Obtain the equation for resolving power of microscope.

Explanation

(i) A microscope is used to see the details of the object under observation. (ii) 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")` · (iii) Smaller the value of `d_("min")` better will be the resolving power of the microscope. (iv) The radius of central maxima is already derived as equation (1),
` N = (1)/(a+b) " " .....(1)`
` r_0 = (1.22 lamda f)/( a)`
(v). In the place of focal length 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) = (1.22 lamda v)/( am ) = (1.22 lamda v)/(a (v/u)) = (1.22 lamda u)/(a) " " [ therefore m = v/u]`
` d_("min") = (1.22 lamda f)/(a) " " [ therefore u= f ]`
On the object side,
` 2 tan beta ~~ 2 sin beta = a/f " " [ therefore a = f 2 sin beta]`
` d_("min") = (1.22 lamda)/(2 sin beta)`
(vi) To reduce the 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.