If the frequency of the source is doubled in Young's double slit experiment, then fringe width $(β)$ will be-

unchanged

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β / 2

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2 β

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3 β

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Solution

The correct option is B $β / 2$
$\begin{matrix} f r i n g e w i d t h β = \frac{λ D}{α} A n d λ = \frac{c}{f} s o, β = \frac{C D}{f d} H e n c e, β \propto \frac{1}{f} \end{matrix}$
If frequency double, fringe width $(β)$ is halved
Option $B$ is correct.

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