KCET · Physics · Ray Optics
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 \(24.2 \mathrm{~m}\)
- C \(2.44 \mathrm{~m}\)
- D \(1.22 \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(2.44 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
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,
\(\begin{aligned}
\mathrm{a}=\mathrm{D} \times \theta &=10 \times 10^{3} \times 2.44 \times 10^{-4} \\
&=2.44 \mathrm{~m}
\end{aligned}\)
\(\theta=\frac{1.22 \lambda}{\mathrm{d}_{\mathrm{e}}}\)
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,
\(\begin{aligned}
\mathrm{a}=\mathrm{D} \times \theta &=10 \times 10^{3} \times 2.44 \times 10^{-4} \\
&=2.44 \mathrm{~m}
\end{aligned}\)
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