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The focal lengths of the objective and eye`-` lens of a microscope are `1cm` and `5 cm` respectively. If the magnifying power for the relaxed eye is `45`, then the length of the tube isA. `6 cm`B. `9 cm`C. `12 cm`D. `15 cm` |
Answer» Correct Answer - D Magnifying power for relaxed eye is `m=(v_(o))/(u_(o))(D)/(f_(e))` Here, `M=-45,` `f_(o)=1 cm` `f_(e)=5 cm,` `D=25 cm` `:. -45 =-(v_(o))/(u_(o)). (25)/(5),` or `(v_(o))/(u_(o))=9,` For objective image is real. Therefore, `u_(o)=v_(o)` `u_(o)=-(v_(o))/(9),f_(o)=+1cm` Substituting in `(1)/(v_(o))-(1)/(u_(o))=(1)/(f_(o)),` we get `(1)/(v_(o))+(9)/(v_(o))=1` or`v_(o)=10cm` `:.` Length of tube `L=v_(o)+f_(e)=10+5=15cm` |
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