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Karnataka Board PUCPUC Science Class 11

Find the Maximum Magnifying Power of a Compound Microscope Having a 25 Diopter Lens as the Objective - Physics

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Question

Find the maximum magnifying power of a compound microscope having a 25 diopter lens as the objective, a 5 diopter lens as the eyepiece and the separation 30 cm between the two lenses. The least distance for clear vision is 25 cm.

Answer in Brief

Solution

For the compound microscope, we have:
Power of the objective lens = 25 D
The focal length of the objective lens is given by

`f_0 = 1/(25 D) = 0.04  m = 4  cm`

Power of the eyepiece = 5 D
The focal length of the eyepiece is given by

`f_e = 1/(5D) =0.2  m =20  cm`

Least distance of clear vision, D = 25 cm
Separation between the objective and the eyepiece, L = 30 cm
Magnifying power is maximum when the image is formed by the eyepiece at the least distance of clear vision, i.e., = 25 cm.
For the eyepiece, we have:

ve = -25 cm and fe = 20 cm

The lens formula is given by

`1/v_e = 1/u_e + 1/f_e`

`=> 1/u_e = 1/v_e -1/f_e`

`=>1/-25 -1/20= -((4+5))/100`

⇒ ue = `(-100)/9` = 11.11 cm

Let uo and vo be the object and image distance for the objective lens.
So, for the objective lens, the image distance will be

V0 = L - u

v0 = 30 -11.11 

⇒ v0 = 18.89 cm   ....(1)

As the image produced is real , v0 =+18.89 cm 

`1/u_0 = 1/v_0 -1/f_0` 

=`1/18.89 - 1/4 = -0.197`

`u_0 =-5.07 cm`  .....(2)

Maximum magnifying power of the compound microscope:

m=`v_0/u_0(1+D/f_e)`

=` -(18.89/-5.07) (1+25/20)`

=3.7225 × 2.25 =8.376

So, the maximum magnifying power of the compound microscope is 8.376.

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Chapter 19: Optical Instruments - Exercise [Page 432]

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HC Verma Concepts of Physics Vol. 1 [English] Class 11 and 12
Chapter 19 Optical Instruments
Exercise | Q 6 | Page 432

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