Explanation

(d) Let $\lambda$ be wavelength of monochromatic light, used to illuminate the slit S, and d be the distance between coherent sources, then width of slits is given by
$W = \frac{D\lambda}{d}$
When D is distance between screen and source.
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Given, $d = 3\ mm,\ \lambda = 5000\ \mathring{\mathrm{A}} = 5 \times 10^{- 7}\ m\ $
$= 5 \times 10^{- 4}\ mm$
$D = 90\ cm = 900\ mm$
$\therefore W = \frac{5 \times 10^{- 4} \times 900}{3}$
$= 15 \times 10^{- 2}\ mm = 0.15\ mm$