Light of wavelength 6328 Å is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be

0.9° 0.18° 0.54° 0.36°

0.36°

$a\,\sin θ=nλ$ $\sin θ=\frac{nλ}{a}=\frac{6328×10^{-10}×1}{0.2×10^{-3}}$ $= 3164 × 10^{–6}\, rad = 0.003164\, radian$ ∴ Angular width = 2θ $=2×0.003164×\frac{180°}{π}=0.36°$