Calculating Film Thickness: Interference Fringes with L = 500nm Incident Light

[phalanx123]

1. Light of wavelength L = 500nm, produced by an extended source, is incident
at an angle of 30o from the normal upon a dielectric film of refractive index n = 2
supported on a solid planar substrate. Reflectivity minima are observed to have an
angular separation of 0.05 radians. What is the thickness of the film?



I got a little confused because I thought that if the film have same thickness and horizontal and the incident light is at a constant angle than there wouldn't be any inteference fringes. They would either all be constructive, destructive or somewhere in between because the condition (phase difference/path difference) for each light ray is the same. Or have I misunderstood the geometry of the question? Is it that the film is laying horizontal and the light all incident at the same angle of 30o and interference fringes were created with angular separation of 0.05 rad between successive minima? I am so stuck on this question could somebody help me please

 

[neelakash]

There is no reason why interference will not occur.

Note that it is amplitude splitting case...

a part of the incident ray is reflected from the film surface

The other part goes inside...gets reflected from the base(supported on the planar hard surface) and emerges from the upper surface of the film.

So, there would be a phase difference: phase difference due to path difference + due to reflection from a hard surface (pi)

Then you get the requisite formula. Use this formula to get the minima...and find the angular difference in the successive minima...

I hope that will do.

From the context, it is clear that the film is assumed planar.Otherwise,they would not ask for the thikness.

 

[phalanx123]

Hi neelakash, I know that inteference will occur in such occasion what I meant was how fringes would occur i.e. both maxima and minima occur at same time. Because what I thought was that if all the light rays are incidenting at the same angle and the film have a constant thickness than the path differences between the reflcted and the refracted-than- reflected ray are same for every incident ray so that if one such pair is destructive than it would be same for all the incident ray and so it would be destructive everywhere. I don't see how fringes would form? Thanks

 

[bsimmo]

Successive minima happen when you move the film+substrate along the perpendicular.

iirc since a phase change occurs on the top surface, for destructive;
Using 2ndcosΘ'=mλ
n being the refractive index of the thin film
d being the thinkness you are looking for
Θ' being the angle wrt normal of the ray in the thin film
m is you minima seperation
λ is wavelength.

Hope that helps.