A compound microscope has an eyepiece of focal length 5 cm and the distance between its objective and the eye piece is 20 cm.
What is the distance of the real intermediate image from the objective, if the final iamge seen by the eye is formed at the distance of distinct vision ?

Options

A. `(95)/(6)` cm
B. `(85)/(6)` cm
C. `(75)/(6)` cm
D. `(65)/(6)` cm

Answer

A

Explanation

The objective forms the real intermediate image `I_(1)` . It acts an object for E and E forms the final magnified virtual image `I_(2)`.
`therefore ` For the eyepiece `v_(e)=D=-25cm and f_(e)=5 cm`
`because (1)/(v_(e))-(1)/(u_(e))=(1)/(f_(e))`
`therefore -(1)/(25)-(1)/(u_(e))=(1)/(5)`
`therefore (1)/(u_(e))= -(1)/(25)-(1)/(5)= -(6)/(25)`
`therefore u_(e)= -(25)/(6) cm`.
`therefore` the distance `(v_(0))` of the real image `I_(1)` from O is
`v_(o)=20 -(25)/(6)=(95)/(6) cm`.