Question

# In a compound microscope the focal length of objective lens is $$1.2\ cm$$ and focal length of eye piece is $$3.0\ cm$$. When object is kept at $$1.25\ cm$$ in from of objective, final image is formed at infinity. Magnifying power of the compound microscope should be :

A
400
B
200
C
150
D
100

Solution

## The correct option is A $$200$$Given $$f_e = 3 cm$$ and            $$f_o = 1.2 cm$$           $$u_o = 1.25 cm$$           $$v_e = \infty$$From , $$\displaystyle \dfrac{1}{u_o} + \dfrac{1}{v_o} = \dfrac{1}{f_o}$$,we get $$\displaystyle \dfrac{1}{v_o} = \dfrac{1}{1.2} - \dfrac{1}{1.25}$$$$v_o = 30$$ cm Magnification of microscope $$\displaystyle m=\dfrac{v_o}{u_o} (\dfrac{D}{f_e})$$$$\displaystyle m = \dfrac{30}{1.25} (\dfrac{25}{3}) = 200$$            .PhysicsNCERTStandard XII

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