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(B) The position of the $n^{th}$ minima in a single-slit diffraction pattern is given by $x_n = \frac{n D \lambda}{a}$,where $a$ is the slit width.Giv

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$0.230$

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(B) The position of the $n^{th}$ minima in a single-slit diffraction pattern is given by $x_n = \frac{n D \lambda}{a}$,where $a$ is the slit width.Given: $D = 50\,cm = 0.5\,m$,$\lambda = 690\,nm = 690 	imes 10^{-9}\,m$,and the distance between the first $(n=1)$ and third $(n=3)$ minima is $\Delta x = x_3 - x_1 = 3.00\,mm = 3 	imes 10^{-3}\,m$.Using the formula: $\Delta x = (3 - 1) \frac{D \lambda}{a} = \frac{2 D \lambda}{a}$.Rearranging for $a$: $a = \frac{2 D \lambda}{\Delta x}$.Substituting the values: $a = \frac{2 	imes 0.5 	imes 690 	imes 10^{-9}}{3 	imes 10^{-3}} = \frac{690 	imes 10^{-9}}{3 	imes 10^{-3}} = 230 	imes 10^{-6}\,m = 0.230\,mm$.

$A$ screen is placed at a distance $50\,cm$ from a single slit,which is illuminated with light of wavelength $690\,nm$. If the distance between the first and third minima is $3.00\,mm$,what is the width of the slit (in $,mm$)?

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