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In Young's double slit experiment the distance between the slits and the screen is doubled. The separation between the slits reduced to half. As a result the fringe width:
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becomes four times
In Young's double slit experiment fringe width
$$
\beta=\frac{\lambda D}{d}
$$
where $D=$ distance between slit and screen $d=$ distance between slit According to the question $D$ is double $\rightarrow 2 D$ and $d$ is half $\rightarrow$ $\frac{d}{2}$
Ther New fringe wilth will be $\frac{\lambda \times 2 D}{\frac{d}{2}}=\frac{4 \lambda D}{d}=4\left(\frac{\lambda D}{d}\right)=\beta$
$\beta^1=4 \beta$ i.e., becomes four times
$$
\beta=\frac{\lambda D}{d}
$$
where $D=$ distance between slit and screen $d=$ distance between slit According to the question $D$ is double $\rightarrow 2 D$ and $d$ is half $\rightarrow$ $\frac{d}{2}$
Ther New fringe wilth will be $\frac{\lambda \times 2 D}{\frac{d}{2}}=\frac{4 \lambda D}{d}=4\left(\frac{\lambda D}{d}\right)=\beta$
$\beta^1=4 \beta$ i.e., becomes four times
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