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(C) From the geometry of the figure,the direct path length is $SP = 2l$.The reflected path consists of two segments,each of length $d = l / \cos(30^{\

Vedclass · Accepted Answer

$l = \frac{(2n - 1)\lambda \sqrt{3}}{4(2 - \sqrt{3})}$

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(C) From the geometry of the figure,the direct path length is $SP = 2l$.The reflected path consists of two segments,each of length $d = l / \cos(30^{\circ}) = l / (\sqrt{3}/2) = 2l/\sqrt{3}$.Thus,the total reflected path length is $2 	imes (2l/\sqrt{3}) = 4l/\sqrt{3}$.The path difference $\Delta x$ between the two rays is $\Delta x = \frac{4l}{\sqrt{3}} - 2l = 2l \left( \frac{2}{\sqrt{3}} - 1 ight)$.Since the ray reflects off a mirror,it undergoes a phase change of $\pi$,which is equivalent to an additional path difference of $\lambda/2$.For constructive interference (maxima),the total path difference must be an odd multiple of $\lambda/2$ (because of the $\pi$ phase shift): $\Delta x + \frac{\lambda}{2} = n\lambda$,or $\Delta x = (n - 1/2)\lambda = \frac{(2n-1)\lambda}{2}$.Equating the two: $2l \left( \frac{2 - \sqrt{3}}{\sqrt{3}} ight) = \frac{(2n-1)\lambda}{2}$.Solving for $l$: $l = \frac{(2n-1)\lambda \sqrt{3}}{4(2 - \sqrt{3})}$.

An interference pattern is observed at $P$ due to the superimposition of two rays coming from a source $S$ as shown in the figure. The value of $l$ for which maxima is obtained at $P$ is: ($R$ is a perfectly reflecting surface)

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