# Minimizing Reflected Light Wavelengths

1. Aug 29, 2013

### InferiorMindX

Hey there, I'm new here mainly because I realized how deficient my ability to think in Physics/Math based scenarios really is, plus if I'm going to stay in Premed I need to pass a Phys 101 course in 1st year.

So, my question is this..

The windows in an office tower are coated with a film to minimize reflected light of wavelength 550 nm. If the glass has an index of refraction of 1.52 and the film coating has an index of refraction of 1.25, how thick should the film be applied?

If someone could help me through and confirm my thinking on this it would be greatly appreciated.

Distance traveled by the ray after reflection = 550 / 2 = 225 nm

Distance traveled in film = 2x

λ (μ(air) / μ(film)) = λ (1 / 1.25) = 550 / 1.25

Total Distance = 2x + (550 / 1.25) / 2 = 2x + 110

Now to go back in the reflection.. (2x + 110)(μ(film) / μ(air)) = 2x(1.25) + 225 nm

These should interfere destructively.
(2) - (1) = 2x(1.25) = λ/2
x = (λ / (4 x 1.25) ) = 550 / 5 = 110 nm

I just don't know if that is how thick the film should be applied on the glass, I am lost.

2. Aug 29, 2013

### rodriguez1gv

Where is your question? It looks like you have found the thickness of your film (110nm) you need to have destructive interference at the surface of the film.

3. Aug 29, 2013

### InferiorMindX

Here it is from my original post.

The only thing that confuses me is that I never use the glasses' index of refraction which is 1.52.

Did I really do everything necessary without it?

4. Aug 29, 2013

### voko

Your basic approach is correct. But you made it more complex that it needs to be and so made some errors.

What you need is destructive interference. That means that the ray reflected at the film/glass interface, when it goes all the way back to the outer edge of the film, must be opposite in phase with the incident ray. What optical distance must the ray cover to be opposite in phase?

5. Aug 30, 2013

### Staff: Mentor

Knowing the index of refraction of the glass, and how it compares to that of the film, allows you to determine the phase shift upon reflection. Here you got lucky, as the first surface (air to film) and second surface (film to glass) have the same phase shift.

6. Aug 31, 2013

### InferiorMindX

Alright, points have been taken into account but my approach on the problem was told wrong on the account of Phase Difference and not using Phase Reversal. Anyone point me in the right direction?

Last edited: Aug 31, 2013