Animated Solution for Physics - Optics: In a double slit experiment, when a thin film of thickness $t$ having refractive index $\mu$ is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of $t$ is ($\lambda$ is the wavelength of the light used)

Imagine standing in a dark room, witnessing one of the most beautiful phenomena in physics: the interference of light. The classic Young's Double Slit Experiment (YDSE) presents a perfect, symmetrical pattern of bright and dark fringes on a screen. The central bright fringe sits proudly at the exact geometric center, a testament to the equal distances traveled by the light rays from both slits.

But what happens when we disrupt this perfect symmetry? What if we place a tiny, almost invisible hurdle in the path of one of the light rays? This is exactly what this fascinating JEE problem explores.

We start with the standard YDSE setup: two coherent slits, $S_1$ and $S_2$, separated by a distance $d$, and a screen placed at a large distance $D$. Normally, the rays reaching the center of the screen travel identical geometric paths, resulting in a net path difference of zero. This constructive interference creates the central maximum.

Now, the twist: a thin transparent film of thickness $t$ and refractive index $\mu$ is introduced right in front of slit $S_1$.

When light enters a denser medium like our thin film, it slows down. Even though the physical thickness of the film is just $t$, the light takes longer to traverse it compared to traveling the same distance in a vacuum (or air).

To account for this, physicists use the concept of Optical Path Length. The optical path is the distance light would have traveled in a vacuum during the same time. It is calculated as $\mu t$.

Since the film replaces a thickness $t$ of air, the extra path length introduced by the film is the new optical path minus the old air path:

Because of this extra optical path, the rays arriving at the geometric center of the screen are no longer in phase. The central maximum cannot survive there! It must migrate to a new location on the screen where the geometric path difference perfectly cancels out the optical path difference introduced by the film.

As a result, the entire fringe pattern shifts towards the slit covered by the film.

The problem provides a beautiful, elegant constraint: the central maximum shifts by exactly one fringe width.

Let's recall what a fringe width ($\beta$) represents. Moving by one full fringe width on the screen corresponds to a change in the path difference of exactly one full wavelength, $\lambda$.

If the entire pattern shifted by one fringe width, it means the extra path difference responsible for this shift must be exactly equal to $\lambda$.

We can now connect our physical intuition with the mathematical constraint. We equate the extra optical path introduced by the film to the path difference corresponding to a shift of one fringe width:

This is our master equation. The beauty of this approach is that we didn't even need to use the messy formulas involving $D$ and $d$; they completely abstract away when we think in terms of phase and wavelengths!

To find the thickness of the film, we simply isolate $t$:

And there we have it. A profound physical disruption, elegantly resolved into a simple, beautiful equation. This perfectly matches option (c). Always remember, in wave optics, thinking in terms of optical paths and wavelengths often provides the most direct and intuitive path to the solution.