Visible light of wavelength $6000 \times 10^{-8} \mathrm{~cm}$ falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at $60^{\circ}$ from the central maximum. If the first minimum is produced at $\theta_{1}$, then $\theta_{1}$ is close to:
Correct Option: , 3
(3) Given, $\lambda=6000 \times 10^{-8} \mathrm{~cm}$
Second diffraction minimum at $60^{\circ}$ i.e., $\theta_{2}=60^{\circ}$
Using, $\mathrm{d} \sin \theta=\mathrm{n} \lambda$
$\mathrm{d} \sin \theta_{2}=2 \lambda$ (for $2 \mathrm{nd}$ minima)
$\Rightarrow d \sin 60^{\circ}=2 \lambda$
$\Rightarrow d \times\left(\frac{\sqrt{3}}{2}\right)=2 \lambda$ $\ldots$ (i)
$\Rightarrow \frac{\lambda}{d}=\frac{\sqrt{3}}{4}$
For first minima, $d \sin \theta_{1}=\lambda$
$\Rightarrow \sin \theta_{1}=\frac{\lambda}{d}=\frac{\sqrt{3}}{4}=0.43 \Rightarrow \theta_{1}<30^{\circ}$
Hence closest option, $\theta_{1} \approx 25^{\circ}$
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