Column I shows four situations of standard Young's double slit arrangement with the screen placed fa

Q: Why is the geometrical path difference $S_2P - S_1P$ negative for $P_1$ and $P_2$?

Because $P_1$ and $P_2$ are located above the central point $P_0$. This means they are closer to the upper slit $S_1$ and farther from the lower slit $S_2$. Therefore, $S_1P < S_2P$, making $S_1P - S_2P$ positive, which implies $S_2P - S_1P$ is negative.

Q: How does the transparent sheet affect the path difference?

Light travels slower in a denser medium. When it passes through a sheet of thickness $t$ and refractive index $\mu$, it covers an equivalent optical path of $\mu t$ in the same time it would cover a distance $t$ in air. This introduces an extra path of $(\mu - 1)t$ to the wave passing through that slit.

Q: Why do we use the formula $I = I_{max} \cos^2(\frac{\pi \Delta x}{\lambda})$?

This is the standard interference intensity formula derived from the superposition of two waves of equal amplitude. The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$. Substituting this into $I = I_{max} \cos^2(\phi/2)$ gives the formula.

Q: In Case B, why does the central maximum shift to $P_1$?

The central maximum occurs where the net path difference is zero. At $P_1$, the geometrical path difference is $-\lambda/4$. The sheet on $S_2$ adds exactly $+\lambda/4$. These cancel out perfectly, making the net path difference zero at $P_1$!

Q: How do we compare intensities without calculating exact values?

You can look at the net path difference as a fraction of $\lambda$. The closer the path difference is to an integer multiple of $\lambda$ (like $0, \lambda, 2\lambda$), the higher the intensity. The closer it is to a half-integer multiple (like $\lambda/2, 3\lambda/2$), the closer the intensity is to zero.

Question

Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits $S_1$ and $S_2$. In each of these cases $S_1P_0 = S_2P_0, S_1P_1 - S_2P_1 = \frac{\lambda}{4}$ and $S_1P_2 - S_2P_2 = \frac{\lambda}{3}$, where $\lambda$ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index $\mu$ and thickness $t$ is pasted on slit $S_2$. The thickness of the sheets are different in different cases. The phase difference between the light waves reaching a point $P$ on the screen from the two slits is denoted by $\delta(P)$ and the intensity by $I(P)$. Match each situation given in Column I with the statement(s) in Column II valid for that situation.

List-I:

P.

Q. $(\mu - 1)t = \frac{\lambda}{4}$

R. $(\mu - 1)t = \frac{\lambda}{2}$

S. $(\mu - 1)t = \frac{3\lambda}{4}$

List-II:

1. $\delta(P_0) = 0$

2. $\delta(P_1) = 0$

3. $I(P_1) = 0$

4. $I(P_0) > I(P_1)$

5. $I(P_2) > I(P_1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In this classic Young's Double Slit Experiment (YDSE) problem, we are given a standard setup with two slits, $S_1$ and $S_2$, and a screen placed far away. The central point on the screen is $P_0$, where the geometrical paths from both slits are equal. Above $P_0$, we have two specific points, $P_1$ and $P_2$.

The problem provides the geometrical path differences for these points. Since $P_1$ and $P_2$ are above the central axis, they are closer to $S_1$ and farther from $S_2$. This means the path from $S_2$ is longer. The problem states $S_1P_1 - S_2P_1 = \frac{\lambda}{4}$, which implies the geometrical path difference $S_2P_1 - S_1P_1 = -\frac{\lambda}{4}$. Similarly, for $P_2$, the geometrical path difference is $S_2P_2 - S_1P_2 = -\frac{\lambda}{3}$.

### The Master Equation

When a transparent sheet of thickness $t$ and refractive index $\mu$ is placed in front of slit $S_2$, it introduces an additional optical path. Light travels slower in the sheet, so it effectively covers an extra distance of $(\mu - 1)t$ compared to traveling in a vacuum.

