Question

In a Fraunhofer diffraction experiment at a single slit using a light of wavelength $400nm$, the first minimum is formed at an angle of ${30}^o$. The direction $\theta$ of the first secondary maximum is given by

• A. ${\sin }^{-1}\frac{2}{3}$
• B. ${\sin }^{-1}\frac{3}{4}$
• C. ${\sin }^{-1}\frac{1}{4}$
• D. ${\tan }^{-1}\frac{2}{3}$

Answer 1

Answer: (C)

${\sin }^{-1}\frac{1}{4}$

Given : $\lambda =400nm$

For minima in diffraction $b\sin \theta =n\lambda$

First minima i.e. $n=1$ is obtained at $\theta ={30}^o$

$\therefore$ $b\sin {30}^o=1\times 400$

$⟹$ $b=800nm$

For maxima in diffraction $b\sin \theta =(n+\frac{1}{2})\lambda$

Let first maxima i.e. $n=0$ is obtained at $\theta$.

$\therefore$ $b\sin \theta =\frac{\lambda }{2}$

Or $800\times \sin \theta =\frac{400}{2}$

Or $\sin \theta =\frac{1}{4}$

$⟹$ $\theta =si{n}^{-1}(\frac{1}{4})$