Answer:
The linear separation at the retina is [tex]5.725\times10^{-6}\ m[/tex]
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
[tex]\alpha=1.22\times\dfrac{\lambda}{n d}[/tex]
[tex]\alpha=1.22\times\dfrac{500\times10^{-9}}{1.33\times2\times10^{-3}}[/tex]
[tex]\alpha=2.29\times10^{-4}[/tex]
We need to calculate the linear separation at the retina
Using formula of linear separation
[tex]D=\alpha l[/tex]
Put the value into the formula
[tex]D=2.29\times10^{-4}\times2.5\times10^{-2}[/tex]
[tex]D=5.725\times10^{-6}\ m[/tex]
Hence, The linear separation at the retina is [tex]5.725\times10^{-6}\ m[/tex].
