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:-
- d
_{min}= (v (1.22λ/D))/m- Where m =magnification.

- d
_{min}= (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 d
_{min}= (1.22λ)/(tanβ) - d
_{min}= (1.22λ)/(2tanβ) if β≈ small, tanβ≈ sinβ≈β **d**_{min}= (1.22 λ)/(2 sin β)- Suppose if the medium between the object and the objective lens has refractive index n.
- d
_{min}= (1.22λ)/ (2n sinβ) where n sinβ = numerical aperture of objective lens. - Therefore d
_{min}= (1.22λ)/ (2n sinβ) is known as__numerical aperture of the objective lens.__ __Resolving Power(R.P) of a microscope____∝ (1/d___{min}__).__- 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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