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(C) The angular resolution limit of the human eye is given by the Rayleigh criterion: $	heta = \frac{1.22 \lambda}{d}$.Here,$\lambda = 500 	imes 10^

Vedclass · Accepted Answer

$50$

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(C) The angular resolution limit of the human eye is given by the Rayleigh criterion: $	heta = \frac{1.22 \lambda}{d}$.Here,$\lambda = 500 	imes 10^{-9} \, m$ is the wavelength of light,$d = 5 	imes 10^{-3} \, m$ is the diameter of the pupil,and $r = 400 	imes 10^{3} \, m$ is the distance (altitude).The linear resolution $x$ is given by $x = r 	heta = \frac{1.22 \lambda r}{d}$.Substituting the values:$x = \frac{1.22 	imes (500 	imes 10^{-9} \, m) 	imes (400 	imes 10^{3} \, m)}{5 	imes 10^{-3} \, m}$.$x = \frac{1.22 	imes 5 	imes 10^{-7} 	imes 4 	imes 10^{5}}{5 	imes 10^{-3}}$.$x = \frac{1.22 	imes 20 	imes 10^{-2}}{5 	imes 10^{-3}} = \frac{24.4 	imes 10^{-2}}{5 	imes 10^{-3}} = 4.88 	imes 10^{1} \approx 50 \, m$.

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