Question

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

Answer 1

Given that the normal near point is 25 cm, focal length of compound microscope is 8.0 mm, focal length of the eyepiece is 2.5 cm and object distance for objective lens is 9 mm or 0.9 cm.

Let v 1 be the image distance for the eyepiece, u 1 be the object distance for the eyepiece and f e be the focal length of the eyepiece.

From lens formula,

1 v 1 − 1 u 1 = 1 f e 1 u 1 = 1 v 1 − 1 f e

Substitute the values in the above expression.

1 u 1 = 1 −25 − 1 2.5 = −1−10 25 = −11 25 u 1 =−2.27 cm

Let v 2 be the image distance for the objective lens, u 2 be the object distance for the objective lens and f o be the focal length of the objective lens, then by lens formula,

1 v 2 − 1 u 2 = 1 f o 1 v 2 = 1 f o + 1 u 2

Substitute the values in the above expression.

1 v 2 = 1 0.8 − 1 0.9 = 0.9−0.8 0.72 v 2 =7.2 cm

Let d be the distance of the objective lens with the eyepiece, then,

d=| u 1 |+ v 2

Substitute the values in the above expression.

d=2.27+7.2 =9.47 cm

Hence, the separation of the objective lens and the eyepiece is 9.47 cm.

Let m be the magnifying power of the microscope, then

m= v 2 | u 2 | ( 1+ d l f e )

Here, d l is the near point of a person.

Substitute the values in the above expression.

m= 7.2 0.9 ( 1+ 25 2.5 ) =8( 1+10 ) =88

Hence, the value of magnifying power is 88.