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

Explanation

Given : Here, `m = -20, m_e = 5, v_e =-20 cm`
For eyepiece, `m_e = (v_e)/(u_e)`
`rArr 5 = (-20)/(5) rArr u_e = (-20)/(5) = - 4cm`
Using lens formula,
`(1)/(v_e) - (1)/(u_e) = (1)/(f_e) = - 1/20 + 1/4 = (1)/(f_e)`
`rArr (-1 + 5)/(20) = (1)/(f_e) rArr f_e = 5cm`
Now, total magnification
` m = m_e xx m_o`
` -20 = 5 xx m_o rArr m_o = -4`
Also ` |v_o| + |u_e| = 14`
`|v_e| + |-4| = 14`
` v_o = 14 -4 =10cm`
`m_o = 1 - (n_o)/(f_o) rArr - 4 = 1 - (10)/(f_e)`