The figure shows a Young's double slit experimental setup. It is observed that when a thin transpare

Q: Why does the central maximum shift towards the slit with the sheet?

The sheet slows down the light, increasing its optical path. To maintain a zero total path difference for the central maximum, the geometric path from the other slit must be longer, which happens when the point shifts towards the slit with the sheet.

Q: What exactly is optical path?

Optical path is the distance light would travel in a vacuum in the same time it takes to travel through a medium. It is given by the product of the refractive index and the geometric thickness ($\mu t$).

Q: Why did we use $a$ instead of $d$ for the slit separation?

The problem statement and the diagram use $a$ for the slit separation. It's just a variable name; the physics remains exactly the same whether you call it $a$ or $d$.

Q: Why are none of the options correct?

The correct derivation yields $t = \frac{n\lambda}{\mu - 1}$. The options provided in the exam likely contained a typo, perhaps confusing the shift formula or the fringe width formula. This happens sometimes in competitive exams!

Q: Does the fringe width change when the sheet is introduced?

No, the fringe width $\beta = \frac{\lambda D}{a}$ depends only on the wavelength, slit separation, and screen distance. The sheet only shifts the entire pattern without changing the spacing between the fringes.

Question

The figure shows a Young's double slit experimental setup. It is observed that when a thin transparent sheet of thickness $t$ and refractive index $\mu$ is put in front of one of the slits, the central maximum gets shifted by a distance equal to $n$ fringe widths. If the wavelength of light used is $\lambda$, $t$ will be

Options:

(A) $\frac{2nD\lambda}{a(\mu-1)}$

(B) $\frac{2D\lambda}{a(\mu-1)}$

(C) $\frac{D\lambda}{a(\mu-1)}$

(D) $\frac{nD\lambda}{a(\mu-1)}$

Sigma Percentile · Accepted Answer

### The Setup: A Twist in the Classic Experiment

Imagine you are in a physics lab, looking at the classic Young's Double Slit Experiment. The setup is familiar: two slits separated by a distance $a$, and a screen at a distance $D$.

But wait, there is a twist! We introduce a thin transparent sheet of thickness $t$ and refractive index $\mu$ right in front of the upper slit.

What happens to the beautiful interference pattern? The entire pattern shifts! Let's dive into the physics of why this happens and how we can calculate the exact thickness of this mysterious sheet.

### The Optical Path: Slowing Down Light

When light travels through a vacuum or air, it moves at its maximum speed. However, when it enters a denser medium like our transparent sheet, it slows down.

Even though the physical distance the light travels through the sheet is just $t$, the **optical path**—the equivalent distance it would have traveled in a vacuum in the same amount of time—is longer.

This introduces an extra path difference between the light waves emerging from the two slits. The additional path difference created by the sheet is given by the elegant formula $\Delta x = (\mu - 1)t$.

### The Shift: Finding the New Center

Because of this extra path difference, the central maximum—the point on the screen where the total path difference is zero—can no longer stay at the geometric center.

It must shift to compensate for the delay introduced by the sheet. To maintain a zero total path difference, the geometric path from the lower slit must be longer.

Therefore, the central maximum shifts upwards. The shift $x$ on the screen is directly proportional to the path difference and is given by $x = \frac{\Delta x \cdot D}{a}$, which becomes $x = \frac{(\mu - 1)t D}{a}$.

### The Grand Equating: Solving for Thickness

Now, let's recall the standard formula for the fringe width, denoted by $\beta$. It is the distance between two consecutive bright fringes, given by $\beta = \frac{\lambda D}{a}$.

The problem gives us a crucial piece of information: the central maximum shifts by exactly $n$ fringe widths. This means we can equate our shift $x$ to $n$ times $\beta$.

Substituting our expressions, we get $\frac{(\mu - 1)t D}{a} = n \frac{\lambda D}{a}$.

Look at this beautiful equation! The terms $D$ and $a$ are present on both sides, meaning the shift is independent of the screen distance and slit separation in this context. They simply cancel out, leaving us with a very neat relation: $(\mu - 1)t = n\lambda$.

### The Verdict: A Bonus Question!

Finally, we isolate $t$, the thickness of the sheet. We get our final answer: $t = \frac{n\lambda}{\mu - 1}$.

If you carefully check the options provided in the question, you will notice that none of them match this correct result! The options likely contained a typo, perhaps confusing the shift formula or the fringe width formula.

This happens sometimes in competitive exams, making it a "bonus" question where all options are incorrect. But the physics remains pristine and beautiful. Keep exploring, and don't let typos shake your confidence in solid derivations!

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Visualized Solution

$The Setup$

$Optical Path Difference$

$Shift of Central Maximum$

$Fringe Width$

$Equating the Shift$

$Simplifying the Equation$

$Final Answer$

$The Way Forward$

The Sigma Insight: Interference and Young's Double-Slit Experiment

The Setup

The Optical Path

The Shift

The Grand Equating

The Verdict

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Optical Path Difference

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Shift of Central Maximum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fringe Width

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating the Shift

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward