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(D) For a single slit diffraction,the condition for the $n^{th}$ secondary maximum is given by $a \sin 	heta = (n + \frac{1}{2}) \lambda$,where $n =

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$1.95 	imes 10^{-6} \ m$

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(D) For a single slit diffraction,the condition for the $n^{th}$ secondary maximum is given by $a \sin 	heta = (n + \frac{1}{2}) \lambda$,where $n = 1, 2, 3, ...$.For the first maximum,$n = 1$,so the condition is $a \sin 	heta = \frac{3\lambda}{2}$.Given: $\lambda = 650 \ nm = 650 	imes 10^{-9} \ m$ and $	heta = 30^\circ$.Substituting the values: $a \sin(30^\circ) = \frac{3 	imes 650 	imes 10^{-9}}{2}$.Since $\sin(30^\circ) = 0.5$,we have $a(0.5) = \frac{1950 	imes 10^{-9}}{2}$.$a = \frac{1950 	imes 10^{-9}}{1} = 1.95 	imes 10^{-6} \ m$.

$A$ slit of width $a$ is illuminated by monochromatic light of wavelength $650 \ nm$. The value of $a$ when the first maximum is formed at a diffraction angle of $30^\circ$ is:

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