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(D) For a single slit diffraction,the condition for the $n^{th}$ minima is given by $a \sin 	heta = n \lambda$.Here,$a = 10^{-3} \ mm = 10^{-6} \ m$,

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$\sin^{-1}\left(\frac{1}{2}\right)$

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(D) For a single slit diffraction,the condition for the $n^{th}$ minima is given by $a \sin 	heta = n \lambda$.Here,$a = 10^{-3} \ mm = 10^{-6} \ m$,$\lambda = 5 	imes 10^{-7} \ m$,and for the first minima,$n = 1$.Substituting the values,we get $\sin 	heta = \frac{n \lambda}{a} = \frac{1 	imes 5 	imes 10^{-7} \ m}{10^{-6} \ m}$.$\sin 	heta = \frac{5 	imes 10^{-7}}{10 	imes 10^{-7}} = 0.5 = \frac{1}{2}$.Therefore,the angle of diffraction is $	heta = \sin^{-1}\left(\frac{1}{2}ight)$.

$A$ parallel beam of monochromatic light of wavelength $5 \times 10^{-7} \ m$ is incident normally on a single narrow slit of width $10^{-3} \ mm$. At what angle of diffraction is the first minima observed?

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