Question

An angular magnification (magnifying power) of 30X is desired usingan objective of focal length 1.25 cm and an eyepiece of focal length5cm. How will you set up the compound microscope?

Answer 1

Given: The focal length of eyepiece is 5 cm, the focal length of objective is 1.25 cm and the desired magnification is 30 times.

The angular magnification of the eyepiece is given as

α e =[ ( 25 f e )+1 ]

Where, f e is the focal length of eyepiece.

By substituting the given values in the above expression, we get

α e =[ ( 25 5 )+1 ] =[ 5+1 ] =6

The angular magnification of the objective is related to that of eyepiece as,

α= α 0 α e

Where, α o is angular magnification of objective.

By substituting the given values in the above expression, we get

30= α o ×6 α o = 30 6 =5

Also,

α o =− v o u o

Where, v o is the image distance for objective lens and u o is the object distance for the objective lens.

By substituting the given values in the above expression, we get

5= v o − u o v o =−5 u o (1)

The lens formula is given as,

1 f o = 1 v o − 1 u o

Where, f o focal length of objective lens.

By substituting the given values in the above expression, we get

1 1.25 = 1 −5 u o − 1 u o = −6 5 u o u o = −6 5 ×1.25 =−1.5 cm

By substituting the value of u o in equation (1), we get

v o =−5×( −1.5 ) =7.5 cm

Thus, the object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.

Apply the lens formula for the eyepiece,

1 f e = 1 v e − 1 u e

Where, v e is the image distance for eyepiece and u e is the object distance for the eyepiece.

By substituting the given values in the above expression, we get

1 5 = 1 −25 − 1 u e 1 u e =− 1 25 − 1 5 = −6 25 u e =−4.17 cm

Separation between the objective lens and the eyepiece is given as,

D=| u e |+| v o |

By substituting the given values in the above expression, we get

D=| −4.17 |+| 7.5 | =4.17+7.5 =11.67 cm

Thus, the separation between the objective lens and the eyepiece should be 11.67 cm.