In a Fraunhofer diffraction at a single slit | vedclass

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(D) For a single slit diffraction,the condition for minima is $d \sin 	heta = n\lambda$ where $n = 1, 2, 3, ...$For the first minimum $(n=1)$,$d \sin

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$\sin ^{-1}\left(\frac{3}{4}\right)$

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(D) For a single slit diffraction,the condition for minima is $d \sin 	heta = n\lambda$ where $n = 1, 2, 3, ...$For the first minimum $(n=1)$,$d \sin 30^{\circ} = \lambda$.Since $\sin 30^{\circ} = 0.5$,we have $d(0.5) = \lambda$,which implies $d = 2\lambda$.The condition for secondary maxima is $d \sin 	heta = (n + \frac{1}{2})\lambda$ where $n = 1, 2, 3, ...$For the first secondary maximum $(n=1)$,$d \sin 	heta = (1 + \frac{1}{2})\lambda = \frac{3}{2}\lambda$.Substituting $d = 2\lambda$ into the equation:$2\lambda \sin 	heta = \frac{3}{2}\lambda$$\sin 	heta = \frac{3}{4}$$	heta = \sin ^{-1}\left(\frac{3}{4}ight)$.

In a Fraunhofer diffraction at a single slit of width $d$ and incident light of wavelength $5500 \text{ Å}$,the first minimum is observed at an angle $30^{\circ}$. The first secondary maxima are observed at an angle $\theta=$

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