Question

Obtain the equation for resolving the power of the microscope.

Answer 1

1. A microscope is used to see the details of the object under observation. The ability of a 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.
   
   Resolving power of microscope
   `"r"_0 = (1.22 lambda "v")/"a"`
2. then the magnification m is,
   m = `"r"_0/"d"_"min"`
   `"d"_"min" = "r"_0/"m" = (1.22 lambda"v")/"am" = (1.22 lambda "v")/("a"("v"//"u")) = (1.22 lambda"u")/"a"`
   [∴ m = v/u]
   `"d"_"min" = (1.22 lambda "f")/"a"` [∴ u ≈f]
   On the other side,
   2 tan β ≈ 2 sin β = `"a"/"f"` ∴ [a = f 2sin β]
   `"d"_"min" = (1.22 lambda)/(2 sin beta)`
3. To further reduce the value of d_min the optical path of the light is increased by immersing the objective of the microscope into a bath containing oil of refractive index n.
   `"d"_"min" = (1.22 lambda)/(2"n" sin beta)`
4. Such an objective is called oil-immersed objective. The term n sin β is called numerical aperture NA.
   `"d"_"min" = (1.22 lambda)/(2 ("NA"))`