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
B
1.5cm
C
2.1cm
D
2.4cm

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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