2. Jul 21, 2013

### collinsmark

The above equation might come in quite handy.

Not quite, no. See my comments below.

Before I comment much, perhaps the easiest way to solve this problem is to look in your textbook or coursework for a particular formula that will help you with this.

That said, the ideal thickness of the film for an anti-reflective layer is $\frac{1}{4} \lambda_{\mathrm{film}}$. The reason for the 1/4 wavelength is that the wave that is reflected from the back of the coating layer will then be 1/2 wavelength out of phase with the wave that gets reflected off the front of the coating. (The wave that travels through the coating and gets reflected off the back of the coating, travels a total of $\frac{1}{4} \lambda_{\mathrm{film}} + \frac{1}{4} \lambda_{\mathrm{film}} = \frac{1}{2} \lambda_{\mathrm{film}}$ within the layer, total). In front of the coating (in the air) there will be two reflected waves. Since the two reflected waves are 180o out of phase, they destructively interfere with each other.

With that piece of knowledge, and one of the things listed in your relevant equations, you'll have enough information to derive the appropriate formula yourself.

3. Jul 22, 2013

### aChordate

The only equation in the book that I see is:

n=λvacuum/λfilm=565nm/?

I am not quite sure what to do with:

nglass = 1.52

or

t=120nm

or how to find the wavelength of the film.

4. Jul 22, 2013

### collinsmark

That's a fine equation. You'll end up using it, either directly or indirectly, by the time this problem is done.

In this particular problem, not much. The index of refraction of the glass (on the other side of the coating layer) doesn't fit into the calculations.

It is important at least qualitatively though. It's important to know qualitatively that the index of refraction of the coating layer is less than the index of refraction of the glass. If it were the other way around, the wave reflected from the layer off the glass gets an inversion upon reflection: it would change the equations. Also, given the choice is better to choose an index of refraction of the coating the be the geometrical mean of the glass and the air indexes of refraction, to promote reflections with equal intensity. But don't worry about that for this problem; this problem has more to do with the coating layer's thickness.

The constraint which you are working with is that the layer of coating has a fixed thickness of 120 nm. You cant change that. The thickness and the coating layer's index of refraction determine the target wavelength.

120 nm = (1/4)λfilm

This link might be of some help regarding where the 1/4 factor comes from:

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/antiref.html#c3

I'm surprised though that you are given this problem without covering anti-reflection in your coursework.

5. Jul 23, 2013

### aChordate

We covered it briefly, but I don't remember my professor saying anything about the 1/4 factor.

So, is this correct?

120nm=1/4λfilm

λfilm=30nm

n=λvacuumfilm=565nm/30nm=18.8

6. Jul 23, 2013

### collinsmark

It's 120 nm=(1/4)λfilm.

In other words, λfilm = (4)(120 nm)

7. Jul 23, 2013

### aChordate

Oh my goodness, that was a silly mistake. I was wondering why it was so low. Thanks for your patience!

λfilm=480

n=1.18 !

8. Jul 23, 2013

Yep!