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

Given,

f₀ = focal length of the objective = 2 cm

f_e = focal length of the eye pieces = 5 cm

M_e = Magnification of the eye pieces

M_o = Magnification of the objective

M_e = 1 + $straight D over straight f subscript straight o$ = $1 space plus space 25 over 5 space equals space 6$
For the objective,

u = -2.4 cm, v = ? and f_o = 2 cm

$1 over straight v minus fraction numerator 1 over denominator negative 2.4 end fraction space equals space 1 half space space space space space space space space space space space space space space space space space space space space space equals space 1 half space minus space fraction numerator 1 over denominator 2.4 end fraction space space space space space space space space space space space space space space space space space space space space space equals space 12 over 24 minus 10 over 24 space space space space space space space space space space space space space space space space space space space space space equals space 2 over 24 space equals space 12 space c m M subscript o space space equals space v over u equals fraction numerator 12 over denominator 2.4 end fraction space equals space 5$

M = magnifying power of the compound microscope.

M = M₀ x M_e = 5 x 6 = 30, is the required magnification.

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