# Transmission through constructive interference

1. Apr 22, 2007

### threewingedfury

A sheet of glass having an index of refraction of 1.32 is to be coated with a film of material having an index of refraction of 1.46 such that green light with a wavelength of 525 nm (in air) is preferentially transmitted via constructive interference.

(a) What is the minimum thickness of the film that will achieve the result in nm?

I was given the equation:
L=lamda/n2-n1

So 525x10^-9/.14 = 37500 nm

But this isn't right because:

With the numbers n2=1.55 and n1=1.40 the thickness is 169 nm

2. Apr 22, 2007

### mezarashi

It doesn't look like the equation you are using is accurate.

First of all, you must calculate the phase shifts involved.

1. Because the film has a higher index than air, the rays reflected by the film will experience a half-wavelength phase shift.

2. Rays will also be reflected off the glass-film interface. These go through a 2d phase-shift, where d is the film thickness. As the glass has a lower index than the film, no half-wavelength shift occurs here.

For constructive interference, the waves must be in-phase, so:

$$m \lambda_f = 2d - \frac{\lambda_f}{2}$$

Last edited: Apr 22, 2007
3. Apr 23, 2007

### threewingedfury

so what would m be? do you add the two indecies together, or subtract them?