A Young's double slit interference arrangement with slits S1 in air and S2 is immersed in water (ref

Q: Why do we multiply the geometrical path by the refractive index?

The refractive index $\mu$ tells us how much slower light travels in a medium compared to a vacuum. Multiplying the geometrical path by $\mu$ gives the 'optical path', which is the equivalent distance the light would have traveled in a vacuum in the same amount of time. This is crucial for determining phase differences!

Q: Why is the geometrical path the same for both rays?

Look at the symmetry of the setup. Both slits $S_1$ and $S_2$ are at the same vertical distance $d$ from the water surface. Since point $P$ is on the surface at a horizontal distance $x$, the Pythagorean theorem gives the exact same hypotenuse $\sqrt{d^2 + x^2}$ for both.

Q: What does $m$ represent in the equation $m\lambda$?

$m$ is an integer ($0, 1, 2, \dots$) that represents the order of the maximum. For constructive interference (a bright fringe), the path difference must be a whole number of wavelengths so that the crests of both waves align perfectly.

Q: Why did we square the equation in the final steps?

We squared the equation to eliminate the square root and isolate $x^2$. The problem asks us to compare our result with a given expression for $x^2$, so algebraically manipulating our equation into that exact form is the most direct way to find the unknown constant $p$.

Q: Would the fringe pattern be the same if both slits were in the air?

Not at all! If both slits were in the air, the optical paths would be identical to the geometrical paths. Since the geometrical paths to any point on the surface are equal, the path difference would be zero everywhere on the surface, resulting in a single, massive central maximum!

Question

A Young's double slit interference arrangement with slits $S_1$ in air and $S_2$ is immersed in water (refractive index = $4/3$) as shown in the figure. The positions of maxima on the surface of water are given by $x^2 = p^2 m^2 \lambda^2 - d^2$, where $\lambda$ is the wavelength of light in air (refractive index = $1$), $2d$ is the separation between the slits and $m$ is an integer. The value of $p$ is

Sigma Percentile · Accepted Answer

The phenomenon of interference is one of the most beautiful demonstrations of the wave nature of light. But what happens when we introduce different media into the classic Young's Double Slit Experiment? The rules of the game change, and we must shift our focus from physical distances to **optical paths**.

### Analyzing the Setup

Imagine the setup: we have two slits, $S_1$ and $S_2$, separated by a total distance of $2d$. However, they are not in the same environment. $S_1$ is situated in the air, while $S_2$ is submerged in water. The water's surface acts as the boundary between these two worlds.

We are tasked with finding the condition for a bright fringe (a maximum) to form exactly on this water surface at a point $P$, which is at a horizontal distance $x$ from the central axis.

### Geometrical vs. Optical Path

To determine whether constructive interference occurs at point $P$, we must calculate the phase difference between the light waves arriving from $S_1$ and $S_2$. This phase difference is directly tied to the **optical path difference**.

First, let's look at the physical, or geometrical, paths. Because both slits are at a vertical distance $d$ from the interface, the triangles formed are perfectly symmetrical. Using the Pythagorean theorem, the geometrical distance from either slit to point $P$ is:

$$S_1P = S_2P = \sqrt{d^2 + x^2}$$

However, light travels slower in water than in air. To account for this, we use the concept of optical path, which is the geometrical path multiplied by the refractive index ($\mu$) of the medium.

For the ray traveling from $S_1$ in air ($\mu = 1$), the optical path is simply:
$$1 \cdot \sqrt{d^2 + x^2}$$

For the ray traveling from $S_2$ in water ($\mu = \frac{4}{3}$), the optical path is:
$$\mu \cdot \sqrt{d^2 + x^2} = \frac{4}{3}\sqrt{d^2 + x^2}$$

### The Master Equation

For a maximum to occur, the optical path difference ($\Delta x$) must be an integer multiple of the wavelength $\lambda$:

$$\Delta x = m\lambda$$

Substituting our optical paths into this condition, we get:

$$\frac{4}{3}\sqrt{d^2 + x^2} - 1 \cdot \sqrt{d^2 + x^2} = m\lambda$$

### Final Calculation

Now, it's just a matter of elegant algebra. Let's factor out the square root term:

$$\left(\frac{4}{3} - 1\right)\sqrt{d^2 + x^2} = m\lambda$$

Simplifying the fraction gives us $\frac{1}{3}$:

$$\frac{1}{3}\sqrt{d^2 + x^2} = m\lambda$$

Multiplying both sides by 3, we isolate the square root:

$$\sqrt{d^2 + x^2} = 3m\lambda$$

To find $x^2$, we square both sides of the equation:

$$d^2 + x^2 = 9m^2\lambda^2$$

Rearranging to solve for $x^2$:

$$x^2 = 9m^2\lambda^2 - d^2$$

The problem provides the general form of this equation as $x^2 = p^2 m^2 \lambda^2 - d^2$. By directly comparing our derived equation with the given form, it is brilliantly clear that:

$$p^2 = 9$$

Taking the square root, we arrive at our final answer:

$$p = 3$$

This problem beautifully illustrates how introducing a different medium alters the optical path, fundamentally shifting the conditions for interference!

Enter Numerical Value:

Visualized Solution

$S_{1}$ and $S_{2}$ Setup$

$Δ x = mλ$

$S_{1} P = S_{2} P = d^{2} + x^{2}$

$Δ x = μ (S_{2} P) - S_{1} P$

$(\frac{4}{3} - 1) d^{2} + x^{2} = mλ$

$d^{2} + x^{2} = 3 mλ$

$x^{2} = 9 m^{2} λ^{2} - d^{2}$

$p = 3$

$The Way Forward$

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

Analyzing the Setup

Geometrical vs

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​P=S2​P=d2+x2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δx=μ(S2​P)−S1​P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (34​−1)d2+x2​=mλ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2+x2​=3mλ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2=9m2λ2−d2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=3

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