The net path difference $\Delta x_{net}$ at any point $P$ on the screen is the sum of the geometrical path difference and this extra optical path:
$$ \Delta x_{net} = (S_2P - S_1P) + (\mu - 1)t $$

The intensity at any point $P$ is governed by the phase difference $\delta(P)$, which is directly proportional to the net path difference. The intensity formula is:
$$ I(P) = I_{max} \cos^2\left(\frac{\pi \Delta x_{net}}{\lambda}\right) $$

### Case by Case Breakdown

**Case A: No Sheet**
Here, $t = 0$, so the net path difference is just the geometrical path difference.
At $P_0$, $\Delta x_{net} = 0$, so $\delta(P_0) = 0$ and $I(P_0) = I_{max}$.
At $P_1$, $\Delta x_{net} = -\frac{\lambda}{4}$, so $I(P_1) = I_{max} \cos^2\left(-\frac{\pi}{4}\right) = \frac{I_{max}}{2}$.
At $P_2$, $\Delta x_{net} = -\frac{\lambda}{3}$, so $I(P_2) = I_{max} \cos^2\left(-\frac{\pi}{3}\right) = \frac{I_{max}}{4}$.
Clearly, $\delta(P_0) = 0$ and $I(P_0) > I(P_1)$. This matches **(A) with p and s**.

**Case B: $(\mu - 1)t = \frac{\lambda}{4}$**
The sheet adds an extra path of $\frac{\lambda}{4}$ to the light from $S_2$.
At $P_1$, the net path difference becomes $\Delta x_{net} = -\frac{\lambda}{4} + \frac{\lambda}{4} = 0$.
Since the net path difference is zero, the phase difference $\delta(P_1) = 0$. The central maximum has effectively shifted to $P_1$! This matches **(B) with q**.

**Case C: $(\mu - 1)t = \frac{\lambda}{2}$**
The sheet adds an extra path of $\frac{\lambda}{2}$.
At $P_0$, $\Delta x_{net} = 0 + \frac{\lambda}{2} = \frac{\lambda}{2}$. This is a condition for destructive interference, so $I(P_0) = 0$.
At $P_1$, $\Delta x_{net} = -\frac{\lambda}{4} + \frac{\lambda}{2} = \frac{\lambda}{4}$, giving $I(P_1) = \frac{I_{max}}{2}$.
At $P_2$, $\Delta x_{net} = -\frac{\lambda}{3} + \frac{\lambda}{2} = \frac{\lambda}{6}$, giving $I(P_2) = I_{max} \cos^2\left(\frac{\pi}{6}\right) = \frac{3I_{max}}{4}$.
Comparing the intensities, we see that $I(P_2) > I(P_1) > I(P_0)$. This matches **(C) with t**.

### Final Calculation

**Case D: $(\mu - 1)t = \frac{3\lambda}{4}$**
The sheet adds an extra path of $\frac{3\lambda}{4}$.
At $P_0$, $\Delta x_{net} = \frac{3\lambda}{4}$, giving $I(P_0) = I_{max} \cos^2\left(\frac{3\pi}{4}\right) = \frac{I_{max}}{2}$.
At $P_1$, $\Delta x_{net} = -\frac{\lambda}{4} + \frac{3\lambda}{4} = \frac{\lambda}{2}$. This creates a perfect dark fringe, so $I(P_1) = 0$.
At $P_2$, $\Delta x_{net} = -\frac{\lambda}{3} + \frac{3\lambda}{4} = \frac{5\lambda}{12}$. The intensity is $I(P_2) = I_{max} \cos^2\left(\frac{5\pi}{12}\right) \approx 0.067 I_{max}$.
Since $I(P_1) = 0$, both $I(P_0)$ and $I(P_2)$ are strictly greater than $I(P_1)$. This perfectly matches **(D) with r, s, and t**.

List-I

List-II

Select Matching Pairs:

Visualized Solution

$Y D S E S e t u p an d P a t h D i f f er e n ce$

$G eo m e t r i c a l P a t h D i f f er e n ces$

$C a se A : N o S h ee t$

$C a se B : (μ - 1) t = \frac{λ}{4}$

$C a se C : (μ - 1) t = \frac{λ}{2}$

$C a seD : (μ - 1) t = \frac{3 λ}{4}$

$F ina l M a t c hin g$

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

Analyzing the Setup

The Master Equation

Case by Case Breakdown

Final Calculation

Similar Questions

Comprehension Passage

The $8^{th}$ bright fringe above the point O oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer ( $μ m$ ), is

The maximum speed in $μ m/s$ at which the $8^{th}$ bright fringe will move is

In the Young's double slit experiment, the distance between the slits varies in time as $d (t) = d_{0} + a_{0} sin ω t$ , where $d_{0}, ω$ and $a_{0}$ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as

List-I

List-II

Select Matching Pairs:

Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CaseB:(μ−1)t=4λ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CaseC:(μ−1)t=2λ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CaseD:(μ−1)t=43λ​

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