Light of wavelength 5000 Å is incident normally on a slit. First minimum of diffraction pattern is formed at a distance of 5 mm from the central maximum. If slit width is 0.2 mm, then distance between slit and screen will be
Answer & explanation
Correct answer: option 3
$a\,\sin θ=nλ$
or $a.\frac{x}{d}=nλ$
$D=\frac{ax}{nλ}=\frac{2×10^{-4}×5×10^{-3}}{1×5000×10^{-10}}=2.0\,metre$