Question

A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope.

Answer 1

First we shall find the image distance for the objective`(v_0)`,

`1/f_0  = 1/v_0  -1/u_0 ; f_0 = 4cm,u_0 =-6cm`

`=> v_0 =12 cm`

Magnification of the microscope is,

`m = m_0 m_e = (v_0)/(u_0)(1+D/f_e) = (12/-6)(1 +25/10)`

= − 7, negative sign indicates that the image is inverted.

The length of the microscope is v_o+ u, u=|u_e| is the object distance for the eyepiece. And u_e can be found using,

`1/f =1/D - 1/u_e`; as D is the image distance for the eyepiece.

`=> 1/10 =1/-25 - 1/u_e => u_e = -7.14 cm`

Hence, u = |u_e| = 7.14 cm.

Length of the microscope v_o+ u= 19.14 cm

Length of the microscope is given as 

`L =(mf_0f_e)/D = (7 xx 4 xx 10)/25 = 11.2 cm`