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(B) For a single slit diffraction,the condition for the $n^{th}$ minimum is given by $d \sin 	heta = n \lambda$. For the first minimum,$n = 1$,so the

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$1.0 	imes 10^{-4} \, 	ext{cm}$

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(B) For a single slit diffraction,the condition for the $n^{th}$ minimum is given by $d \sin 	heta = n \lambda$. For the first minimum,$n = 1$,so the equation becomes $d \sin 	heta = \lambda$. Given: $\lambda = 5000 \,\mathring{A} = 5000 	imes 10^{-8} \, 	ext{cm} = 5 	imes 10^{-5} \, 	ext{cm}$ and $	heta = 30^o$. Substituting the values: $d \sin 30^o = 5 	imes 10^{-5} \, 	ext{cm}$. Since $\sin 30^o = 0.5$,we have $d 	imes 0.5 = 5 	imes 10^{-5} \, 	ext{cm}$. Therefore,$d = \frac{5 	imes 10^{-5}}{0.5} = 10 	imes 10^{-5} \, 	ext{cm} = 1.0 	imes 10^{-4} \, 	ext{cm}$.

The first diffraction minima due to a single slit diffraction is at $\theta = 30^o$ for a light of wavelength $5000 \,\mathring{A}$. The width of the slit is

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