Find the minimum magnification of a microscope, whose objective's numerical apertune is `sin alpha = 0.24`, at which the resolving power of the objective is totally employed if the diameter of the eye's pupil is `d_(0) = 4.0mm`.

Explanation

Suppose `d_(min)=` minimum separation resolved by the microscope `psi =` angle subtended at the eye by this object when the object is at the least distance of distinct vision `l_(0)(=25cm)`.
`psi' =` minimum angular separation resolved by the eye `= (1.22lambda)/(d_(0))`
From the previous problem
`d_(min) = (0.61lambda)/(sin alpha)`
and `psi = (d_(min))/(l_(0)) = (0.61lambda)/(l_(0)sin alpha)`
`T =` magnifying `= ("angle subtended at the eye by the image")/("angle subtended at the eye by the object")`
when the object is at the least disatnce of distinct vision
`ge (psi')/(psi) = 2 ((l_(0))/(d_(0)))sin alpha`
Thus `T_(min) = 2((l_(0))/(d_(0)))sin alpha = 2 xx (25)/(0.4) xx 0.24 = 30`