A compound microscope consists of an objective lens of focal length $2.0 c m$ and an eyepiece of focal length $6.25 c m$ separated by a distance of $15 c m$ . How far from the objective should an object be placed in order to obtain the final image at:
(a) the least distance of distinct vision $(25 c m)$ , and
(b) at infinity? What is the magnifying power of the microscope in each case?

Open in App

Solution

Focal length of the objective lens $f_{1} = 2 c m$
Focal length of the eyepiece, $f_{2} = 6.25 c m$
Distance between the objective lens and the eyepiece, $d = 15 c m$
Least distance of distinct vision, $d^{'} = 25 c m$
Image distance for the eyepiece, $v_{2} = - 25 c m$
bject distance for the eyepiece = $u_{2}$
According to the lens formula
$\frac{1}{f_{2}} = \frac{1}{v_{2}} - \frac{1}{u_{2}}$
$\frac{1}{6.25} = \frac{1}{- 25} - \frac{1}{u_{2}}$
$u_{2} = - 5 c m$
Image distance for the objective lens $v_{1} = d + u_{2} = 15 - 5 = 10 c m$
According to the lens formula
$\frac{1}{f_{1}} = \frac{1}{v_{1}} - \frac{1}{u_{1}}$
$\frac{1}{2} = \frac{1}{10} - \frac{1}{u_{1}}$
$u_{2} = - 2.5 c m$
The magnifying power of a compound microscope is given by the relation
$m = \frac{v_{1}}{| u_{1} |} (1 + \frac{d^{'}}{f_{2}})$
$m = \frac{10}{2.5} (1 + \frac{25}{6.25}) = 20$
b)The final image is formed at infinity
Image distance for the eyepiece $v_{2} = \infty$
According to the lens formula
$\frac{1}{f_{2}} = \frac{1}{v_{2}} - \frac{1}{u_{2}}$
According to the lens formula
$\frac{1}{6.25} = \frac{1}{\infty} - \frac{1}{u_{2}}$
$u_{2} = - 6.25 c m$
Image distance for the objective lens $v_{1} = d + u_{2} = 10 - 6.25 = 8.75$
Object distance for the objective len = $u_{2}$
According to the lens formula
$\frac{1}{f_{1}} = \frac{1}{v_{1}} - \frac{1}{u_{1}}$

$\frac{1}{2} = \frac{1}{8.75} - \frac{1}{u_{1}}$
$u_{2} = - 2.59 c m$
The magnifying power of a compound microscope is given by the relation:
$\frac{v_{1}}{| u_{1} |} (\frac{d^{'}}{| u_{2} |}) = 13.51$

Suggest Corrections

Similar questions

Q. A compound microscope consists of an objective lens of focal length2.0 cm and an eyepiece of focal length 6.25 cm separated by adistance of 15cm. How far from the objective should an object beplaced in order to obtain the final image at (a) the least distance ofdistinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Q. A compound microscope has an objective of focal length

2.0 c m

and an eyepiece of focal length

6.25 c m

separated by

15 c m

. If the final image is formed at the least distance of distinct vision

(25 c m)

, the distance of the object from the objective is :

Q. The focal lengths of objective and eyepiece of a compound microscope are

5 c m

6.25 c m

respectively. When an object is placed in front of the objective at a distance of

6.25 c m

, the final image is formed at least distance of distinct vision. The length of a microscope is :

Q. The focal length of the objective lens of a compound microscope is

0.02 m

and that of the eyepiece is

0.04 m

. The distance between the objective lens and the eyepiece is

0.20 m

. If the final image is formed at the least distance of distinct vision, then calculate the magnifying power of the compound microscope.

A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placed at a distance of 0.5 cm from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on a screen 30 cm behind the eyepiece?

Join BYJU'S Learning Program