#### Question

An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

#### Solution

Focal length of the objective lens,`f_o` = 1.25 cm

Focal length of the eyepiece, *f*_{e} = 5 cm

Least distance of distinct vision, *d* = 25 cm

Angular magnification of the compound microscope = 30X

Total magnifying power of the compound microscope, *m* = 30

The angular magnification of the eyepiece is given by the relation:

`m_e = (1+d/f_e)`

`= (1+25/5) = 6`

The angular magnification of the objective lens (*m*_{o}) is related to *m*_{e} as:

`m_om_e = m`

`m_o = m/m_e`

`= 30/6 = 5`

We also have the relation:

`m_o = ("Image distance for the objective lens"(v_o))/("Object distance for the objective lens"(-u_o))`

`5 = v_o/-u_o`

`:. v_o = -5u_o` ....(1)

Applying the lens formula for the objective lens:

`1/f_o = 1/v_o - 1/u_o`

`1/1.25 = 1/(-5u_o) - 1/u_o = (-6)/(5u_o)`

`:. u_o = -6/5 xx 1.25 = - 1.5 cm`

And `v_o = -5u_o`

`= -5xx(-1.5) = 7.5 cm`

The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.

Applying the lens formula for the eyepiece:

`1/v_e - 1/u_e = 1/f_e`

Where,

`v_e` = Image distance for the eyepiece = −*d* = −25 cm

`u_e` = Object distance for the eyepiece

`1/u_e = 1/v_e - 1/f_e`

`= (-1)/25 - 1/5 = - 6/25`

`:. u_e = -4.17 cm`

Separation between the objective lens and the eyepiece = `|u_e| + |v_o|`

= 4.17 + 7.5

= 11.67

Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.