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In a Fraunhofer diffraction at single slit of width $d$ with incident light of wavelength 5500 $Å$, the first minimum is observed, at angle $30^{\circ}$. The first secondary maximum is observed at an angle $\theta=$
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The correct answer is:
$\sin ^{-1} \frac{3}{4}$
Slit width $=d$
$\lambda=5500 Å=5.5 \times 10^{-7} \mathrm{~m}, \theta_n=30^{\circ}$
For first secondary minima, $d \sin \theta_n=\lambda$
$d=\frac{\lambda}{\sin \theta_n}=\frac{5.5 \times 10^{-7}}{\sin 30^{\circ}}=11 \times 10^{-7} \mathrm{~m}$
For first secondary maxima, $d \sin \theta_n=\frac{3 \lambda}{2}$
i. e. $\sin \theta_n=\frac{3 \lambda}{2 d}=\frac{3 \times 5.5 \times 10^{-7}}{2 \times 11 \times 10^{-7}}$
$\sin \theta_n=\frac{3}{4}$ or $\theta_n=\sin ^{-1}(3 / 4)$
$\lambda=5500 Å=5.5 \times 10^{-7} \mathrm{~m}, \theta_n=30^{\circ}$
For first secondary minima, $d \sin \theta_n=\lambda$
$d=\frac{\lambda}{\sin \theta_n}=\frac{5.5 \times 10^{-7}}{\sin 30^{\circ}}=11 \times 10^{-7} \mathrm{~m}$
For first secondary maxima, $d \sin \theta_n=\frac{3 \lambda}{2}$
i. e. $\sin \theta_n=\frac{3 \lambda}{2 d}=\frac{3 \times 5.5 \times 10^{-7}}{2 \times 11 \times 10^{-7}}$
$\sin \theta_n=\frac{3}{4}$ or $\theta_n=\sin ^{-1}(3 / 4)$
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