Explain giving reason, how the magnifying power of a compound microscope depends on the (i) wavelength of incident light, and (ii) focal length of the objective lens. 

 

Answers (1)
S safeer

The magnifying power of the compound microscope is given by:

m=m_0m_e

When the final image is formed at near point

m=\left ( \frac{L}{f_0} \right )\left ( 1+\frac{D}{f_e} \right )

When the final image is formed at infinity 

m=\left ( \frac{L}{f_0} \right )\left ( \frac{D}{f_e} \right )

where L \rightarrow Tube length

        f_0\rightarrow  the focal length of the objective lens

        f_e\rightarrow   the focal length of the eyepiece

and the resolving power of a microscope

\approx \frac{2nsin\beta}{1.22\lambda}

nsin\beta\ \text{is numerical aperture}

From the above equations, it is clear that   

(i) the magnifying power of a compound microscope varies inversely with the wavelength of the incident light.

(ii) the magnifying power of a compound microscope is inversely proportional to the focal length of the objective lens.

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