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In Young's experiment, fringes are obtained on a screen placed a distance $75 \mathrm{~cm}$ from the slits. When the separation between two narrow slits is doubled, then the fringe width is decrease. In order to obtain the initial fringe width, the screen should be moved through.
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Verified Answer
The correct answer is:
75 cm away from the slits
Fringe width, $\beta=\frac{\lambda D_1}{d_1}=\frac{\lambda D_2}{d_2}$
$$
\begin{aligned}
& \therefore \frac{d_2}{d_1}=\frac{D_2}{D_1} \\
& \therefore 2=\frac{D_2}{D_1} \\
& \therefore D_2=2 D_1=2 \times 75=150 \mathrm{~cm}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \frac{d_2}{d_1}=\frac{D_2}{D_1} \\
& \therefore 2=\frac{D_2}{D_1} \\
& \therefore D_2=2 D_1=2 \times 75=150 \mathrm{~cm}
\end{aligned}
$$
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