Chapter 24, Problem 78GP

To determine

To draw: The diagram for the given surface.

To explain: Conditions required for maxima and minima considering all possible values of index of refraction. Also, the relative size of minima when compared to maxima and zero.

Explanation of Solution

Given:

A light is transmitted through a thin film that is layered on a flat piece of glass.

Formula used:

Thickness is given by,

  $t=\frac{m\lambda }{2n}$

Where,

m is a multiple.

n is refractive index.

  $\lambda$ is wavelength.

Calculation:

The ray diagram for the given surface condition is shown below.

Here, $n_f$ is refractive index of film, $n_w$ is refractive index of water and $t$ is the thickness of the film.

In figure, ray 1 is incident on the film surface while ray 2 is partially refracted and then reflected again back into the same medium whose thickness is t . The ray will enter into the water by refraction. For an interference to be maximum, path difference will be an even multiple of half of the wavelength. For an interference to be minimum, path difference is an odd multiple of half of the wavelength.

Since the refractive index of film is greater than glass, so shift produced by first reflection is $\frac{{\lambda }_f}{2}$

Total shift will be,

  $d_2-d_1=2n_{f}t+\frac{{\lambda }_f}{2}$

For ray-1, maxima:

  $\begin{array}{l}2n_{f}t+\frac{1}{2}{\lambda }_f=m\lambda \\ 2n_{f}t=m{\lambda }_f-\frac{1}{2}{\lambda }_f\end{array}$

The thickness will be,

  $\begin{array}{l}t=\frac{\left(m-\frac{1}{2}\right){\lambda }_f}{2n_f}\\ m=1,2,3,\mathrm{.....}\end{array}$

Minima:

  $\begin{array}{l}2n_{f}t+\frac{1}{2}{\lambda }_f=\left(m+\frac{1}{2}\right){\lambda }_f\\ 2n_{f}t=m{\lambda }_f\end{array}$

The thickness will be,

  $\begin{array}{l}t=\frac{m\lambda }{2n_f}\\ m=0,1,2,3,\mathrm{....}\end{array}$

Now, when the two rays are compared at the film-glass interface, then total shift of second ray is,

  $d_2-d_1=2n_{f}t$

For ray-2, maxima:

  $\begin{array}{l}2n_{f}t=m{\lambda }_f\\ t=\frac{m{\lambda }_f}{2n_f}\\ m=0,1,2,3,\mathrm{......}\end{array}$

Minima:

  $\begin{array}{l}2n_{f}t=\left(m-\frac{1}{2}\right){\lambda }_f\\ t=\frac{\left(m-\frac{1}{2}\right){\lambda }_f}{2n_f}\\ m=1,2,3,\mathrm{......}\end{array}$

From above derived conditions of maxima and minima, it is observed that for a film of zero thickness means t is very less, there will only be a maximum in case of second reflection.