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

Given data : Magnifying power of compound microscope = m = 100
The focal length of objective `f_0 = 0.5 cm` (or)
` 5 xx 10^(-3) m`
Tube length = 6.5 cm
To find : The focal length of eye piece `f_e` = ?
Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eye piece.
So ` v_0 + f_e = 6.5 = 6.5 cm " " ....(1)`

Formula : Magnifying power = `(v_0)/(u_0) xx (D)/(F_e)` (for normal adjustment)
` rArr m = - [ 1 - (v_0)/(f_0) ] (D)/(F_e) [ therefore (v_0)/(u_0) =1 - (v_0)/( f_0)]`
``rArr 100 = - [ 1 - (v_0)/(0.5) ] xx (25)/(f_e) `
(Taking D = 25 cm)
` rArr 100 f_e = - 1 ( 1- 2 v_0) xx 25`
` rArr 2v_0 - 4f_e = 1 " " ...(2)`
Solving (1) & (2)
` v_0 = 4.5 cm `
` f_e = 2 cm `
The focal length of eye piece = 2 cm