Two luminous point sources separated by a certain distance are at 10 km from an observer. If the aperture of his eye is `2.5xx10^(-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

Options

A. 12.2 m
B. 24.2 m
C. 2.44 m
D. 1.22 m

Answer

C

Explanation

According to Rayleigh's criterion,
`theta=(1.22lambda)/(d_(e))`

where, `lambda=` wavelength of light,
`d_(e)=` diameter of the pupil of the eye.
`therefore theta=(1.22xx500xx10^(-9))/(2.5xx10^(-3))= 2.44 xx 10^(-4)` rad
But `theta=(a)/(D)`
`therefore` Distance of separation,
`a=D xx theta =10 xx10^(3)xx2.44 xx 10^(-4)=2.44m`