Two-point white dots are 2 ,mm apart on a bla | vedclass

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(D) The distance between the two point dots is $x = 2 \,mm = 2 	imes 10^{-3} \,m$. The diameter of the pupil is $d = 3 \,mm = 3 	imes 10^{-3} \,m$.

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) The distance between the two point dots is $x = 2 \,mm = 2 	imes 10^{-3} \,m$. The diameter of the pupil is $d = 3 \,mm = 3 	imes 10^{-3} \,m$. The wavelength of light is $\lambda = 500 \,nm = 5 	imes 10^{-7} \,m$. According to the Rayleigh criterion for the resolution of two points, the angular separation $	heta$ is given by $	heta = \frac{1.22 \lambda}{d}$. Also, the angular separation can be expressed as $	heta = \frac{x}{D}$, where $D$ is the maximum distance. Equating the two expressions: $\frac{x}{D} = \frac{1.22 \lambda}{d}$. Rearranging for $D$: $D = \frac{x d}{1.22 \lambda}$. Substituting the values: $D = \frac{(2 	imes 10^{-3} \,m) 	imes (3 	imes 10^{-3} \,m)}{1.22 	imes (5 	imes 10^{-7} \,m)}$. $D = \frac{6 	imes 10^{-6}}{6.1 	imes 10^{-7}} = \frac{60}{6.1} \approx 9.836 \,m$. Rounding to the nearest integer, we get $D \approx 10 \,m$.

Two-point white dots are $2 \,mm$ apart on a black paper. They are viewed by an eye of pupil diameter $3 \,mm$. What is the maximum distance at which these dots can be resolved by the eye (in $\,m$)? $(\lambda = 500 \,nm)$

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