A vessel ABCD of 10 cm width has two small slits S1 and S2 sealed with identical glass plates of equ

Q: Why is the initial path difference $SS_1 - SS_2$ and not the other way around?

Since the source $S$ is located below the central axis, the distance to the upper slit $S_1$ is geometrically longer than the distance to the lower slit $S_2$. You can verify this using the Pythagorean theorem.

Q: Why can we use the approximation $\sin \alpha \approx \tan \alpha$ here?

The vertical distance $y_1 = 40$ cm is much smaller than the horizontal distance $D_1 = 200$ cm. This makes the angle $\alpha$ small enough that the small-angle approximation holds perfectly.

Q: Why does the central bright fringe shift upwards initially?

The light reaching $S_1$ has traveled further and is 'lagging' behind the light at $S_2$. To compensate and have zero net path difference, the light from $S_1$ must travel a shorter distance after the slits. This happens when the fringe shifts upwards towards $S_1$.

Q: How does the liquid change the path difference at $Q$?

Light travels slower in a denser medium. The liquid increases the 'optical path' of the ray passing through $S_2$ by $(\mu - 1)D_2$, effectively making it travel a longer equivalent distance in a vacuum.

Q: Why do we equate $(\mu - 1)D_2$ directly to the initial path difference?

At point $Q$, the geometric distances from $S_1$ and $S_2$ are equal. So the only path difference after the slits is the optical path added by the liquid. For the central fringe to form at $Q$, this added optical path must exactly cancel the initial path difference from before the slits.

Question

A vessel $ABCD$ of 10 cm width has two small slits $S_1$ and $S_2$ sealed with identical glass plates of equal thickness. The distance between the slits is 0.8 mm. $POQ$ is the line perpendicular to the plane $AB$ and passing through $O$, the middle point of $S_1$ and $S_2$. A monochromatic light source is kept at $S$, 40 cm below $P$ and 2 m from the vessel, to illuminate the slits as shown in the figure alongside. Calculate the position of the central bright fringe on the other wall $CD$ with respect to the line $OQ$. Now, a liquid is poured into the vessel and filled upto $OQ$. The central bright fringe is found to be at $Q$. Calculate the refractive index of the liquid.

Sigma Percentile · Accepted Answer

## Shifting Fringes: The Dance of Optical Paths

Young's Double Slit Experiment (YDSE) is a beautiful demonstration of the wave nature of light. But what happens when we mess with the symmetry? In this classic JEE Advanced problem, we explore two ways to break the symmetry: moving the source off-axis and introducing a denser medium in the path of one of the rays.

### Analyzing the Setup (The Initial Path Difference)

Normally, the light source is placed exactly on the central axis, ensuring that light reaches both slits simultaneously. However, in our setup, the source $S$ is placed $y_1 = 40 	ext{ cm}$ below the central axis $POQ$.

Because $S$ is below the axis, the light ray traveling to the upper slit $S_1$ has to cover a slightly longer geometric distance than the ray traveling to the lower slit $S_2$. This creates an **initial path difference** even before the light enters the vessel.

We can calculate this path difference by finding the angle $\alpha$ that the rays make with the horizontal.

$$	an \alpha = \frac{y_1}{D_1} = \frac{40 	ext{ cm}}{200 	ext{ cm}} = \frac{1}{5}$$

Since the angle is very small, we can safely use the small-angle approximation $\sin \alpha \approx 	an \alpha = \frac{1}{5}$. The initial path difference $\Delta x_1$ is given by:

$$\Delta x_1 = d \sin \alpha = 0.8 	ext{ mm} 	imes \frac{1}{5} = 0.16 	ext{ mm}$$

This means the light reaching $S_1$ is 'lagging' behind the light reaching $S_2$ by $0.16 	ext{ mm}$.

### Finding the Central Bright Fringe

The central bright fringe is defined as the point on the screen where the **net path difference is zero**. To achieve this, the path difference introduced *after* the slits must perfectly cancel out the initial path difference introduced *before* the slits.

Since the ray from $S_1$ traveled further initially, it must travel a shorter distance after the slits to catch up. This means the central fringe will shift upwards towards $S_1$. Let this new position be $R$, located at a distance $y_2$ above $Q$.

The path difference after the slits is $\Delta x_2 = d \sin 	heta$, where $	heta$ is the angular position of $R$. Equating the two path differences:

$$d \sin 	heta = 0.16 	ext{ mm}$$

$$\sin 	heta = \frac{0.16 	ext{ mm}}{0.8 	ext{ mm}} = \frac{1}{5}$$

Again, using the small-angle approximation $	an 	heta \approx \sin 	heta = \frac{y_2}{D_2}$:

$$\frac{y_2}{10 	ext{ cm}} = \frac{1}{5} \implies y_2 = 2 	ext{ cm}$$

So, the central bright fringe is formed **2 cm above point Q**.

### The Liquid Twist (Optical Path in a Medium)

Now for the second part of the problem. A liquid is poured into the vessel up to the level $OQ$. This means the lower slit $S_2$ is now submerged, while $S_1$ remains in the air.

We are told that the central bright fringe shifts back to the point $Q$. Let's analyze the paths at $Q$. Geometrically, the distances from $S_1$ and $S_2$ to $Q$ are equal ($S_1Q = S_2Q = D_2$).

However, light travels slower in a denser medium. The ray from $S_2$ travels through the liquid, which increases its **optical path** by an amount $(\mu - 1)D_2$.

### Final Calculation

For the central fringe to form at $Q$, this newly introduced optical path difference must perfectly balance the initial path difference of $0.16 	ext{ mm}$ that we calculated earlier.

$$(\mu - 1)D_2 = \Delta x_1$$

$$(\mu - 1) 	imes 100 	ext{ mm} = 0.16 	ext{ mm}$$

$$\mu - 1 = \frac{0.16}{100} = 0.0016$$

$$\mu = 1.0016$$

By carefully tracking the geometric and optical paths of the light rays, we successfully navigated through both the off-axis source and the introduction of a refractive medium. This is the true essence of mastering wave optics!

Visualized Solution

$Visualizing the Setup$

$Initial Path Difference$

$Calculating Angle α$

$Computing Δ x_{1}$

$Condition for Central Fringe$

$Position of Central Fringe$

$Introducing the Liquid$

$Optical Path Difference$

$Calculating Refractive Index$

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

Analyzing the Setup (The Initial Path Difference)

Finding the Central Bright Fringe

The Liquid Twist (Optical Path in a Medium)

Final Calculation

Similar Questions

Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Computing Δx1​

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

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

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

.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; } Calculating Refractive Index