Two luminous point sources separated by a cer | vedclass

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(C) According to Rayleigh's criterion, the angular resolution limit is given by $	heta = \frac{1.22 \lambda}{d_e}$, where $\lambda$ is the wavelength

Vedclass · Accepted Answer

$2.44$

Vedclass · Answer

(C) According to Rayleigh's criterion, the angular resolution limit is given by $	heta = \frac{1.22 \lambda}{d_e}$, where $\lambda$ is the wavelength of light and $d_e$ is the diameter of the pupil of the eye.Given: $\lambda = 500 \,nm = 500 	imes 10^{-9} \,m$, $d_e = 2.5 	imes 10^{-3} \,m$, and distance $D = 10 \,km = 10^4 \,m$.Substituting the values:$	heta = \frac{1.22 	imes 500 	imes 10^{-9}}{2.5 	imes 10^{-3}} = 2.44 	imes 10^{-4} \,rad$.The separation distance $a$ is related to the angular resolution by $	heta = \frac{a}{D}$.Therefore, $a = D 	imes 	heta = 10^4 \,m 	imes 2.44 	imes 10^{-4} \,rad = 2.44 \,m$.

Two luminous point sources separated by a certain distance are at $10 \,km$ from an observer. If the aperture of his eye is $2.5 \times 10^{-3} \,m$ and the wavelength of light used is $500 \,nm$, the distance of separation between the point sources just seen to be resolved is (in $\,m$)

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