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QJEEAdv22Ph90a4: JEE Advanced 2022 Physics - Class 12 JEEAdv22Ph90a4 Solution: A double slit setup is shown in the figure. One of the slits is …

A double slit setup is shown in the figure. One of the slits is in medium 2 of refractive index n2n_{2} . The other slit is at the interface of this medium with another medium 1 of refractive index n1(≠n2)n_{1}\left(\neq n_{2}\right) . The line joining the slits is perpendicular to the interface and the distance between the slits is dd . The slit widths are much smaller than dd . 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\theta from the line joining them, so that θ\theta equals the angle of refraction of the beam. Consider two approximately parallel rays from the slits received by the detector.

JEE Advanced 2022 Paper 2 Online Physics - Wave Optics Question 7 English

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

Multiple Choice Options

  • A The phase difference between the two rays is independent of dd .
  • B The two rays interfere constructively at the detector.
  • C The phase difference between the two rays depends on n1n_{1} but is independent of n2n_{2} .
  • D The phase difference between the two rays vanishes only for certain values of dd and the angle of incidence of the beam, with θ\theta being the corresponding angle of refraction.

Step-by-Step Verified Solution

JEE Advanced 2022 Paper 2 Online Physics - Wave Optics Question 7 English Explanation

tan⁡θ=ABd\tan \theta =\frac{\mathrm{AB}}{d}

And, AB=dtan⁡θ\mathrm{AB} =d \tan \theta ........ (i)

BC=ABsin⁡α\mathrm{BC} =\mathrm{AB} \sin \alpha

=dtan⁡θsin⁡α=d \tan \theta \sin \alpha

AD=ABsin⁡θ\mathrm{AD} =\mathrm{AB} \sin \theta

Path difference in vacuum

Δx=n1BC−n2ADΔx=n1(AB)sin⁡α−n2(ABsin⁡θ)

Hence, AB(n1sin⁡α−n2sin⁡θ)=0\mathrm{AB}\left(n_1 \sin \alpha-n_2 \sin \theta\right)=0

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