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(D) For a single slit diffraction,the condition for the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda$,where $a$ is the slit width,$\lambda$

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(D) For a single slit diffraction,the condition for the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda$,where $a$ is the slit width,$\lambda$ is the wavelength,and $n = 1, 2, 3, ...$.For the second minimum $(n = 2)$,the angle is $	heta_2 = 60^{\circ}$.Thus,$a \sin 60^{\circ} = 2 \lambda$.$a \left( \frac{\sqrt{3}}{2} ight) = 2 \lambda \implies \frac{\lambda}{a} = \frac{\sqrt{3}}{4}$.For the first minimum $(n = 1)$,the condition is $a \sin 	heta_1 = 1 \lambda$.$\sin 	heta_1 = \frac{\lambda}{a} = \frac{\sqrt{3}}{4}$.$\sin 	heta_1 \approx \frac{1.732}{4} = 0.433$.$	heta_1 = \arcsin(0.433) \approx 25.6^{\circ}$.Rounding to the nearest integer,$	heta_1 \approx 25^{\circ}$.

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