In the figure, Young's double-slit experiment | vedclass

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(A) Given that $Q$ is the position of the $1^{st}$ bright fringe on the right side of the central point $O$. $P$ is the $11^{th}$ fringe on the other

Vedclass · Accepted Answer

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

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(A) Given that $Q$ is the position of the $1^{st}$ bright fringe on the right side of the central point $O$. $P$ is the $11^{th}$ fringe on the other side, measured from $Q$. This means $P$ is the $10^{th}$ bright fringe on the left side of the central point $O$ (since $11 - 1 = 10$). For a bright fringe, the path difference is given by $\Delta x = n\lambda$. Here, $n = 10$ and $\lambda = 6000 	imes 10^{-10} 	ext{ m}$. From the geometry of the figure, the path difference between the waves reaching $P$ from $S_1$ and $S_2$ is $S_1B = S_2P - S_1P$. Thus, $S_1B = n\lambda = 10 	imes 6000 	imes 10^{-10} 	ext{ m} = 60000 	imes 10^{-10} 	ext{ m} = 6 	imes 10^{-6} 	ext{ m}$.

In the figure, Young's double-slit experiment is shown. $Q$ is the position of the first bright fringe on the right side of $O$. $P$ is the $11^{th}$ fringe on the other side, as measured from $Q$. If the wavelength of the light used is $6000 \times 10^{-10} \text{ m}$, then $S_1B$ will be equal to:

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