Question

A compound microscope consist of an objective lens of focal length of 0.5cm and an eyepiece of focal length 1cm. The two lenses are separated by distance 14cm. What is the position of the object if the final virtual image is formed at a distance of 25cm from the eyepiece. Calculate also the linear magnification of the instrument.

Answer 1

The  position of the object from the lens is 0.52 cm and the linear magnification of the instrument is 1356.

What is the object distance?

The object distance is the distance of the object from the lens.

Focal length of the objective lens f₁ = 0.5 cm

Focal length of the eyepiece, f₂ = 1 cm

Distance between objective lens and eyepiece, d = 14cm

Least distance of distinct vision, d' = 25 cm

Image distance for the eyepiece, v = -25cm

Using the lens formula:

$\frac{1}{f_{2}} = \frac{1}{v_{2}} - \frac{1}{u_{2}}\\\\\frac{1}{1} = \frac{1}{-25} - \frac{1}{u_{2}}\\\\u_{2} = -0.96 cm$

Image distance for the objective lens, $v_{1}$ = 14 + (- 0.96) cm = 13.04 cm

Using the lens formula:

​$\frac{1}{f_{1}} = \frac{1}{v_{1}} - \frac{1}{u_{1}}\\\\\frac{1}{0.5} = \frac{1}{13.04} - \frac{1}{u_{1}}\\\\u_{1} = -0.52 cm$

Linear magnification of the compound microscope is given by:

$m = \frac{v_{1}}{|u_{1}|}(1 + \frac{d'}{f_{2}})\\\\m = \frac{13.04}{0.52}(1 + \frac{25}{1})\\\\m=1356$

Therefore, the position of the object is 0.52 cm and the linear magnification of the instrument is 1356.

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