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(D) The condition for minima in a single-slit diffraction pattern is given by $a \sin 	heta = n \lambda$, where $a$ is the slit width, $\lambda$ is t

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$0.2$

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(D) The condition for minima in a single-slit diffraction pattern is given by $a \sin 	heta = n \lambda$, where $a$ is the slit width, $\lambda$ is the wavelength, and $n = 1, 2, 3, ...$ For small angles, $\sin 	heta \approx 	an 	heta = \frac{y_n}{D}$, where $y_n$ is the position of the $n^{th}$ minimum and $D$ is the distance to the screen. Thus, $y_n = \frac{n \lambda D}{a}$. The distance between the first $(n=1)$ and third $(n=3)$ minima is $\Delta y = y_3 - y_1 = \frac{3 \lambda D}{a} - \frac{1 \lambda D}{a} = \frac{2 \lambda D}{a}$. Given: $\Delta y = 3 \,mm = 3 	imes 10^{-3} \,m$, $D = 0.5 \,m$, and $\lambda = 6000 	ext{ Å} = 6000 	imes 10^{-10} \,m = 6 	imes 10^{-7} \,m$. Substituting the values: $3 	imes 10^{-3} = \frac{2 	imes (6 	imes 10^{-7}) 	imes 0.5}{a}$. $a = \frac{2 	imes 6 	imes 10^{-7} 	imes 0.5}{3 	imes 10^{-3}} = \frac{6 	imes 10^{-7}}{3 	imes 10^{-3}} = 2 	imes 10^{-4} \,m$. Converting to $mm$: $a = 0.2 \,mm$.

$A$ screen is placed $0.5 \,m$ away from a single slit which is illuminated by a monochromatic light of wavelength $6000 \text{ Å}$. If the distance between the first and third minima in the diffraction pattern on the screen is $3 \,mm$, then the slit width is: (in $\,mm$)

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