Question

Two point white dots are $1 \, mm$ apart on a black paper. They are viewed by the eye of pupil diameter $3 \, mm$ . Approximately, what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light = $500 \, nm$ ]

• A. $6m$
• B. $3 \, m$
• C. $1 \, m$
• D. $5 \, m$

Answer 1

Answer: (D)

Resolution limit $= \frac{\text{1.22} \lambda }{\textit{d}}$
Again resolution limit $= \text{sin} \theta = \theta = \frac{y}{\textit{D}}$
$\therefore $ $\frac{y}{\textit{D}} = \frac{1 \cdot 2 2 \lambda }{\textit{d}}$

or $\textit{D} = \frac{y \textit{d}}{\text{1.22} \lambda }$
or $\textit{D} = \frac{\left(\left(\text{10}\right)^{- 3}\right) \times \left(3 \times \left(\text{10}\right)^{- 3}\right)}{\left(\text{1.22}\right) \times \left(5 \times \left(\text{10}\right)^{- 7}\right)} = \frac{\text{30}}{\text{6.1}} \approx 5 \text{ m}$ .