Question

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 are just seen to be resolved is

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

Answer 1

Answer: (B)

2.44 m

Answer 2

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 $\,\,\,\theta=\frac{a}{D}$ Distance of separation, $a=D \times \theta =10 \times 10^{3} \times 2.44 \times 10^{-4}$ $=2.44 \,m$