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 $\sin \theta=\frac{3 \lambda}{2 d}=\frac{3}{2}$
$\left(\frac{400 \times 10^{-9}}{800 \times 10^{-9}}\right)=\frac{3}{4}$
$\theta=\sin ^{-1}\left(\frac{3}{4}\right)$