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Q 9.12 A person with a normal near point (25 cm) using a compound microscope with an objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope

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Inside a microscope,

For the eyepiece lens,

$\frac{1}{f_{eyepiece}}=\frac{1}{v_{eyepiece}} - \frac{1}{u_{eyepiece}}$

we are given

$v_{eyepiece}= -25cm$

$f_{eyepiece}= 2.5cm$

$\frac{1}{2.5}=\frac{1}{-25} - \frac{1}{u_{eyepiece}}$

$\frac{1}{u_{eyepiece}}=\frac{1}{-25} - \frac{1}{2.5}=-\frac{11}{25}$

$u_{eyepiece}=- \frac{25}{11}=-2.27cm$

we can also find this value by finding image distance in the objective lens.

So, in the objective lens

$\frac{1}{f_{objective}}=\frac{1}{v_{objective}} - \frac{1}{u_{objective}}$

we are given

$f_{objective}= 0.8$

$u_{objective}= -0.9$

$\frac{1}{0.8}=\frac{1}{v_{objective}} - \frac{1}{-0.9}$

$\frac{1}{v_{objective}} = \frac{0.1}{0.72}$

$v_{objective} = 7.2cm$

Distance between object lens and eyepiece = $|u_{eyepiece}| + v_{objective}$ = 2.27 + 7.2 = 9.47 cm.

Now,

Magnifying power :

$m = \frac{v_{obejective}}{|u_{objective}|}(1+\frac{d}{f_{eyepiece}})$

$m = \frac{7.2}{0.9}(1+\frac{25}{2.5})= 88$

Hence magnifying power for this case will be 88.

Posted by

Pankaj Sanodiya

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