In a compound microscope, the focal lengths of two lenses are `1.5 cm` and `6.25 cm` an object is placed at `2 cm` form objective and the final image is formed at `25 cm` from eye lens. The distance between the two lenses is

Options

A. `6.00 cm`
B. `7.75 cm`
C. `9.25 cm`
D. `11.0 cm`

Answer

D

Explanation

Here, `f_(o) = 1.5 cm, f_(e) = 6.25 cm, u_(o) = -2 cm`
`v_(e) = -25 cm`
For objective
`(1)/(v_(o))- (1)/(u_(0)) = (1)/(f_(o)) :. (1)/(v_(o)) = (1)/(-2) = (1)/(1.5` or `v_(o) = 6 cm`
For eye piece,
`:. (1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e)) , (1)/(-25) - (1)/(u_(e)) = (1)/(6.25)`
`- (1)/(u_(e)) = (1)/(6.25) + (1)/(25)` or `u_(e) = -5 cm`
Distance between two lenses `= |v_(e)| + |u_(e)|`
`= 6 cm + 5 cm = 11 cm`