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Question
In a compound microscope an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye-piece has a focal length of 5 cm and the final image is formed at the near point, find the magnifying power of the microscope.
Solution
Given: μ0 = -1.5 cm, f0 = 1.25 cm
We have,
`1/f_0 = 1/v_0 - 1/u_0`
`1/1.25 = 1/v_0 - 1/(-1.5)`
⇒ `v_0 = 7.5` cm
`m = v_0/u_0(1 + D/f_e)`
= `7.5/(-1.5)(1 + 25/5)`
⇒ m = -30
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