Two luminous point sources separated by a certain distance are at $10 \mathrm{~km}$ from an observer. If the aperture of his eye is $2.5 \times 10^{-3} \mathrm{~m}$ and the wavelength of light used is $500 \mathrm{~nm}$, the distance of separation between the point sources just seen to be resolved is

• A

  $12.2 \mathrm{~m}$
• B

  $24.2 \mathrm{~m}$
• Correct answer
  C

  $2.44 \mathrm{~m}$
• D

  $1.22 \mathrm{~m}$

Solution

According to Rayleigh's criterion, $\theta=\frac{1.22 \lambda}{\mathrm{d}_{\mathrm{e}}}$ where $\lambda=$ wavelength of light, $\mathrm{d}_{\mathrm{e}}=$ 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$\quad \theta=\frac{a}{D}$ $\therefore$ Distance of separation, $\begin{aligned} \mathrm{a}=\mathrm{D} \times \theta &=10 \times 10^{3} \times 2.44 \times 10^{-4} \\ &=2.44 \mathrm{~m} \end{aligned}$

Hence, the correct answer is option: 'C'.