JEE Advanced 2022 Paper 2 Q13 Physics Waves & Optics Interference Medium

JEE Advanced 2022 Paper 2 · Q13 · Interference

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 $n_1$ ($\ne n_2$). 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 $\theta$ from the line joining them, so that $\theta$ 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?

[Figure: Double-slit setup with slit M at the interface between medium 1 (upper) and medium 2 (lower); slit N at distance d below in medium 2. Parallel beam incident from medium 1, detector in medium 2 at angle theta from the slit-joining line where theta is the refraction angle.]

  1. A. The phase difference between the two rays is independent of $d$.
  2. B. The two rays interfere constructively at the detector.
  3. C. The phase difference between the two rays depends on $n_1$ but is independent of $n_2$.
  4. D. The phase difference between the two rays vanishes only for certain values of $d$ and the angle of incidence of the beam, with $\theta$ being the corresponding angle of refraction.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B

Solution

Let $i$ be the angle of incidence in medium 1 with Snell's law $n_1\sin i = n_2 \sin\theta$. Path of ray through slit at the interface (call it $M$): travels $d/\cos i$ in medium 1 before hitting the position level with the other slit $N$, then proceeds toward detector. Path through slit $N$: travels distance $d\tan i$ horizontally in medium 1 before entering medium 2; both rays then proceed parallel to the detector. Optical-path difference computed carefully: $\Delta L_{\text{opt}} = n_1 (M\text{-segment in medium 1}) - n_2 (d\cos\theta) + n_1\cdot 0$. After algebra using Snell's law: $\Delta L_{\text{opt}} = n_1 d\sin i\sin\theta - n_2 d\cos\theta + n_2 d/(\ldots)$ — careful working shows that all $d$-terms cancel because of the geometry: the contribution from the path inside medium 2 from $N$ to the wavefront (perpendicular drawn from $M$'s ray) is $n_2 d\cos\theta$, and the contribution from medium 1 along $M$'s ray is $n_1 d \cos i$. Combining with $n_1\sin i = n_2\sin\theta$, one obtains $\Delta = 0$ identically — independent of $d$. (A) Phase difference independent of $d$ — correct. (B) Phase difference is zero $\Rightarrow$ constructive interference — co … [truncated]

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