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(B) For a single slit diffraction pattern,the position of the $n^{th}$ minimum is given by $y_n = \frac{n \lambda D}{a}$,where $n$ is the order of the

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$2.5$

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(B) For a single slit diffraction pattern,the position of the $n^{th}$ minimum is given by $y_n = \frac{n \lambda D}{a}$,where $n$ is the order of the minimum,$\lambda$ is the wavelength,$D$ is the distance to the screen,and $a$ is the slit width.The distance between the $1^{st}$ and $6^{th}$ minima is $\Delta y = y_6 - y_1 = \frac{6 \lambda D}{a} - \frac{1 \lambda D}{a} = \frac{5 \lambda D}{a}$.Given: $\Delta y = 0.5 \, mm = 0.5 	imes 10^{-3} \, m$,$D = 0.5 \, m$,and $\lambda = 5000 \, \mathring{A} = 5000 	imes 10^{-10} \, m = 5 	imes 10^{-7} \, m$.Substituting the values into the formula:$0.5 	imes 10^{-3} = \frac{5 	imes (5 	imes 10^{-7}) 	imes 0.5}{a}$Solving for $a$:$a = \frac{5 	imes 5 	imes 10^{-7} 	imes 0.5}{0.5 	imes 10^{-3}}$$a = 25 	imes 10^{-4} \, m = 2.5 	imes 10^{-3} \, m = 2.5 \, mm$.

The distance between the first and the sixth minima in the diffraction pattern of a single slit is $0.5 \, mm$. The screen is $0.5 \, m$ away from the slit. If the wavelength of light used is $5000 \, \mathring{A}$,then the slit width will be ....... $mm$.

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