# Thin film / interference question

## Homework Statement

Monochromatic laser light is shown through a double-slit apparatus with a slit separation of onto a screen, but no
interference pattern is seen because the thin transparent coating (index of refraction n=1.35) on the screen is creating a thin-film destructive interference effect. This is known because when the light is shown on an old uncoated screen, they saw an interference pattern with the first bright fringe deflected at an angle of 3º.Find the minimum thickness of coating that needs to be scrubbed off so that the interference pattern for that light will
appear as brightly on the screen as if it were uncoated (i.e. no destructive interference at all).

## Homework Equations

$$dsin(\theta) = m\lambda \\ t = \frac{\lambda}{2n} (constructive)\\ t = \frac{\lambda}{4n} (destructive)$$

## The Attempt at a Solution

We know that m = 1 at theta = 3 degrees since that's where the first interference pattern is, so the first equation becomes lambda = d*sin(theta). Plug in lambda = d*sin(theta) into t = lambda / 4*n to find the thickness at which destructive interference occurs, so t = d*sin(theta)/(4*n) (where n = 1.35) and get 106.6 nm. If 106.6nm of the coating is removed, then constructive interference on the film will occur, so 106.6nm has to be removed.

Related Introductory Physics Homework Help News on Phys.org
ehild
Homework Helper
Welcome to PF!

Your derivation looks correct , but what is the distance between the slits?
And it is not sure that the coating has lambda/(4n) thickness.

Welcome to PF!

Your derivation looks correct , but what is the distance between the slits?
And it is not sure that the coating has lambda/(4n) thickness.
Hi, thanks!

Woops, sorry I forgot to include the thickness. The thickness is 1.1x10^-5 m. And you make a good point that the coating might not have lambda/(4n) thickness, but the problem doesn't say how thick the layer is. It only says the minimum thickness so I assume that it is greater than 106.6nm

ehild
Homework Helper
Your result is correct that 106.6 nm should be removed (supposing the refractive index of the screen material is greater than 1.35. Hopefully, it is.)