# Thin film interference and anti-reflective coating

1. Oct 5, 2014

### mayer

Hi,
I actually have two questions that are somewhat related so I am including them in one post. I add much more detail about the system then probably necessary because I'd like to see if I have a correct understanding of how the system works in addition to the question itself. Please bear with me haha. The actual question I underlined. Thank You very much.

1) Say we have a three layer system consisting of air,water, and plastic, in order. Air has the least index of refraction and plastic has the most. If you shine the laser on the water/air interface, you get two waves, one reflected(by 180 degrees) and the other refracted. The refracted ray will go down and hit the water/plastic interface and produce two waves as well, one reflected(by 180 degrees) and one refracted. Now this system can produce either constructive or destructive interference depending on the thickness of the water layer, correct?

2) My second question refers to anti-reflective coating on eyeglasses. So after a light wave is incident upon the air(n=1)/coating(n=1.2) interface, there is a reflection(by 180 degrees) and a refraction. The refracted portion of the wave goes down to the coating/glass(n=1.5) interface and reflects(by 180 degrees) and refracts again. This new reflected wave goes back to the air/coating interface and, if the thickness of the coat is "some integer" + 1/4 lambda( so that the path length the wave travels forwards and back is "some integer" + 1/2 lambda), the refracted wave will cancel out the first reflected wave. I read that this causes more light to penetrate through the glasses and less to be reflected. I understand how less would be reflected, but why would there be more light penetrating in? How is that portion of light that was deconstructively eliminated seemingly being reallocated to the light that penetrates through?

2. Oct 5, 2014

### Simon Bridge

1. correct.
2. you just calculated that no light gets reflected - if not all the light gets transmitted, then where does it go?

3. Oct 5, 2014

### mayer

Thanks for the reply! Hmm I think the rest of my trouble with #2 could be condensed to this "is the destructive interference of the reflected waves the equivalent of saying that no light was reflected?
Edit:
Wait, that was a bit redundant haha, i still cant completely wrap my mind around how the cancellation of the relected wave results in more waves going through. In other words, in answer to your question, in my mind, it hasn't suddenly just disappeared but, rather, subtracted.

Last edited: Oct 5, 2014
4. Oct 5, 2014

### Simon Bridge

It's because the theory you just learned is incomplete.
Treat it as a way of doing a calculation rather than following a physical process.
Later you will learn about transmission and reflection coefficients.

5. Oct 5, 2014

### mayer

Ahh icic, All questions answered. Thanks!