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.  .