In Fraunhofer diffraction by a single slit,th | vedclass

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(D) For the $n$-th order secondary maxima in single-slit diffraction,the condition is given by:$a \sin 	heta = (n + \frac{1}{2}) \lambda$Given:Slit w

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$0.0125$

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(D) For the $n$-th order secondary maxima in single-slit diffraction,the condition is given by:$a \sin 	heta = (n + \frac{1}{2}) \lambda$Given:Slit width $a = 0.01 \ cm = 10^{-4} \ m$Wavelength $\lambda = 5000 \ \mathring{A} = 5 	imes 10^{-7} \ m$Order of maxima $n = 2$Substituting the values:$\sin 	heta = (2 + \frac{1}{2}) \frac{\lambda}{a}$$\sin 	heta = \frac{2.5 	imes 5 	imes 10^{-7}}{10^{-4}}$$\sin 	heta = 12.5 	imes 10^{-3} = 0.0125$Since $	heta$ is very small,$\sin 	heta \approx 	heta$.Therefore,$	heta = 0.0125 \ 	ext{rad}$.

In Fraunhofer diffraction by a single slit,the width of the slit is $0.01 \ cm$. If the wavelength of the light incident normally on the slit is $5000 \ \mathring{A}$,the angular distance of the second maxima from the midline of the central maxima is . . . . . . $\text{rad}$.

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