Question

What is the linear separation D at the retina (micrometers)? The minimum diameter of the pupil is about 2 mm, the eye is most sensitive to wavelengths of about 500 nm (air), the index of refraction of the aqueous humor is n = 1.33, and the distance between the cornea and retina is about 2.5 cm.

Answer 1

Answer:

The linear separation at the retina is $5.725\times10^{-6}\ m$

Explanation:

Given that,

Minimum diameter of the pupil = 2 mm

Wavelength = 500 nm

Index of refraction = 1.33

Distance between the cornea and retina = 2.5 cm

We need to calculate the angular separation

Using formula of angular separation

$\alpha=1.22\times\dfrac{\lambda}{n d}$

$\alpha=1.22\times\dfrac{500\times10^{-9}}{1.33\times2\times10^{-3}}$

$\alpha=2.29\times10^{-4}$

We need to calculate the linear separation at the retina

Using formula of linear separation

$D=\alpha l$

Put the value into the formula

$D=2.29\times10^{-4}\times2.5\times10^{-2}$

$D=5.725\times10^{-6}\ m$

Hence, The linear separation at the retina is $5.725\times10^{-6}\ m$.