Question

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

• A. 6 m
• B. 3 m
• C. 5 m
• D. 1 m

Answer 1

The correct option is C 5 m

$R.P.=\frac{2\mu \sin \theta }{1.22\lambda }$

$\frac{1}{{10}^{-3}}=\frac{2\sin \theta }{1.22\times 5\times {10}^{-7}}$

$2\theta =6\times {10}^{-4}$

$\frac{d}{D}=6\times {10}^{-4}$

$\therefore D=\frac{3\times {10}^{-3}}{6\times {10}^{-4}}=5m$