Thin-Film Interference and Antireflective coating

[aChordate]

Homework Statement

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?

Homework Equations

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

thin film interference
n=λ_vacuum/λ_film

The Attempt at a Solution

n-glass = 1.52

n-layer=?

λ_vacuum=565nm

λ_film=120nm ?

n_coating= 565nm/120nm = 4.71

 

[collinsmark]

Homework Statement

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?

Homework Equations

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

thin film interference
n=λ_vacuum/λ_film

The above equation might come in quite handy.

The Attempt at a Solution

n-glass = 1.52

n-layer=?

λ_vacuum=565nm

λ_film=120nm ?

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 180^o 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.

 

[aChordate]

I am really confused about this question. I've read the chapter and what you wrote.

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.

 

[collinsmark]

I am really confused about this question. I've read the chapter and what you wrote.

The only equation in the book that I see is:

n=λvacuum/λfilm=565nm/?

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

I am not quite sure what to do with:

nglass = 1.52

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 can't change that. The thickness and the coating layer's index of refraction determine the target wavelength.

or

t=120nm

or how to find the wavelength of the film.

120 nm = (1/4)λ_film

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.

 

[aChordate]

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

So, is this correct?

120nm=1/4λ_film

λ_film=30nm

n=λ_vacuum/λ_film=565nm/30nm=18.8

 

[collinsmark]

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

So, is this correct?

120nm=1/4λ_film

λ_film=30nm

It's 120 nm=(1/4)λ_film.

In other words, λ_film = (4)(120 nm)

 

[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 !

 

[collinsmark]

λfilm=480

n=1.18 !

Yep!