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
Answer & explanation
Correct answer: option 4
$a\,\sin θ=nλ$
$\sin θ=\frac{nλ}{a}=\frac{6328×10^{-10}×1}{0.2×10^{-3}}$
$= 3164 × 10^{–6}\, rad = 0.003164\, radian$
∴ Angular width = 2θ
$=2×0.003164×\frac{180°}{π}=0.36°$