Question

A ray of light is coming from the source S. If a thin film of thickness $t$ and refractive index $\mu$ is placed in its path, the increase in the length of the optical path is:
A. $\mu t$
B. $\dfrac{\mu }{t}$
C. $\left( {\mu - 1} \right)t$
D. None of these

Answer 1

To solve this question, we need to consider Young’s double-slit experiment, first without the thin film, then with the thin film. We have to calculate the path difference in both the cases and compare both of them. By comparing them, we will get the final answer.

Complete step by step answer:
Let us consider Young’s double-slit experiment, where the ray of light coming from the source S passes through the two slits ${S_1}$ and ${S_2}$ having a width of $d$. After passing through the slits, these lights interfere at a point P on a screen situated at a distance of $D$ from the two slits. Let the wavelength of the ray of light passing through the source S be $\lambda$.
The path difference of the rays interfering at the point P on the screen is given by
$\Delta x = {S_2}P - {S_1}P$(1)
We know that the path difference in the Young’s double slit experiment is given by
$\Delta x = \dfrac{{\lambda d}}{D}$(2)
From (1) and (2) we can write
${S_2}P - {S_1}P = \dfrac{{\lambda d}}{D}$____(3)
Now, let us suppose that the thin film of thickness $t$ and refractive index $\mu$ is placed in the path of the ray of light coming from the slit ${S_1}$. We know that the path difference for a light ray travelling through a transparent medium of refractive index $\mu$ is given by
$PD = \mu t$
So the path length covered by the ray of light from the slit ${S_1}$ to the point P becomes
${S_1}P' = {S_1}P - t + \mu t$
$\Rightarrow {S_1}P' = {S_1}P + \left( {\mu - 1} \right)t$__(4)
So the new path difference becomes
$\Delta x' = {S_2}P - {S_1}P'$
From (4)
$\Delta x' = {S_2}P - \left( {{S_1}P + \left( {\mu - 1} \right)t} \right)$
$\Rightarrow \Delta x' = {S_2}P - {S_1}P - \left( {\mu - 1} \right)t$
Putting (3) in the above expression, we get
$\Rightarrow \Delta x' = \dfrac{{\lambda d}}{D} - \left( {\mu - 1} \right)t$
Thus, there is a change in the path difference of $\left( {\mu - 1} \right)t$ for the given ray of light. This means that the increase in the optical path of the ray of light is equal to $\left( {\mu - 1} \right)t$.
Hence, the correct answer is option C.

Note: The main reason for the increase in the path length for the light traveling in a thin film is due to the lower speed of light in the film. We know that the refractive index of a medium is the ratio of the speed of light in a vacuum to the speed of light in that medium. Using this definition we can get the required result.