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A microscope has an objective of focal length `1.5 cm` and eye piece of focal length `2.5 cm`. If the distance between objective and eyepiece is `25 cm`. What is the approximate value of magnification produced for relaxed eye ?A. 75B. 110C. 140D. 25 |
Answer» Correct Answer - C Length of the tube is `L=v_(o)+f_(e)` `v_(o)=L-f_(e)` Now applying `(1)/(v_(o))-(1)/(u_(o))=(1)/(f_(o))` we have `(1)/(22.5)-(1)/(u_(o))=(1)/(1.5)` `:. |u_(o)| ~~1.6cm` `:. |M|= (v_(o))/(u_(o))xx(D)/(f_(e))` `=((22.5)/(1.6))((25)/(2.5)) ~~140~~140` |
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