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A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. What is the focal length of the eyepiece.

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#### Solution

`"v"_circ + "f"_"e"` = 6.5 cm ....(1)

m = `"v"_circ/"u"_circ xx "D"/"f"_"e"` [taking D = 25 cm]

`=> "m" = - [1 - "v"_circ/"f"_circ] "D"/"f"_"e"`

`100 = - [1 - "v"_circ/0.5] 25/"f"_"e"`

`100 "f"_"e" = - (1 - 2"v"_circ) xx 25`

`2"v"_circ - 4 "f"_"e" = 1` .....(2)

Solving equation (1) and (2) we can get,

v_{0} = 4.5 cm and f_{e} = 2 cm

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