CBSE Class 12 · 2009 · Geometrical OpticsLong Answer

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Draw the labelled ray diagram for the formation of an 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.

Solution

The objective (

m_{o} = \frac{L}{f _{o}}

) forms a real magnified image, further magnified by the eyepiece (

m_{e} = 1 + \frac{D}{f _{e}}

for the final image at the near point). The total magnification is

m = m_{o} m_{e} = \frac{L}{f _{o}} (1 + \frac{D}{f _{e}})

Since

m \propto \frac{1}{f _{o}} \cdot \frac{1}{f _{e}}

, \emph{both} focal lengths must be small for large magnification; a small

f_{o}

(shorter than

f_{e}

) also widens the field of view.

CBSE Class 12 Board · 2009 · Subjective

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