A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image

Name: A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image
Uploaded: Fri, 2019-05-31 11:52
Description: A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image

Q 9.11 A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15cm. 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 (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Answers (1)

In a compound microscope, first, the image of an object is made by the objective lens and then this image acts as an object for eyepiece lens.

Given

the focal length of objective lens = $f_{objective}$ = 2 cm

focal length of eyepiece lense = $f_{eyepiece}$ = 6.25cm

Distance between the objective lens and eyepiece lens = 15 cm

Now in Eyepiece lense

Image distance = $v_{final}$ = -25 cm (least distance of vision with sign convention)

focal length = $f_{eyepiece}$ = 6.25 cm

$\frac{1}{f_{eyepiece}} = \frac{1}{v_{final}} - \frac{1}{u}$

$\frac{1}{u} = \frac{1}{v_{final}} -\frac{1}{f_{eyepiece}}$

$\frac{1}{u} = \frac{1}{-25} -\frac{1}{6.25} = -\frac{1}{5}$

$u = -5 cm.$

Now, this object distance $u$ is from the eyepiece lens since the distance between lenses is given we can calculate this distance from the objective lens.