A double slit setup is shown in the figure. One of the slits is in medium 2 of refractive index n2.

Q: Why do we need to consider the ray crossing the interface before the slit?

Because Slit 2 is submerged in Medium 2! The incident beam is in Medium 1. For light to reach Slit 2, it MUST cross the interface and refract first.

Q: Why is the outgoing angle exactly $\theta$?

The problem states the detector is placed at an angle $\theta$ which equals the angle of refraction. Since the detector is far away, the rays reaching it are parallel and travel at this exact angle.

Q: What is an optical path and why do we multiply by the refractive index?

Light travels slower in a medium. The optical path ($n \times \text{distance}$) represents the equivalent distance the light would have traveled in a vacuum in the same amount of time. It's crucial for calculating phase differences!

Q: How did we know the horizontal distance from the normal is $d \tan\theta$?

Simple right-angle geometry! The ray travels from the interface to Slit 2 at an angle $\theta$ to the vertical. Since Slit 2 is at depth $d$, the horizontal displacement is $d \tan\theta$.

Q: Does this mean there are no fringes on the screen?

There are fringes! The phase difference is zero *specifically* at the angle $\theta$. At other angles, the path difference won't perfectly cancel out, creating a pattern of bright and dark fringes. The detector just happens to be sitting exactly on the central maximum!

Question

A double slit setup is shown in the figure. One of the slits is in medium 2 of refractive index $n_2$. The other slit is at the interface of this medium with another medium 1 of refractive index $n_1(
eq n_2)$. The line joining the slits is perpendicular to the interface and the distance between the slits is $d$. The slit widths are much smaller than $d$. A monochromatic parallel beam of light is incident on the slits from medium 1. A detector is placed in medium 2 at a large distance from the slits, and at an angle $	heta$ from the line joining them, so that $	heta$ equals the angle of refraction of the beam. Consider two approximately parallel rays from the slits received by the detector.
Which of the following statement(s) is (are) correct?

Options:

(A) The phase difference between the two rays is independent of $d$.

(B) The two rays interfere constructively at the detector.

(C) The phase difference between the two rays depends on $n_1$ but is independent of $n_2$.

(D) The phase difference between the two rays vanishes only for certain values of $d$ and the angle of incidence of the beam, with $	heta$ being the corresponding angle of refraction.

Sigma Percentile · Accepted Answer

### The Deceptive Setup

At first glance, this problem looks like a standard Young's Double Slit Experiment, but the geometry holds a beautiful secret. We have a horizontal interface separating Medium 1 (refractive index $n_1$) and Medium 2 (refractive index $n_2$). Slit $S_1$ is positioned exactly on this interface, while Slit $S_2$ is submerged at a depth $d$ inside Medium 2.

A parallel beam of monochromatic light is incident from Medium 1. The crucial realization here is understanding how light reaches the submerged slit $S_2$. It doesn't just magically appear there; it must first strike the interface at some point $P$, refract, and then travel through Medium 2 to hit $S_2$.

### Tracing the Rays: A Tale of Two Paths

Let's trace two specific rays from the incident beam. Ray 1 travels through Medium 1 and hits $S_1$ directly. Ray 2 hits the interface at point $P$ and refracts. According to Snell's Law, $n_1 \sin\alpha = n_2 \sin	heta$, where $\alpha$ is the angle of incidence and $	heta$ is the angle of refraction.

Since Ray 2 must hit $S_2$ (which is at depth $d$), simple right-angle geometry tells us that point $P$ must be located at a horizontal distance of $d 	an	heta$ from $S_1$. The physical distance Ray 2 travels in Medium 2 from $P$ to $S_2$ is $\frac{d}{\cos	heta}$.

### The Geometry of Wavefronts

To find the phase difference at the detector, we need to calculate the total optical path difference. The detector is placed at an angle $	heta$, which perfectly matches the angle of refraction. This means the rays emerging from both slits and heading towards the detector are parallel, traveling at this exact angle $	heta$.

Imagine an incident wavefront passing through point $P$. From this wavefront, Ray 2 travels straight to $S_2$ in Medium 2. Its optical path is simply:

$$ 	ext{Path}_2 = n_2 \left( \frac{d}{\cos	heta} ight) $$

What about Ray 1? Starting from the same incident wavefront, it first travels an extra distance in Medium 1 to reach $S_1$. This distance is $d 	an	heta \sin\alpha$. Then, after leaving the slits, Ray 1 has to travel an extra distance in Medium 2 to catch up with the outgoing wavefront of Ray 2. This distance is $d \cos	heta$. Adding these up, we get its total optical path:

$$ 	ext{Path}_1 = n_1 (d 	an	heta \sin\alpha) + n_2 (d \cos	heta) $$

### The Grand Cancellation

Now for the magic. Let's find the path difference $\Delta x$ by subtracting $	ext{Path}_1$ from $	ext{Path}_2$:

$$ \Delta x = \frac{n_2 d}{\cos	heta} - \left( n_1 d 	an	heta \sin\alpha + n_2 d \cos	heta ight) $$

We can simplify this beautifully by substituting $n_1 \sin\alpha$ with $n_2 \sin	heta$ using Snell's Law:

$$ \Delta x = \frac{n_2 d}{\cos	heta} - n_2 d \frac{\sin^2	heta}{\cos	heta} - n_2 d \cos	heta $$

Factoring out $n_2 d$, we get:

$$ \Delta x = n_2 d \left( \frac{1 - \sin^2	heta}{\cos	heta} - \cos	heta ight) $$

Look closely at the expression inside the bracket. Since $1 - \sin^2	heta = \cos^2	heta$, the first term becomes $\frac{\cos^2	heta}{\cos	heta}$, which is just $\cos	heta$.

$$ \Delta x = n_2 d (\cos	heta - \cos	heta) = 0 $$

The optical path difference is exactly zero! This means the phase difference is always zero, completely independent of the distance $d$ or the refractive indices. Because the phase difference is zero, the two rays will always interfere constructively at the detector.

Select Answer:

Visualized Solution

\text{Understanding the Setup}

\text{Incident Plane Wave}

\text{Refraction at the Interface}

\text{Rays to the Detector}

\text{Optical Path of Ray 2}

\text{Optical Path of Ray 1}

\text{Evaluating the Path Difference}

\text{Constructive Interference}

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

The Deceptive Setup

Tracing the Rays

The Geometry of Wavefronts

The Grand Cancellation

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