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 Rayleighs 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,     \[a=D\times \theta =10\times {{10}^{3}}\times 2.44\times {{10}^{-4}}\]        \[=2.44\,\,m\]