Question

Light of wavelength 500 nm illuminates a soap film ( n = 1.33). What is the minimum thickness of film that will give an interference maximum when the light is incident normally on it?

Answer 1

Answer:

94 nm

Explanation:

We are given that

Wavelength of light,$\lambda=500 nm=500\times 10^{-9} m$

$1 nm=10^{-9} m$

n=1.33

We have to find the minimum thickness of film that will give an interference maximum when the light is incident normally on it.

We know that minimum thickness of film,d=$\frac{\lambda}{4n}$

By using the formula

$d=\frac{500\times 10^{-9}}{4\times 1.33}$

d=$9.4\times 10^{-8}=94\times 10^{-9} m$

d=94 nm