Question

A thin film of soap with n = 1.34 hanging in the air reflects dominantly red light with λ = 642 nm. What is the minimum thickness of the film? Tries 0/20 Now this film is on a sheet of glass, with n = 1.55. What is the wavelength of the light in air that will now be predominantly reflected?

Answer 1

Answer:

The minimum thickness of the film and the wavelength of the light in air are $1.197\times10^{-7}\ m$ and 371 nm.

Explanation:

Given that,

Refractive index of soap= 1.34

Refractive index  of glass= 1.55

Wavelength = 642 nm

(I). We need to calculate the minimum thickness

Using formula of thickness

$t=\dfrac{(2m+1)\lambda}{4n}$

Where, m = 0 for constrictive

Put the value into the formula

$t=\dfrac{(642\times10^{-9})}{4\times1.34}$

$t=1.197\times10^{-7}\ m$

(II). We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{2nt}{m}$

Where, m = 1

Put the value into the formula

$\lambda=\dfrac{2\times1.55\times1.197\times10^{-7}}{1}$

$\lambda=371\ nm$

Hence, The minimum thickness of the film and the wavelength of the light in air are $1.197\times10^{-7}\ m$ and 371 nm.