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The length of the compound microscope is `14 cm.` 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 beA. `1.8cm`B. `1.5 cm`C. `2.1 cm`D. `2.4 cm` |
Answer» Correct Answer - A `L_(oo)=v_(o)+f_(e)implies14=v_(o)+5impliesv_(o)=9cm` Magnifying power of microscope for relaxed eye `m=(v_(o))/(u_(o)).(D)/(f_(e))` or `25=(9)/(u_(o)).(25)/(5)`or `u_(o)=(9)/(5)=1.8cm` |
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