Two-point white dots are $2 \mathrm{~mm}$ apart on a black paper. They are viewed by eye of pupil diameter…
Two-point white dots are $2 \mathrm{~mm}$ apart on a black paper. They are viewed by eye of pupil diameter $3 \mathrm{~mm}$. What is the maximum distance at which these dots can be resolved by the eye? $(\lambda=500 \mathrm{~nm})$
$5 \mathrm{~m}$
$1 \mathrm{~m}$
$6 \mathrm{~m}$
$10 \mathrm{~m}$
Solution
Distance between two point white dots,
$
x=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}
$
Diameter of pupil, $d=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}$ and $\quad \lambda=500 \mathrm{~mm}=5 \times 10^{-7} \mathrm{~m}$
If $D$ be the maximum distance at which these two point dots can be resolve, then
$
\begin{aligned}
\frac{x}{D} & =\frac{1.22 \lambda}{d} \\
\Rightarrow \quad D & =\frac{x d}{1.22 \lambda}=\frac{2 \times 10^{-3} \times 3 \times 10^{-3}}{1.22 \times 5 \times 10^{-7}} \\
D & =9.8 \simeq 10 \mathrm{~m}
\end{aligned}
$