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# An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1 point 25 cm and an eyepiece of focal length 5cm. How will you set up the compound microscope?

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

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

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