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(C) The condition for the resolution of two point objects by an optical system is given by the Rayleigh criterion: $	heta = \frac{1.22 \lambda}{a}$,w

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(C) The condition for the resolution of two point objects by an optical system is given by the Rayleigh criterion: $	heta = \frac{1.22 \lambda}{a}$,where $	heta$ is the angular separation,$\lambda$ is the wavelength of light,and $a$ is the diameter of the aperture (pupil).From the geometry of the problem,the angular separation is also given by $	heta = \frac{x}{d}$,where $x$ is the distance between the dots and $d$ is the distance of the observer from the dots.Equating the two expressions for $	heta$: $\frac{1.22 \lambda}{a} = \frac{x}{d}$.Rearranging to solve for $d$: $d = \frac{x \cdot a}{1.22 \lambda}$.Given values: $x = 1 \ mm = 1 	imes 10^{-3} \ m$,$a = 3 \ mm = 3 	imes 10^{-3} \ m$,$\lambda = 500 \ nm = 500 	imes 10^{-9} \ m$.Substituting these values into the formula:$d = \frac{(1 	imes 10^{-3} \ m) 	imes (3 	imes 10^{-3} \ m)}{1.22 	imes 500 	imes 10^{-9} \ m}$$d = \frac{3 	imes 10^{-6}}{610 	imes 10^{-9}} = \frac{3000}{610} \approx 4.918 \ m$.Rounding to the nearest integer,the maximum distance is approximately $5 \ m$.

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

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