Question

Two white dots are $1\text{mm}$ apart on a black paper. They are viewed by eye of pupil diameter $3\text{mm}$. What is the maximum distance from the black paper at which these dots can be resolved by the eye? $(\lambda =500\text{nm}$)

• A. $2\text{m}$
• B. $3\text{m}$
• C. $4\text{m}$
• D. $5\text{m}$

Answer 1

The correct option is D $5\text{m}$

Limit resolution of eye

$\theta =\frac{1.22\lambda }{b}$

where $b=\text{pupil diameter}$

$\therefore \theta =\frac{1.22\times 500\times {10}^{-9}}{3\times {10}^{-3}}$

$\Rightarrow \theta =2.03\times {10}^{-4}\text{rad}$

If $x$ is the maximum distance at which dots are just resolved and $D$ is the distance between the dots, then

$\theta =\frac{D}{x}=\frac{{10}^{-3}}{x}$

$\Rightarrow \frac{{10}^{-3}}{x}=2.03\times {10}^{-4}$

$\Rightarrow x=\frac{{10}^{-3}}{2.03\times {10}^{-4}}\approx 5\text{m}$