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(A) For a single slit diffraction, the condition for $n^{th}$ order minima is given by $b \sin 	heta = n \lambda$.For small angles, $\sin 	heta \app

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$6$

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(A) For a single slit diffraction, the condition for $n^{th}$ order minima is given by $b \sin 	heta = n \lambda$.For small angles, $\sin 	heta \approx 	heta$, so $	heta = \frac{n \lambda}{b}$.Given: Wavelength $\lambda = 600 \,nm = 600 	imes 10^{-9} \,m$, Slit width $b = 0.4 \,mm = 4 	imes 10^{-4} \,m$, and order $n = 2$.The angular position of the second-order minimum is $	heta = \frac{2 	imes 600 	imes 10^{-9}}{4 	imes 10^{-4}} = 3 	imes 10^{-3} \,rad$.The total angular divergence between the second-order minima on both sides of the central maximum is $2	heta$.Total divergence $= 2 	imes (3 	imes 10^{-3} \,rad) = 6 	imes 10^{-3} \,rad$.

$A$ parallel beam of monochromatic light of wavelength $600 \,nm$ passes through a single slit of $0.4 \,mm$ width. The angular divergence corresponding to the second-order minima would be $...... \times 10^{-3} \,rad$.

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