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Question

(A) The angular resolution of the human eye is given by $\Delta 	heta = \frac{1.22 \lambda}{D}$,where $D$ is the diameter of the pupil.Given radius $

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$30$

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(A) The angular resolution of the human eye is given by $\Delta 	heta = \frac{1.22 \lambda}{D}$,where $D$ is the diameter of the pupil.Given radius $r = 0.25 \ cm$,so diameter $D = 2r = 0.50 \ cm = 5 	imes 10^{-3} \ m$.Wavelength $\lambda = 500 \ nm = 500 	imes 10^{-9} \ m = 5 	imes 10^{-7} \ m$.The angular resolution is $\Delta 	heta = \frac{1.22 	imes 5 	imes 10^{-7}}{5 	imes 10^{-3}} = 1.22 	imes 10^{-4} \ rad$.The minimum separation $d$ at a distance $L = 25 \ cm = 0.25 \ m$ is $d = L 	imes \Delta 	heta$.$d = 0.25 	imes 1.22 	imes 10^{-4} \ m = 0.305 	imes 10^{-4} \ m = 30.5 \ \mu m$.Rounding to the nearest given option,the answer is $30 \ \mu m$.

Assuming the human pupil to have a radius of $0.25 \ cm$ and a comfortable viewing distance of $25 \ cm$,the minimum separation between two objects that the human eye can resolve at $500 \ nm$ wavelength is..... $\mu m$.

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