Class 12 Physics Wave Optics Diffraction in resolving power of Microscope

Diffraction in resolving power of Microscope

  • In case of microscope resolving power helps to magnify the image.
  • Resolution helps to distinguish between the small similar particles of an object when they are so closely placed with each other.
  • When the distance between 2 points is comparable to the wavelength then diffraction occurs.
  • Image formed due to the diffraction pattern vθ = v (22λ/D).
  • The minimum distance for the object to be resolved is v (22 λ)/(D)andbeyond this point the object cannot be resolved.
  • Minimum separation between objects to get resolved is given as:-
  • dmin = (v (1.22λ/D))/m  
    • Where m =magnification.
  • dmin= (1.22λ/D) f;
    • Where f=focal length.
  • tanβ = (D/2f)
    • Whereβ = angle made by the objective lens at its focus, D=diameter and f=focus.
  • tanβ = 1.22f λ
  • Therefore dmin = (1.22λ)/(tanβ)
  • dmin= (1.22λ)/(2tanβ) if β≈ small, tanβ≈ sinβ≈β
  • dmin= (1.22 λ)/(2 sin β)
  • Suppose if the medium between the object and the objective lens has refractive index n.
  • dmin= (1.22λ)/ (2n sinβ) where n sinβ = numerical aperture of objective lens.
  • Therefore dmin= (1.22λ)/ (2n sinβ) is known as numerical aperture of the objective lens.
  • Resolving Power(R.P) of a microscope ∝ (1/dmin).
  • This implies resolving power decreases as the distance increases.
  • =>P. =(2n sin β)/(1.22 λ)
  • Conclusion:-Resolving power can be increased by choosing medium of higher refractive index.

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