Question

Define the magnifying power of a compound microscope when the final image is formed at infinite. Why both the objective and the eyepiece of a compound microscope must has short focal lengths? Explain.

Answer 1

• The magnifying power of a compound microscope can defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object.

• This can be represented as :

$M=\frac{V_{o}}{u_{o} } ( 1+ \frac{D}{f_{e} } )$

• This formula is used when both the final image and the object are at the least distance of distinct vision.

• When the final image is formed at infinity, then it is represented as :

Magnification due to eyepiece= $M_{e} = \frac{D}{f_{e} }$  

Magnification due to objective lens =   $M_{o} = \frac{L}{f_{o} }$

Total Magnification = $M_{o} * M_{e}$

                               =$\frac{L}{f_{o} } * \frac{D}{f_{e} }$

• Since, the magnification of both the eyepiece and objective lens is indirectely proportional to their respective focal lengths,

$M_{e}$∝$\frac{1}{f_{e} }$ and $M_{o}$∝$\frac{1}{f_{o} }$

• Therefore , for high magnifying power, both the objective and the eyepiece of a compound microscope must has short focal lengths

Answer 2

The total magnification, when the image is formed at infinity is  M = Mo x Me = L / fo x D / fe

Explanation:

Magnifying power of compound power is defined as:

"The ratio of angle made at eye by image Ie formed  at infinity to the angle made by object Io, if placed at a distinct of distant  vision from an unaided eye."

Magnification due to eyepiece, Me=D/fe

• Where D is the near point of human eye.
• Fe is the focal length of the eyepiece lens.

The angular magnification for the objective lens is given by

Magnification due to objective lens: Mo=L/fo

• Where L is the length of the tube.
• Fo is the focal length of the objective lens.

Thus, the total magnification, when the image is formed at infinity, is

M = Mo x Me = L / fo x D / fe