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If in a Young's double slit experiment the slit separation is doubled and the distance of the screen from the slits is reduced to half, then the fringe widths become how many times their original value?
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Verified Answer
The correct answer is:
$\frac{1}{4}$
Fringe width in Young's double slit experiment is given as
$\beta=\frac{D \lambda}{d}$
where, $D=$ distance between source and screen
and $\quad d=$ slit width.
$\begin{array}{ll}
\therefore & \frac{\beta_2}{\beta_1}=\frac{d_1}{d_2} \times \frac{D_2}{D_1} \\
\Rightarrow & \frac{\beta_2}{\beta_1}=\frac{d_1}{2 d_1} \times \frac{D_1 / 2}{D_1} \quad\left[\because d_2=2 d_1 \text { and } D_2=\frac{D_1}{2}\right] \\
\Rightarrow & \frac{\beta_2}{\beta_1}=\frac{1}{4} \Rightarrow B_2=\frac{\beta_1}{4}
\end{array}$
$\beta=\frac{D \lambda}{d}$
where, $D=$ distance between source and screen
and $\quad d=$ slit width.
$\begin{array}{ll}
\therefore & \frac{\beta_2}{\beta_1}=\frac{d_1}{d_2} \times \frac{D_2}{D_1} \\
\Rightarrow & \frac{\beta_2}{\beta_1}=\frac{d_1}{2 d_1} \times \frac{D_1 / 2}{D_1} \quad\left[\because d_2=2 d_1 \text { and } D_2=\frac{D_1}{2}\right] \\
\Rightarrow & \frac{\beta_2}{\beta_1}=\frac{1}{4} \Rightarrow B_2=\frac{\beta_1}{4}
\end{array}$
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