Two point white dots are 1 , mm apart on blac | vedclass

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(C) The condition for the resolution of two point objects by an optical system (like the human eye) is given by the Rayleigh criterion: $	heta = \fra

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$5$

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(C) The condition for the resolution of two point objects by an optical system (like the human eye) is given by the Rayleigh criterion: $	heta = \frac{1.22 \lambda}{d}$.Here,$	heta$ is the angular separation,$\lambda$ is the wavelength of light,and $d$ is the diameter of the pupil.Also,for small angles,the angular separation $	heta$ can be expressed as $	heta = \frac{y}{D}$,where $y$ is the linear separation between the dots and $D$ is the distance from the eye.Equating the two expressions: $\frac{y}{D} = \frac{1.22 \lambda}{d}$.Rearranging to solve for $D$: $D = \frac{y \cdot d}{1.22 \cdot \lambda}$.Given values: $y = 1 \, mm = 1 	imes 10^{-3} \, m$,$d = 3 \, mm = 3 	imes 10^{-3} \, m$,and $\lambda = 500 \, nm = 500 	imes 10^{-9} \, m$.Substituting these values: $D = \frac{1 	imes 10^{-3} 	imes 3 	imes 10^{-3}}{1.22 	imes 500 	imes 10^{-9}}$.$D = \frac{3 	imes 10^{-6}}{610 	imes 10^{-9}} = \frac{3000}{610} \approx 4.918 \, m$.Rounding to the nearest integer,we get $D \approx 5 \, m$.

Two point white dots are $1 \, mm$ apart on black paper. They are viewed by an eye with a pupil diameter of $3 \, mm$. Approximately,what is the maximum distance at which these dots can be resolved by the eye? (Take $\lambda = 500 \, nm$)

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