Q 12-10-147JEE MainJEE Main 2018 (15 Apr, Shift 1)Easy
Light of wavelength $550\ \text{nm}$ falls normally on a slit of width $22.0 \times 10^{-5}\ \text{cm}$. The angular position of the second minima from the central maximum will be (in radians):
Answer: (D) $\dfrac{\pi}{6}$
Minima of single-slit diffraction: $a\sin\theta = n\lambda$.
For $n = 2$ with $a = 22.0\times10^{-5}\ \text{cm} = 2.2\times10^{-6}\ \text{m}$:
$$\sin\theta = \frac{2\times550\times10^{-9}}{2.2\times10^{-6}} = 0.5 \quad\Rightarrow\quad \theta = \frac{\pi}{6}$$
Solution by Sreeraj P, M.Sc Physics