A compound microscope has an objective of focal length `2.0 cm` and an eye-piece of focal length `6.25cm` and distance between the objective and eye-piece is `15cm`. If the final image is formed at the least distance vision `(25 cm)`, the distance of the object form the objective is

Options

A. `1.5 cm`
B. `2.5 cm`
C. `3.0 cm`
D. `4.0 cm`

Answer

B

Explanation

Image by eye-piece:
`(1)/(v_(e))-(1)/(u_(e))=(1)/(f_(e)) rArr (1)/(-25)-(1)/(-u_(e)) = (1)/(25//4)`
`(1)/(u_(e)) = (4)/(25)+(1)/(25) rArr u_(e) = 5cm`
`L = v_(0)+|u_(e)|`
`15 = v_(0) +5 rArr v_(0) = 15-5 = 10cm`
Image by objective:
`(1)/(v_(0))-(1)/(u_(0))=(1)/(f_(0)) rArr (1)/(10)-(1)/(-u_(0)) = (1)/(2)`
`(1)/(u_(0))=(1)/(2)-(1)/(10)=(4)/(10)`
`u_(0) = 5//2cm = 2.5 cm `