Q. 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
Explanation
(c) According to Rayleigh’s criterion,
$\theta = 1.22\lambda/d_{e}$
[image:10_image44.jpeg]
Where $\lambda$ = wavelength of light,
$d_{e} =$ diameter of the pupil of the eye.
$\therefore\ \theta = \frac{1.22 \times 500 \times 10^{- 9}}{2.5 \times 10^{- 3}} = 2.44 \times 10^{- 4}\ rad$
But $\theta = \frac{a}{D}$
$\therefore$ distance of separation,
$a = D \times \theta = 10 \times 10^{3} \times 2.44 \times 10^{- 4} = 2.44\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.