question_answer

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

A)  12.2 m                 

B)  24.2 m

C)  2.44 m                                 

D)  1.22 m

Correct Answer: C

Solution :

According to Rayleigh's criterion,                 \[\theta =\frac{1.22\lambda }{{{d}_{e}}}\] where \[\lambda =\] wavelength of light, \[{{d}_{e}}=\] diameter of the pupil of the eye \[\therefore \]    \[\theta =\frac{1.22\times 500\times {{10}^{-9}}}{2.5\times {{10}^{-3}}}=2.44\times {{10}^{-4}}\]radian But         \[\theta =\frac{a}{D}\] \[\therefore \] Distance of separation,                 \[a=D\times \theta =10\times {{10}^{3}}\times 2.44\times {{10}^{-4}}\]                 \[=2.44\,m\]