Draw the labelled ray diagram for the formation of image by a compound microscope.Derive the expression for the total magnification of a compound microscope. Explain why both the objective and the eyepiece of a compound microscope must have short focal lengths.
CBSE DELHI - SET 1 2009
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Expression for total magnification: The (linear) magnification due to the objective, namely h’/ h, equals
$\text{m}_{\circ} = \frac{\text{h}'}{\text{h}} = \frac{\text{L}}{\text{f}_{\circ}}$
where we have used the result
$\tan\beta = \bigg(\frac{\text{h}}{f}_{\circ}\bigg) = \bigg(\frac{\text{h}'}{\text{L}}\bigg)$
$\text{L}\cong$ Distance between the second focal point of the objective and the first focal point of the eye piece = tube length of the compound microscope. [When the final image is formed at infinity, the angular magnification due to the eyepiece is
$\text{m}_{e} = \big(\text{D}/f_{e}\big)]$
Alternate Answer
The (angular) magnification $m_e$ ,due to eyepiece, when the final image is formed at the near point, is
$\text{m}_{e} = \bigg(1 + \frac{\text{D}}{f}_{e}\bigg)$
The total magnification when the image is formed at infinity, is
$\text{m} = \text{m}_{\circ}\text{m}_{e} = \bigg(\frac{\text{L}}{f}_{\circ}\bigg)\bigg(\frac{\text{D}}{f}_{e}\bigg)$
Alternate Answer
The total magnification, when the final image is formed at the near point, is
$\text{m} = \text{m}_{\circ}\text{m}_{e} = \frac{\text{L}}{f}_{\circ}\bigg(1 + \frac{\text{D}}{f}_{\circ}\bigg)$
Reason: With short focal lengths of objective and eyepiece the total magnification of the compound microscope increases.
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