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Question

The length of the compound microscope is $$14cm$$. The magnifying power for relaxed eye is $$25$$. If the focal length of eye lens is $$5 cm$$, then the object distance for objective lens will be  


A
1.8cm
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B
1.5cm
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C
2.1cm
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D
2.4cm
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Solution

The correct option is A $$1.8cm$$
$$L_{\infty}=v_0+f_e \Rightarrow 14=v_0 +5 \Rightarrow v_0=9\ cm$$
Magnifying power of microscope for relaxes eye 
$$m=\dfrac{v_0}{u_0}.\dfrac{D}{f_e}$$ or $$25=\dfrac{9}{u_0}.\dfrac{25}{5}$$ or $$u_0=\dfrac{9}{5}=1.8\ cm$$

Physics

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