Question

In a Fraunhoffer diffraction at single slit of width 'd' with incident light of wavelength $5500\mathring{A}$, the first minimum is observed, at angle ${30}^{\circ }$. The first secondary maximum is observed at an angle $\theta =$

• A. $si{n}^{-1}\frac{1}{\sqrt{2}}$
• B. $si{n}^{-1}\frac{1}{4}$
• C. $si{n}^{-1}\frac{3}{4}$
• D. $si{n}^{-1}\frac{\sqrt{3}}{2}$

Answer 1

Answer: (C)

$si{n}^{-1}\frac{3}{4}$

First minimum occurs at $\theta$ then $\sin \theta =\frac{\lambda }{a}$

$\therefore \sin {30}^{\circ }=\frac{\lambda }{a}$

$\Rightarrow \frac{1}{2}=\frac{\lambda }{a}$ ........$\left(1\right)$

First secondary maxima occurs at $\theta$ then $\sin {\theta }^{\prime }=\{n+\left(\frac{1}{2}\right)\}\left(\frac{\lambda }{a}\right)$ where $n=1$

$\therefore \sin {\theta }^{\prime }=\{1+\left(\frac{1}{2}\right)\}\left(\frac{\lambda }{a}\right)$

$=\left(\frac{3}{2}\right)\times \left(\frac{1}{2}\right)=\left(\frac{3}{4}\right)$

${\theta }^{\prime }={\sin }^{-1}\left(\frac{3}{4}\right)$