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(C) According to Rayleigh's criterion,the angular resolution $\Delta 	heta$ of a circular aperture of diameter $D$ is given by $\Delta 	heta = 1.22

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$6$

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(C) According to Rayleigh's criterion,the angular resolution $\Delta 	heta$ of a circular aperture of diameter $D$ is given by $\Delta 	heta = 1.22 \frac{\lambda}{D}$.The angular separation between two objects at a distance $z$ separated by a distance $s$ is $\Delta 	heta = \frac{s}{z}$.Equating the two expressions for $\Delta 	heta$,we get $\frac{s}{z} = 1.22 \frac{\lambda}{D}$.Rearranging to solve for $D$,we get $D = 1.22 \frac{\lambda z}{s}$.Given values: $\lambda = 600 	imes 10^{-9} \ m$,$z = 10 	imes 10^3 \ m$,$s = 0.12 \ m$.Substituting these values: $D = 1.22 	imes \frac{600 	imes 10^{-9} 	imes 10^4}{0.12} = 1.22 	imes \frac{6 	imes 10^{-3}}{0.12} = 1.22 	imes 0.05 \ m = 0.061 \ m$.Converting to centimeters: $D = 0.061 	imes 100 \ cm = 6.1 \ cm \approx 6 \ cm$.

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