Question

White light reflected at a perpendicular incidence from a uniform soap film has an interference maximum at 666 nm, and a minimum at 555 nm, with no minima between 666 nm and 555 nm. If n=1.34 for the film, what is the film thickness?

Answer 1

The condition for maximum in a thin film is:

$2dn+\dfrac{\lambda_1}{2}=m\lambda_1$

Where m is a whole number and do is the thickness of the film.

The condition for minimum will be:

$2dn+\dfrac{\lambda_2}{2}=(2m+1)\dfrac{\lambda_2} {2}$

Solving these two equations simultaneously, obtain:

$\dfrac{2dn} {\lambda_1} +0.5 =\dfrac{2dn} {\lambda_2}$

Thus:

$0.004d +0.5 =0.0048d\\ 0.0008d=0.5\\ d=625 nm.$

Answer. d=625 nm.