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(C) The angular resolution of the eye lens is given by $	heta = \frac{1.22 \lambda}{D}$.From the geometry,$	heta \approx \sin 	heta = \frac{y}{x}$,

Vedclass · Accepted Answer

$5$

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(C) The angular resolution of the eye lens is given by $	heta = \frac{1.22 \lambda}{D}$.From the geometry,$	heta \approx \sin 	heta = \frac{y}{x}$,where $y = 1 \, mm = 10^{-3} \, m$ is the separation between the dots,$x$ is the distance from the eye,and $D = 3 \, mm = 3 	imes 10^{-3} \, m$ is the diameter of the eye lens.Equating the two expressions for $	heta$:$\frac{y}{x} = \frac{1.22 \lambda}{D}$$x = \frac{y \cdot D}{1.22 \lambda}$Substituting the values:$x = \frac{(10^{-3} \, m) 	imes (3 	imes 10^{-3} \, m)}{1.22 	imes (500 	imes 10^{-9} \, m)}$$x = \frac{3 	imes 10^{-6}}{6.1 	imes 10^{-7}} = \frac{30}{6.1} \approx 4.92 \, m \approx 5 \, m$.Thus,the maximum distance is approximately $5 \, m$.

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