A microscople has numberical apertune `sin alpha = 0.12`, where `alpha` is the aperture angle subtended by the entrance pupil of the microscope. Assuming the diameter of an eye's pupil to be equal to `d_(0) = 4.0 mm` determine the microscope magnification at which
(a) the diameter of the beam of light coming from the microscope is equal to the diameter of the eye's pupil,
(b) the illuminance of the image on the retina is independent of magnification (consider the case when the beam of light passing through teh sysytem ''microscope-eye'' is bounded by the mounting of the objective).

Explanation

It is implied in the problem that final image of the object is at infinity (otherwise light coming out of the eyepiece will not have a definite diameter).
(a) We see that `s'_(0)2 beta = |s_(0)|2 alpha`, then
`beta = (|s_(0)|)/(s'_(0)) alpha`
Then, from the figure
`d = 2f_(e) beta = 2f_(e) alpha // (s'_(0))/(|s_(0)|)`
But when the final image is at infinity, the magnification `T` in a microscope is given by
`T=(s'_(0))/(|s_(0)|).(l)/(f_(e)) (l =` least distance of distinct vision) So `d = 2l alpha//T`
So `d = d_(0)` when `T = T_(0) = (2lalpha)/(d_(0)) = 15` on putting the values.

(b) If `T` is the magnification prodduced by the microscope, then the area of the image produced on the retina (when we observe an object through a microscope) is : `T^(2) ((f)/(s))^(2) A`
Where `u =` distance of the image produced by the microscope from the eye lens, `f =` foca length of the eye lens and `A =` area of rthe object. if `Phi =` luminous flux reaching the objective from the object and `d le d_(0)` so that the entire flux is admitted into the eye),
`= (Phi)/(T^(2)(f//s)^(2)A)`
But if `d ge d_(0)^(2)`, then only a fraction `(d_(0) |d)^(2)` of light is admitted into the eye and the illuminance becomes
`(Phi)/(A((f)/(s))^(2)) ((d_(0))/(d))^(2) = (Phid_(0)^(2))/(A((f)/(s))^(2)(2l alpha)^(2))`
independent of `T`. the condition for this is then
`d ge d_(0)` or `T le T_(0) = 15`.