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?
Answer:
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 the 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
a)
Now in the Eyepiece lens
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.
the distance of $u$ from objective lens = $d+u=$ $15 - 5=10cm$ . This length will serve as the image distance for the objective lens.
$v = 10 cm$
so in the objective lens
$\frac{1}{f_{objective}} = \frac{1}{v} - \frac{1}{u_{initial}}$
$\frac{1}{u_{initial}} = \frac{1}{v} - \frac{1}{f_{objective}}$
$\frac{1}{u_{initial}} = \frac{1}{10} - \frac{1}{2} = -\frac{4}{10}=-\frac{2}{5}$
$u_{intial}$ = -2.5 cm
Hence the object distance required is -2.5 cm.
Now, the magnifying power of a microscope is given by
$m=\frac{v}{|u_{initial}|}(1+\frac{d}{f_{eyepiece}})$ where $d$ is the least distance of vision
so putting these values
$m=\frac{10}{2.5}(1+\frac{25}{6.25}) = 20$
Hence the lens can magnify the object 20 times.
b) When the image is formed at infinity
in the eyepiece lens,
$\frac{1}{f_{eyepiece}} = \frac{1}{v_{final}} - \frac{1}{u}$
$\frac{1}{6.25} = \frac{1}{infinity} - \frac{1}{u}$
from here $u =$ - 6.25., this distance from objective lens = $d+u$ = 15 - 6.25 = 8.75 = $v$
in the optical lens:
$\frac{1}{f_{objective}} = \frac{1}{v} - \frac{1}{u_{initial}}$
$\frac{1}{2} = \frac{1}{-6.25} - \frac{1}{u_{initial}}$
$\frac{1}{u_{initial}} =- \frac{6.75}{17.5}$
$u_{initial}=-2.59cm$
Now,
$m=\frac{v}{|u_{initial}|}(1+\frac{d}{f_{eyepiece}})$ where $d$ is the least distance of vision
putting the values, we get,
$m=\frac{8.75}{2.59}(1+\frac{25}{6.25})= 13.51$
Hence magnifying power, in this case, is 13.51.
