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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
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B
200
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C
150
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D
100
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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 $$\displaystyle m=\dfrac{v_o}{u_o} (\dfrac{D}{f_e})$$

$$\displaystyle  m = \dfrac{30}{1.25} (\dfrac{25}{3}) = 200 $$            .

Physics
NCERT
Standard XII

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