Question

Light of wavelength $550\, nm$ falls normally on a slit of width $22.0 \times 10^{-5} \, cm$. The angular position of the second minima from the central maximum will be (in radians) :

• A. $\frac{\pi}{12}$
• B. $\frac{\pi}{8}$
• C. $\frac{\pi}{6}$
• D. $\frac{\pi}{4}$

Answer 1

Answer: (C)

$\frac{\pi}{6}$

Answer 2

Given: wavelength of light $=550\, nm$;
width of slit $=22.0 \times 10^{-5}\, cm$
We know that $n \lambda=d \sin \theta$; where $n$ is 2 (for second minima), $\lambda$ is wavelength, $d$ is width of slit.
Therefore, $\sin \theta=\frac{n \lambda}{d}$
$\Rightarrow \theta=\sin ^{-1}\left(\frac{n \lambda}{d}\right)$
Substituting the values, we get
$\theta=\sin ^{-1}\left(\frac{2 \times 550 \times 10^{-9} m }{22.0 \times 10^{-5} cm }\right)$
$=\sin ^{-1}\left(\frac{2 \times 550 \times 10^{-9} m }{22.0 \times 10^{-5} \times 10^{-2} m }\right)$
$\theta=\sin ^{-1}\left(\frac{1}{2}\right)$
$\Rightarrow \theta=\frac{\pi}{6}$
Thus, angular position of second minima from central maximum will be $\pi / 6$.