A double slit setup is shown in the figure. One of the slits is in medium 2 of refractive index n₂. The other slit is at the interface of this medium with another medium 1 of refractive index n₁ ($n_1\ne n_2)$. 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 θ from the line joining them, so that θ equals the angle of refraction of the beam. Consider two approximately parallel rays from the slits received by the detector.

Which of the following statements (s) is (are) correct ?

The correct answers are: "The phase difference between the two rays is independent of d" and "The two rays interfere constructively at the detector."

Setting up the Geometry from the Figure:

From the diagram, Medium 1 (refractive index n₁) is on top and Medium 2 (refractive index n₂) is on the bottom. The interface is the horizontal boundary between them. Slit S₁ sits exactly at this interface, and Slit S₂ is at distance d directly below S₁, inside Medium 2. A monochromatic parallel beam (plane wave) in Medium 1 strikes the interface at angle of incidence α. By Snell's Law:

$n_{1}sin\alpha =n_{2}sin\theta$

The detector in Medium 2 is placed at angle θ (the refraction angle) from the line joining the slits (the vertical axis).

Step 1: Coordinate System

Let us place S₁ at the origin (0, 0) on the interface, with the x-axis pointing downward (into Medium 2) — perpendicular to the interface — and the y-axis running along the interface. Then:
  • S₁ is at (0, 0) — at the interface
  • S₂ is at (d, 0) — inside Medium 2

The incident plane wave in Medium 1 travels at angle α to the x-axis (normal). A ray from this plane wave that reaches S₂ does NOT enter directly from above S₂; instead, it must first cross the interface at some point. Tracing the refracted ray in Medium 2 backward from S₂ (direction at angle θ to x-axis), it hits the interface at the point P = (0, −d tan θ).

Step 2: Phase Accumulated by Ray 2 at S₂ (Incident Part)

The incident plane wave in Medium 1 has phase:

$\phi =\frac{2\pi n_1}{{\lambda }_0}(xcos\alpha +ysin\alpha )$

Taking the phase at S₁ = (0,0) as the reference (φ₁ = 0), the phase at the interface entry point P = (0, −d tan θ) for Ray 2 is:

${\phi }_{\mathrm{inc}}=\frac{2\pi n_1}{{\lambda }_0}(0-dtan\theta ·sin\alpha )=-\frac{2\pi n_{1}dtan\theta sin\alpha }{{\lambda }_0}$

Substituting Snell's Law (n₁ sin α = n₂ sin θ):

${\phi }_{\mathrm{inc}}=-\frac{2\pi n_{2}dsin\theta tan\theta }{{\lambda }_0}$

After crossing the interface at P = (0, −d tan θ), the ray travels through Medium 2 to S₂ at (d, 0). The geometric distance is:

$\mathrm{PS}=\sqrt{d^2+d^2{tan}^2\theta }=\frac{d}{cos}$

Phase accumulated in Medium 2 on this leg:

${\phi }_{\mathrm{leg}}=\frac{2\pi n_{2}d}{{\lambda }_{0}cos\theta }$

Total phase at S₂ relative to S₁ (from the incident wave):

$\Delta {\phi }_{\mathrm{at\; S₂}}={\phi }_{\mathrm{inc}}+{\phi }_{\mathrm{leg}}=\frac{2\pi n_{2}d}{{\lambda }_0}\left[-sin\theta tan\theta +\frac{1}{cos}\right]$

Simplifying the bracket:

$-sin\theta tan\theta +\frac{1}{cos}=\frac{-}{{sin}^2}=\frac{{cos}^2}{\theta }=cos\theta$

Therefore:

$\Delta {\phi }_{\mathrm{at\; S₂}}=\frac{2\pi n_{2}dcos\theta }{{\lambda }_0}$

Step 3: Phase Difference for Paths from Slits to Detector

The detector is far away in Medium 2, in direction (cos θ, sin θ) from the x-axis. For a distant detector, the path length difference between the two rays from S₁ and S₂ is found by projecting the displacement S₂ − S₁ = (d, 0) onto the detector direction:

$\Delta L_{\mathrm{to\; detector}}=(d,0)·(cos\theta ,sin\theta )=dcos\theta$

This means S₂ is closer to the detector by d cos θ, so Ray 2 travels less in Medium 2 by this amount. The phase contribution (Ray 2 minus Ray 1) for this leg:

$\Delta {\phi }_{\mathrm{to\; detector}}=-\frac{2\pi n_{2}dcos\theta }{{\lambda }_0}$

Step 4: Net Phase Difference Between the Two Rays

Adding both contributions:

$\Delta \phi =\Delta {\phi }_{\mathrm{at\; S₂}}+\Delta {\phi }_{\mathrm{to\; detector}}=\frac{2\pi n_{2}dcos\theta }{{\lambda }_0}-\frac{2\pi n_{2}dcos\theta }{{\lambda }_0}=0$

The two terms cancel exactly and completely.

Step 5: Evaluating Each Option

Option 1 — "The phase difference is independent of d" ✓ CORRECT:
The net phase difference is Δφ = 0 for all values of d. The two terms that contain d cancel identically. Therefore the phase difference is not just independent of d — it is identically zero for every value of d.

Option 2 — "The two rays interfere constructively at the detector" ✓ CORRECT:
Since Δφ = 0, the two rays are perfectly in phase. This is the condition for constructive interference. The physical reason is elegant: the extra path gained during the incidence leg (because Ray 2 enters medium 2 at a displaced point and travels an extra distance inside it before reaching S₂) is exactly equal in optical path length to the shorter path from S₂ to the detector compared to S₁. These two effects cancel precisely because the detector is placed at angle θ — the same angle of refraction — making the geometry self-consistent via Snell's Law.

Option 3 — "Phase difference depends on n₁ but is independent of n₂" ✗ INCORRECT:
The phase difference is zero — it depends on neither n₁ nor n₂. Snell's Law was used in the derivation to replace n₁ sin α with n₂ sin θ, but the final result of Δφ = 0 is independent of all material parameters.

Option 4 — "The phase difference vanishes only for certain values of d and angle of incidence" ✗ INCORRECT:
The phase difference is zero for all values of d and for any angle of incidence α (as long as θ is the corresponding refraction angle where the detector is placed). It is not restricted to special values.

Summary:
The beautiful physical insight here is that placing the detector exactly at the refraction angle θ creates a perfect geometric symmetry: the "extra" optical path accumulated when the incident wavefront reaches S₂ (via refraction into Medium 2) is exactly compensated by the shorter path from S₂ to the detector. The cancellation is exact, universal (independent of d, n₁, n₂, and the specific value of α), and results in permanent constructive interference at the detector.