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(C) For the first minima in single slit diffraction,the condition is given by $a \sin 	heta = n \lambda$. For the first minima,$n = 1$,so $a \sin 	h

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$1.3 \; \mu m$

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(C) For the first minima in single slit diffraction,the condition is given by $a \sin 	heta = n \lambda$. For the first minima,$n = 1$,so $a \sin 	heta = \lambda$. Given $\lambda = 6500 \; \mathring{A} = 6500 	imes 10^{-10} \; m$ and $	heta = 30^\circ$. Substituting the values: $a \sin(30^\circ) = 6500 	imes 10^{-10} \; m$. Since $\sin(30^\circ) = 0.5$,we have $a(0.5) = 6500 	imes 10^{-10} \; m$. $a = 2 	imes 6500 	imes 10^{-10} \; m = 13000 	imes 10^{-10} \; m = 1.3 	imes 10^{-6} \; m$. Since $1 \; \mu m = 10^{-6} \; m$,we get $a = 1.3 \; \mu m$.

$A$ slit of width $a$ is illuminated by white light. For red light $(\lambda = 6500 \; \mathring{A})$,the first minima is obtained at $\theta = 30^\circ$. Then the value of $a$ will be

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