Question

In a Fraunhofer 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. $\sin^{-1} \big(\frac{1}{\sqrt2}\big)$
• B. $\sin^{-1} \big(\frac{1}{4}\big)$
• C. $\sin^{-1} \big(\frac{3}{4}\big)$
• D. $\sin^{-1} \frac{\sqrt3}{2}$

Answer 1

The Correct Option is C

Condition for $n$th secondary minimum is that path difference
$= a \sin \, \theta_n = n \lambda$
nth secondary maximum is part difference
$ = a \sin \, \theta_n = (2n + 1) \frac{\lambda}{2}$
For 1st minimum, $\lambda = 5500 \mathring {A}, \theta_n = 30^\circ$
$ a \sin \, 30^\circ = \lambda ... (i)$
For 2nd maximum path difference
$ = a \sin \, \theta_n = (2n + 1)\frac{\lambda}{2} ...(ii)$
Dividing E (i) by E (ii), we get
$ \frac{\frac{1}{2}}{\sin \, \theta_n} = \frac{2}{3}.$
$\Rightarrow \sin \, \theta_n = \frac{3}{4}$
$\Rightarrow \, \theta_n = \sin^{-1} \big(\frac{3}{4}\big)$