Question

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. 75
B. 110
C. 140
D. 25

Answer 1

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`