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(B) For diffraction at a single slit,the condition for the $n^{	ext{th}}$ minimum is given by $b \sin 	heta = n \lambda$.For small angles,$\sin 	he

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$3$

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(B) For diffraction at a single slit,the condition for the $n^{	ext{th}}$ minimum is given by $b \sin 	heta = n \lambda$.For small angles,$\sin 	heta \approx 	an 	heta = \frac{x}{D}$,where $x$ is the distance from the central maximum and $D$ is the distance of the screen.Thus,$x_n = \frac{n \lambda D}{b}$.Given for $n=2$,$x_2 = 3\,cm$ and for $n=4$,$x_4 = 6\,cm$.Using the formula: $x_n = n \left( \frac{\lambda D}{b} ight)$.For $n=2$: $3 = 2 \left( \frac{\lambda D}{b} ight) \Rightarrow \frac{\lambda D}{b} = 1.5\,cm$.For $n=4$: $6 = 4 \left( \frac{\lambda D}{b} ight) \Rightarrow \frac{\lambda D}{b} = 1.5\,cm$.The width of the central maximum is the distance between the first minima on either side,which is $w = 2x_1$.Since $x_1 = 1 \left( \frac{\lambda D}{b} ight) = 1.5\,cm$,the width of the central maximum is $w = 2 	imes 1.5 = 3\,cm$.

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