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Wave Optics — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsWave Optics3 Marks
Question
Define resolving power of a microscope. Write its formula. How will it be affected when
(i) the wavelength of incident light decreases and
(ii) the aperture of the objective decreases ?
✓

Answer

Resolving power of Microscope : The resolving power of a microscope is its ability to show two nearby closely lying objects as just distinguished and it is equal to the reciprocal of its limit of resolution.
Image
"The resolving power of a microscope is defined as the reciprocal of the minimum distance between two point-objects which are seen distinctly separated through the microscope." According to Abbe, this minimum distance is given by :
$s=\frac{1.22 \lambda}{2 n \sin \alpha}$, ... (1)
where $ \lambda $ is the wavelength of the light, $n$ is the refractive index of the transparent medium between the objects and the objective of the microscope and $ \alpha $ is half the angle of the cone of light from any one object. Thus
Resolving Power :
$ R.P.=\frac{2n \sin \alpha}{1.22\lambda} $ ... (2)
From above formula (2) it is obvious that :
(i) when $ \lambda $ decreases, resolving power will be increased.
(ii) when aperture of objective is decreased the value of $ \alpha $ will be decreased. Hence R.P. will be decreased.

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