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Q9.27 An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25cm and an eyepiece of focal length 5cm. How will you set up the compound microscope?

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Given,

magnifying power = 30

objective lens focal length

$f_{objeective}$ = 1.25cm

eyepiece lens focal length

$f_{eyepiece}$ = 5 cm

Normally, image is formed at distance d = 25cm

Now, by the formula;

Angular magnification by eyepiece:

$m_{eyepiece}=1+\frac{d}{f_{eyepiece}}=1+\frac{25}{5}=6$

From here, magnification by the objective lens :

$m_{objective}=\frac{30}{6}=5$ since ( $m_{objective}*m_{eyepiece}=m_{total}$ )

$m_{objective}=-\frac{v}{u}=5$

$v=-5u$

According to the lens formula:

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{1.25}=\frac{1}{-5u}-\frac{1}{u}$

from here,

$u = -1.5cm$

hence object must be 1.5 cm away from the objective lens.

$v= -mu=(-1.5)(5)=7.5$

Now for the eyepiece lens:

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

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

$\frac{1}{u}=-\frac{6}{25}$

$u = -4.17 cm$

Hence the object is 4.17 cm away from the eyepiece lens.

The separation between objective and eyepiece lens

$u_{eyepiece} +v_{objectivve}=4.17 + 5.7 = 11.67 cm$

Posted by

Pankaj Sanodiya

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