Explanation

(c) Let S be a slit illuminated by monochromatic light of wavelength$(\lambda)$, let $S_{1},\ S_{2}$ be coherent sources and distance between them be d and distance between source and screen is D. then, fringe width (W) is given by
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$W = \frac{D\lambda}{d}$
When, $d_{2} = \frac{d}{2},D_{2} = 2D$
$\therefore W_{2} = \frac{(2D)\lambda}{d/2} = 4\frac{D\lambda}{d} = 4W$
The fringe width is quadrupled.