Explanation

(c) Let $\lambda$ be wavelength of monochromatic light, d the distance between coherent sources, and D the distance between screen and source, then fringe width is
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$\beta = \frac{D\lambda}{d}$
Given, $d_{1} = d,\ D_{1} = D,\ d_{2} = 10d,\ D_{2} = \frac{D}{2}$
$\therefore\ \beta_{2} = \frac{\frac{D}{2}\lambda}{10d} = \frac{D\lambda}{20d}$
$\Rightarrow \beta_{2} = \frac{\beta}{20}$