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Question
Obtain the equation for resolving the power of the microscope.
Solution
- 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 dmin. Smaller the value of dmin better will be the resolving power of the microscope.
Resolving power of microscope
`"r"_0 = (1.22 lambda "v")/"a"` - 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)` - To further reduce the value of dmin 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)` - 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"))`
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