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(D) In Young's double slit experiment, the path difference $\Delta x$ at any point $P$ is given by $S_2P - S_1P = d \sin 	heta$. From the geometry of

Vedclass · Accepted Answer

$5.4 	imes 10^{-6} 	ext{ m}$

Vedclass · Answer

(D) In Young's double slit experiment, the path difference $\Delta x$ at any point $P$ is given by $S_2P - S_1P = d \sin 	heta$. From the geometry of the figure, $S_2P - S_1P = S_2B = \Delta x$.For a bright fringe, the path difference is $\Delta x = n\lambda$, where $n$ is the order of the fringe.Point $Q$ is the $2^{nd}$ bright fringe on one side of the central maximum $O$, so its position is $y_Q = 2\beta$, where $\beta$ is the fringe width.Point $P$ is the $11^{th}$ bright fringe on the other side of $Q$. The distance of $P$ from $O$ is $y_P = (11 - 2)\beta = 9\beta$ on the opposite side of $O$.Thus, the path difference at point $P$ is $\Delta x = 9\lambda$.Given $\lambda = 6000 	ext{ Å} = 6000 	imes 10^{-10} 	ext{ m} = 6 	imes 10^{-7} 	ext{ m}$.Therefore, $S_2B = 9 	imes 6 	imes 10^{-7} 	ext{ m} = 54 	imes 10^{-7} 	ext{ m} = 5.4 	imes 10^{-6} 	ext{ m}$.

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