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

Correct Answer - `2 cm ; 5 cm`
Here, `m = -20, m_e = 5`
`:. m_o = (m)/(m_e) = (-20)/(5) = -4`
As `m_e = 1 +m (d)/(f_e)`
`5 = 1 + (20)/(f_e) :. f_e = (20)/(4) = 5 cm`
As `m_e = (f_e)/(u_e - f_e)`
`:. 5 = (5)/(u_e + 5) u_e = (5 - 25)/(5) = -4 cm`
Distance between objective and eye piece
=`v_o + |u_e| = 14`
`v_o = 14 - |u_e|= 14 - 4 = 10 cm`
Magnification produced by objective lens
`m_o = (v_o)/(u_o)`
`-4 = (10)/(u_o) :. u_o = -2.5 cm`
From `(1)/(f_o)=(1)/(v_o)-(1)/(u_o)`
=`(1)/(10)-(1)/(-2.5) = (1 + 4)/(10) = (5)/(10)`
`f_o = (10)/(5) = 2 cm`.