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Question

The focal lengths of the objective and the eye-piece of a compound microscope are $$2.0\ cm$$ and $$3.0\ cm$$ respectively. The distance between the objective and the eye-piece is $$15.0\ cm$$. The final image formed by the eye-piece is at infinity. The two lenses are thin. The distances in $$cm$$ of the object and the image produced by the objective measured from the objective lens are respectively


A
2.4 and 12.0
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B
2.4 and 15.0
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C
2.3 and 12.0
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D
2.3 and 3.0
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Solution

The correct option is A $$2.4$$ and $$12.0$$
When final image is formed at infitinity, 
length of the tube $$=v_0+f_e$$
$$\Rightarrow 15=v_0+3 \Rightarrow v_0=12\ cm$$
For objective lens $$\dfrac{1}{f_0}=\dfrac{1}{v_0}-\dfrac{1}{u_0}$$
$$\Rightarrow \dfrac {1}{(+2)}=\dfrac {1}{(+12)}-\dfrac {1}{u_0}\Rightarrow u_0=-2.4\ cm$$

Physics
NCERT
Standard XII

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