Q 12-10-019NEETJEE MainMedium
In single-slit diffraction with light of wavelength $600$ nm, the first minimum is at an angle of $30°$. The width of the slit is
Answer: (D) $1.2\ \mu$m
First minimum: $a\sin\theta = \lambda \Rightarrow a = \dfrac{600 \times 10^{-9}}{0.5} = 1.2\ \mu$m.
Solution by Sreeraj P, M.Sc Physics