Question

In a Fraunhofer diffraction, at single slit of width $a$ with incident light of wavelength $5500\mathring{\text{A}}$, the first minimum is observed at an angle of ${30}^{\circ }$. The first secondary maximum will be observed at an angle $\theta$, equal to,

• A. ${45}^{\circ }$
• B. ${\sin }^{-1}\left(\frac{1}{4}\right)$
• C. ${\sin }^{-1}\left(\frac{3}{4}\right)$
• D. ${60}^{\circ }$

Answer 1

The correct option is C ${\sin }^{-1}\left(\frac{3}{4}\right)$

Slit width $=d$
$\lambda =5500\mathring{\text{A}}=5.5\times {10}^{-7}\text{m},\theta ={30}^{\circ }$

For first secondary minima, $d\sin {\theta }_n=\lambda$
$d=\frac{\lambda }{\sin \theta }=\frac{5.5\times {10}^{-7}}{\sin {30}^{\circ }}=11\times {10}^{-7}\text{m}$

For first secondary maxima, $d\sin {\theta }_n=\frac{3\lambda }{2}$
i.e. $\sin {\theta }_n=\frac{3\lambda }{2a}=\frac{3\times 5.5\times {10}^{-7}}{2\times 11\times {10}^{-7}}$

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

Hence, $\left(C\right)$ is the correct answer.