Chapter 24, Problem 41P

To determine

The smallest thicknessof a soap film.

Answer to Problem 41P

  $169nm$

Explanation of Solution

Given info:

Wavelength, $\lambda =480nm$

The refractive index, $n=1.42$

Formula used:

The condition for a dark fringe in case of thin film interference: $2t=m{\lambda }_n$

Where, $m$ is an integer.

  $t$ is the thickness of the film.

  ${\lambda }_n$ is wavelength of light in the film.

The wavelength of light in the film is related to the wavelength of light in air by the relation:

  ${\lambda }_n=\frac{\lambda }{n}$

Where,

  $\lambda$ is wavelength of light in air.

  $n$ is refractive index of the film.

Calculation:

Calculate the thickness by using the formula, $t=\frac{1}{2}\frac{m\lambda }{n}$

Substitute all the given values:

  $t=\frac{1}{2}\frac{\left(1\right)\left(480nm\right)}{1.42}=169nm$

Conclusion:

The required thickness is $169nm$ .