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In Young's double slit experiment the separation between the slits is doubled without changing other setting of the experiment to obtain same fringe width, the distance ' $\mathrm{D}$ ' of the screen from slit should be made
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The correct answer is:
$2 \mathrm{D}$
Fringe width,
$\mathrm{W}=\frac{\lambda \mathrm{D}}{\mathrm{d}}$
Given: $\mathrm{d}=2 \mathrm{~d}$
$\Rightarrow \mathrm{W}^{\prime}=\frac{\lambda \mathrm{D}^{\prime}}{2 \mathrm{~d}}$
Since given,
$\begin{aligned}
& \mathrm{W}=\mathrm{W}^{\prime} \\
\therefore & \frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{\lambda \mathrm{D}^{\prime}}{2 \mathrm{~d}} \\
\therefore & \mathrm{D}^{\prime}=2 \mathrm{D}
\end{aligned}$
$\mathrm{W}=\frac{\lambda \mathrm{D}}{\mathrm{d}}$
Given: $\mathrm{d}=2 \mathrm{~d}$
$\Rightarrow \mathrm{W}^{\prime}=\frac{\lambda \mathrm{D}^{\prime}}{2 \mathrm{~d}}$
Since given,
$\begin{aligned}
& \mathrm{W}=\mathrm{W}^{\prime} \\
\therefore & \frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{\lambda \mathrm{D}^{\prime}}{2 \mathrm{~d}} \\
\therefore & \mathrm{D}^{\prime}=2 \mathrm{D}
\end{aligned}$
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