The resolving power of a microscope is

Explanation

The diagram related to the calculation of resolution of microscope .A microscope is used to see the details of the object under observation.The ability of microscope depends not only in magnifying the object but also in resolving two points on the object separated by a small distance `d_(min)` smaller the value of `d_(min)` better will be the resolving power of the microscope The radius of central maxima is, `r_0=(1.22lambdav)/a`
In the place of local f we have the image distance v.if the difference between the two points on the object to be resolved is `d_(min)` then the magnification m is ,`m=(r_0)/(d_(min)`
`d_(min)=(r_0)/m=(1.22lambdav)/am=(1.22lambdav)/(a(v//u))=(1.22lambdau)/a[therefore m=v//u]`
`d_(min)=(1.22lambdaf)/a[thereforeu=f]`
On the object side, `2tan beta approx 2sin beta=a/f therefore[a=f2sinbeta]`
`d_(min)=(1.22lambda)/(2sin beta)`
to further reduce value of `d_(min)` the optical path of the light is increased by immersing the objective of the microscope in to a bath containing oil of refractive index n.
`d_(min)=(1.22 lambda)/(2n sin beta)`
Such an objective is called oil immersed objective.The term `n sin beta` is called numerical aperture NA.
`d_(min)=(1.22 lambda)/(2(NA))`