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A compound microscope consists of an objective of focal length 2cm eye-piece of focal length 5cm. When an object in kept 2.4cm from an objective, final image formed is virtual and 25 cm form an eye-piece. Determine magnifying power of this compound microscope in this set up, i.e., in normal use. When an object in kept 2.4cm from an objective, final image formed is virtual and 25 cm form an eye-piece. Determine magnifying power of this compound microscope in this set up, i.e., in normal use.

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Here fo = 2cm, fe = 5cm, uo = 2.4cm, ve = D = 25cm.

In normal use, magnifying power

M = \(-\frac{v_o}{u_o}(1+\frac{D}{f_e})\)

For objective, \(\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}\)

\(\frac{1}{2}=\frac{1}{v_o}-\frac{1}{(-2.4)}\)

\(\frac{1}{v_e}=\frac{1}{2}-\frac{1}{2.4}=\frac{6-5}{12}\)

ve = 12cm

So, M = \(-\frac{12}{24}(1+\frac{25}{5})\)

= −5 × 6 = −30

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