There is a microscope whose objective's numerical aperture is `sin alpha = 0.24`, where `alpha` is the half-angle subtended by the objective's rim. Find the minimum spearation resolved by this microscope when an object is illuminated by light with wavelength `lambda = 0.55 mu m`.

Explanation

Let `A` and `B` be two points in the field of a microscope which is represented by the lens `CD`. Let `A', B'` be their image points which are at equal distance from the axis of the lans `CD`. Then all paths from `A` to `A'` are equal and the exterme difference of paths from `A` to `B'` is equal to
`ADB' - ACB'`
`= AD + DB' - (AC+CB')`
`= AD + DB' - BD - DB'`
`+BC + CB' - AC - CB'`
(as `BD + DB' = BC + CB')`
`= AD - BD + BC - AC`
`= 2Ab cos (90^(@)- alpha) = 2AB sin alpha`
from the theory of diffraction by circular apertures this disatnce must be equal to `1.22lambda`
when `B'` coincider with the minimum of the diffraction due to `A` and `A'` with the minimum of the diffraction due to `B`. Thus
`Ab = (1.22lambda)/(2sin alpha) = 0.61 (lambda)/(sin alpha)`
Here `2 alpha` is the angle subtended by the objective of the microscope at the object. Substituting the values
`Ab = (0.60 xx 0.55)/(0.24)mum = 140mum`.