First diffraction minima due to a single slit of width $1.0 × 10^{–5} cm$ is at 30°. Then wavelength of light used is
Answer & explanation
Correct answer: option 2
As $a\, \sin θ = nλ$ or $λ=\frac{a\, \sin θ}{n}$
or $λ=1.0×10^{-5}×\sin 30°=1.0×10^{-5}×\frac{1}{2}$
$= 0.5 × 10^{–5} = 5 × 10^{–6} cm$
$=500 Å$