# 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.

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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