Question

The total magnification produced by a compound microscope is `20`. The magnification produced by the eye piece is `5`. The microscope is focussed on a certain object. The distance between the objective and eye piece is observed to be `14 cm`. If least distance of distinct vision is `20 cm`, calculate the focal length of objective and eye piece.

Answer 1

From `m = m_0 xx m_e`,
`m_0 = (m)/(m_e) = (20)/(5) = 4`
Now, `m_0 = (v_0)/(u_0) = (L)/(f_0) = 4`,
`f_0 = (L)/(4) = (14)/(4) = 3.5 cm`
Also, `m_e = 1 + (d)/(f_e) = 5`
`(d)/(f_e) = 5 - 1 = 4, f_e = (d)/(4) = (20)/(4) = 5 cm`.