Light of wavelength λ is incident on a slit. First minima of the diffraction pattern is found to lie at a distance of 6 mm from the central maximum on a screen placed at a distance of 2 m from the slit. If slit width is 0.2 mm, then wavelength of the light used will be
Answer & explanation
Correct answer: option 2
$a\,\sin θ=nλ$
$a.\frac{x}{D}=nλ$ or $λ=\frac{ax}{nD}$
or $λ=\frac{2×10^{-4}×6×10^{-3}}{1×2}=6000Å$