Two luminous point sources seperated by a certain distance are at 10 km from an observer. If the aperture of his eyes is `2.5xx10^(-3)m` and the wavelength of light used is 500 nm, the distance of seperation between the point sources just seen to be resolved is

Options

A. 12.2 m
B. 24.4 m
C. 2.44 m
D. 1.22 m

Answer

C

Explanation

According to Rayleigh's criterion,
`theta=(1.22lamda)/(d_(e))`
Where `lamda=` wavelength of light `d_(e)=` diameter of the pupil of the eye
`theta=(1.22xx500xx10^(-9))/(2.5xx10^(-3))=2.44xx10^(-4)` radian

`:.thata=a/D`
or `a=D xx theta=10xx10^(3)xx2.44xx10^(-4)=2.44m`