The y-coordinate of the point where the total phase difference between the interfering waves is zero, is
(A) y = 0                          (B) $y=+\frac{3}{4}\mathrm{m}$                   (C) $\mathrm{y}=-\frac{3}{4}\mathrm{m}$                       (D) $-\frac{1}{\sqrt{3}}\mathrm{m}$

Dear Student,

Please find below the solution to the asked query:

Here, the optical path length in the first medium is $d \sin \theta$.

So, the actual length is,

$d_{geo}=\frac{d_{op}}{n}=\frac{d \sin \theta }{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{3}{4}d \sin {30}^0=\frac{3d}{4}$

Therefore, the the ray passing through S₂ travels a distance by d_geo​ more than the distance travelled by the ray passing from S₁. Now, after passing from the slits, the ray from S₁ should travel a distance equal to d_geo​ more than the ray from S₂. So, the point will come below O, and the y is negative.

$$\begin{gathered} d_{geo}=d\frac{y}{D}\Rightarrow d\frac{y}{1}=\frac{3d}{4}\Rightarrow y=\frac{3}{4} m \\ Since, y is on -ve y - axis, \\ y=-\frac{3}{4} m \end{gathered}$$

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