Question

Light of wavelength $550nm$ 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)

Given: wavelength of light $=550nm$;
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 $\pi /6$.