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Q.
In a Young's double slit experiment the distance between slits is increased five times where as their distance from screen in halved, then the fringe width is
Wave Optics
Solution:
Fringe width, $\beta=\frac{\lambda D}{d}$ ...(i)
According to the question,
$D^{\prime}=\frac{D}{2} \text { and } d^{\prime}=5 d$
$\beta^{\prime}=\frac{D^{\prime} \lambda}{d^{\prime}}=\frac{\left(\frac{D}{2}\right) \lambda}{5 d}=\frac{1}{10} \frac{D \lambda}{d}$
$\beta^{\prime}=\frac{\beta}{10}$ [from Eq. (i)]