Question

Two point white dots are $1\text{mm}$ apart on a black paper. The are viewed by the eye of pupil diameter $3\text{mm}$. Approximately, the minimum distance(in $mt$ at which these dots can be resolved by the eye is
Take wavelength to be $500\text{nm}$

Answer 1

We know,
$\sin {\theta }_{\text{dark}}=\frac{\lambda }{a}$
$\sin \theta =\frac{1.22\lambda }{D}$
As $\theta$ is small we can write it as
$\theta =\frac{1.22\lambda }{D}$ From diagram,
$\frac{1\times {10}^{-3}}{x}=\frac{1.22\times 500\times {10}^{-9}}{3\times {10}^{-3}}$
$\begin{array}{cc}x=& \frac{{10}^{-3}\times 3\times {10}^{-3}}{1.22\times 500\times {10}^{-9}}\\ =& 5m\end{array}$