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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
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Verified Answer
The correct answer is:
$2.44 \mathrm{~m}$
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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