Question

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 535 nm ? The index of refraction of the film is 1.33, and there is air on both sides of the film.

Answer 1

Explanation:

It is given that,

The thinnest soap film appears black when illuminated with light with a wavelength of 535 nm, $\lambda=5.35\times 10^{-7}\ m$

Refractive index, $\mu=1.33$

We need to find the thickness of soap film. The soap film appear black means there is an destructive interference. The condition for destructive interference is given by :

$2t=m\dfrac{\lambda}{\mu}$

t = thickness of film

m = 0,1,2....

$\mu$ = refractive index

$t=m\dfrac{\lambda}{2\mu}$

$t=\dfrac{\lambda}{2\mu}$

For thinnest thickness, m = 1

$t=1\times \dfrac{5.35\times 10^{-7}\ m}{2\times 1.33}$

$t=2.01\times 10^{-7}\ m$

Hence, this is the required solution.