# Question #6f7cd

Oct 3, 2017

The Rayleigh criterion for diffraction-limited vision for the minimum resolvable detail of objects located at a distance $D$, using a circular aperture (an iris) of diameter of ${d}_{e}$ and light of wavelength $\lambda$ is given by the expression

$\sin {\theta}_{D} = 1.22 \frac{\lambda}{\theta} _ D$

Inserting given values we get
$\sin \theta = 1.22 \times \frac{500 \times {10}^{-} 9}{2.5 \times {10}^{-} 3} = 2.44 \times {10}^{-} 4$

We know that for very small value of $\sin \theta$

$\sin \theta \approx \theta$
$\implies \theta \approx 2.44 \times {10}^{-} 4 \text{ radian}$

$a$ is the distance of separation of the objects being resolved. From the figure, for small angles

$a = \theta \times D$

Inserting calculated and given values we get

$a = \left(2.44 \times {10}^{-} 4\right) \times \left(10 \times {10}^{3}\right) = 2.44 \text{ m}$