Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope have short focal lengths? Explain.

Question

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Answer 1

Magnifying power is defined as the angle subtends at the eye by the image to the angle subtended by the object.
$m = m_{0} \times m_{e} = \frac{L}{f _{0}} \times \frac{D}{f _{e}}$
To increase the magnifying power both objective and eye piece must have short focal lengths such as,
$m = \frac{L}{f _{0}} \times \frac{D}{f _{e}}$

Answer 2

Magnifying power is defined as the angle subtends at the eye by the image to the angle subtended by the object.

m = m_{0} \times m_{e} = \frac{L}{f _{0}} \times \frac{D}{f _{e}}

To increase the magnifying power both objective and eye piece must have short focal lengths such as,

m = \frac{L}{f _{0}} \times \frac{D}{f _{e}}