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(C) The width of the central maximum in a single-slit diffraction pattern is given by $y = \frac{2 \lambda D}{a}$,where $\lambda$ is the wavelength,$D

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$3y$

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(C) The width of the central maximum in a single-slit diffraction pattern is given by $y = \frac{2 \lambda D}{a}$,where $\lambda$ is the wavelength,$D$ is the distance to the screen,and $a$ is the slit width.Initially,$y = \frac{2 \lambda_1 D}{a_1}$,where $\lambda_1 = 400 \ nm$ and $a_1 = a$.In the second case,the new slit width is $a_2 = a/2$ and the new wavelength is $\lambda_2 = 600 \ nm$.The new width $y'$ is given by $y' = \frac{2 \lambda_2 D}{a_2} = \frac{2 	imes 600 	imes D}{a/2} = \frac{4 	imes 600 	imes D}{a} = \frac{2400 D}{a}$.Comparing this with the initial width $y = \frac{800 D}{a}$,we get $\frac{y'}{y} = \frac{2400}{800} = 3$.Therefore,$y' = 3y$.

$A$ single slit of width $a$ 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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