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(B) The angular width of the central maximum of a single slit diffraction pattern is given by $2	heta = \frac{2\lambda}{a}$,where $a$ is the slit wid

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$0.2$

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(B) The angular width of the central maximum of a single slit diffraction pattern is given by $2	heta = \frac{2\lambda}{a}$,where $a$ is the slit width.The angular separation between adjacent maxima in a double-slit interference pattern is $\Delta	heta = \frac{\lambda}{d}$,where $d$ is the distance between the slits.We are given that $10$ maxima of the double-slit pattern lie within the central maximum of the single-slit pattern. This means the angular width of the central maximum $(2\lambda/a)$ must contain $10$ interference fringes. Since the central maximum spans from $-\lambda/a$ to $+\lambda/a$,the total number of fringes $N$ is given by the condition that the width of $10$ fringes equals the width of the central maximum: $10 	imes (\frac{\lambda}{d}) = \frac{2\lambda}{a}$.Simplifying the equation: $\frac{10}{d} = \frac{2}{a}$.Given $d = 1 \; mm = 10^{-3} \; m$,we solve for $a$:$a = \frac{2d}{10} = \frac{d}{5} = \frac{1 \; mm}{5} = 0.2 \; mm$.

Two slits are made $1 \; mm$ apart and the screen is placed $1 \; m$ away. What should the width of each slit be to obtain $10$ maxima of the double-slit pattern within the central maximum of the single-slit pattern (in $; mm$)? Blue-green light of wavelength $500 \; nm$ is used.

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