A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. What is the focal length of the eyepiece.

Explanation

Magnifying power, m = 100
Focal length of the object, `f_(o) = 0.5 cm`
Tube length, l = 6.5 cm ,
Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eyepiece.
`v_(o) + f_(e) = 6.5cm`
The magnifying power for normal adjustment is given by
`m=((v_(o))/(u_(o)))xx(D)/(f_(e))`
`=-[1-(v_(o))/(f_(o))](D)/(f_(e))`
`100=-[1-(v_(o))/(0.5)]xx(25)/(f_(e))`
`2 v_(o) - 4f_(e) = 1`
On solving equations (1) and (2), we get
`v_(o) = 4.5 cm and f_(e) = 2cm`
Thus, the focal lenght of the eyepiece is 2 cm.