A double slit setup is shown in the figure. One of the slits is in...

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 ${n}_{1}\left(≠ {n}_{2}\right)$ . 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.

$A double slit setup is shown in the figure. One of the slits is in medium 2 of refractive index n _ 2 . The other slit$

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

Options

AThe phase difference between the two rays is independent of $d$ .
BThe two rays interfere constructively at the detector.
CThe phase difference between the two rays depends on ${n}_{1}$ but is independent of ${n}_{2}$
DThe 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.

Correct answer

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

Step-by-step solution

$A double slit setup is shown in the figure. One of the slits is in medium 2 of refractive index n _ 2 . The other slit$

From the above figure,

$AB = dtan\theta$

and $BC = ABsin\alpha = \left(d\right)\left(tan\theta\right)\left(sin\alpha\right)Also,$ AD = ABsin\theta $$ ⇒ $Path difference (in vaccum)$ = {n}_{1}BC - {n}_{2}AD $$ = {n}_{1}\left(AB\right)sin\alpha - {n}_{2} $\left(ABsin\theta\right)From snell's law$ {n}_{1}sin\alpha = {n}_{2}sin\theta $. Therefore, path difference$ = AB\left({n}_{1}sin\alpha - {n}_{2}sin\theta\right) $= 0$

$⇒ \left(A\right)$ , \left(B\right) $are correct\left(C\right)$ , \left(D\right) are incorrect

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