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(B) Given: Slit separation $d = 1 \,mm = 10^{-3} \,m$, screen distance $D = 1 \,m$, wavelength $\lambda = 500 \,nm = 5 	imes 10^{-7} \,m$.The width o

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$2$

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(B) Given: Slit separation $d = 1 \,mm = 10^{-3} \,m$, screen distance $D = 1 \,m$, wavelength $\lambda = 500 \,nm = 5 	imes 10^{-7} \,m$.The width of the central maximum of a single slit diffraction pattern is given by $w = \frac{2 \lambda D}{a}$, where $a$ is the width of each slit.The fringe width of the double slit interference pattern is $\beta = \frac{\lambda D}{d}$.We are given that $10$ maxima of the double slit pattern fit within the central maximum of the single slit pattern. The central maximum spans from the first minimum on one side to the first minimum on the other, covering a width of $2 \beta_{total}$ or simply the range of $10$ fringes. Specifically, the condition is $10 	imes \beta = \frac{2 \lambda D}{a}$.Substituting $\beta = \frac{\lambda D}{d}$:$10 	imes \frac{\lambda D}{d} = \frac{2 \lambda D}{a}$Simplifying the equation:$\frac{10}{d} = \frac{2}{a}$$a = \frac{2d}{10} = \frac{d}{5}$Given $d = 1 \,mm = 10 	imes 10^{-4} \,m$:$a = \frac{10 	imes 10^{-4} \,m}{5} = 2 	imes 10^{-4} \,m$.Thus, the value is $2$.

Two slits are $1 \,mm$ apart and the screen is located $1 \,m$ away from the slits. $A$ light of wavelength $500 \,nm$ is used. The width of each slit to obtain $10$ maxima of the double slit pattern within the central maximum of the single slit pattern is $\ldots \ldots \ldots \times 10^{-4} \,m$.

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