Light of wavelength 6000 Å is normally incident on a slit. Angular position of second minimum from central maximum us 30°. Width of the slit should be
Answer & explanation
Correct answer: option 3
$a\,\sin θ=nλ$
$a=\frac{nλ}{\sin θ}=\frac{2×6000×10^{-10}}{\sin 30°}$
$=\frac{2×6000×10^{-10}}{1/2}=4×6000×10^{-10}$
$2410^{-10}10^{-7}m=24× 10^{–5} cm$