1.

The value of numerical aperature of the objective lens of a microscope is 1.25. If light of wavelength 5000 Åis used, the minimum separation between two points, to be seen as distinct, will be :(A) 0.48 μm (B) 0.38 μm(C) 0.24 μm (D) 0.12 μm

Answer»

The minimum separation between two points, to be seen as DISTINCT, will be:

(C) 0.24 μm

This can be calculated as FOLLOWS:

  • We know that Numerical Aperture of the microscope is represented as -

        NA = ( 0.61λ ) / d

  • Here, d is known as the minimum separation between two points to be seen as distinct.
  • Thus, d = ( 0.61λ ) / NA

                     = ( 0.61 X 5000 x 10⁻¹⁰ ) / 1.25

                     = 2.4 x 10 ⁻⁷ m

                     = 0.24 μm



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