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(C) The width of the central diffraction maximum is given by $W = \frac{2 \lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance to the

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$3 Y$

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(C) The width of the central diffraction maximum is given by $W = \frac{2 \lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance to the screen,and $d$ is the slit width.Initially,$W = Y = \frac{2 	imes 400 	imes D}{d}$.When half the slit is covered,the new slit width $d' = \frac{d}{2}$.The new wavelength $\lambda' = 600 \ nm$.The new width $W'$ is given by $W' = \frac{2 \lambda' D}{d'} = \frac{2 	imes 600 	imes D}{d/2} = \frac{4 	imes 600 	imes D}{d}$.Taking the ratio: $\frac{W'}{Y} = \frac{4 	imes 600 	imes D / d}{2 	imes 400 	imes D / d} = \frac{2400}{800} = 3$.Therefore,$W' = 3 Y$.

$A$ single slit of width $d$ is illuminated by violet light of wavelength $400 \ nm$ and the width of the diffraction pattern is measured as $Y$. When half of the slit width is covered and illuminated by yellow light of wavelength $600 \ nm$,the width of the diffraction pattern is

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