Question

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

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

Answer 1

Answer: (C)

For first minima $\sin \theta =\frac{n}{d}$
$\Rightarrow \sin 30^{\circ }=\frac{400\times 10^{-9}m}{d}$
or $d=\frac{400\times 10^{-9}m}{(0.5)}=800\times 10^{-9}m$.
Again $\sin \theta =\frac{3\lambda }{2d}=\frac{3}{2}$
$\left(\frac{400\times 10^{-9}}{800\times 10^{-9}}\right)=\frac{3}{4}$
$\theta ={\sin }^{-1}\left(\frac{3}{4}\right)$