How is the working of a telescope different from that of a microscope? The focal lengths of the objective and eyepiece of a microscope are $1.25\ \mathrm{cm}$ and $5\ \mathrm{cm}$ respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of $30$ in normal adjustment.

A telescope magnifies distant large objects (objective has large $f$); a microscope magnifies tiny near objects (both lenses short $f$). The object must be placed $1.5\ \mathrm{cm}$ in front of the objective.

CBSE Class 12 · 2012 · Geometrical OpticsLong Answer

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How is the working of a telescope different from that of a microscope? The focal lengths of the objective and eyepiece of a microscope are

1.25 cm

and

5 cm

respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of

30

in normal adjustment.

Solution

\textbf{Difference:} a telescope views far-off large objects and its objective has a \emph{large} focal length and aperture; a microscope views very small nearby objects and both its lenses have \emph{short} focal lengths.
Normal adjustment (

D = 25 cm

m_{e} = 1 + \frac{D}{f _{e}} = 1 + \frac{25}{5} = 6

, so

m_{o} = \frac{m}{m _{e}} = \frac{30}{6} = 5 \Rightarrow \frac{v _{o}}{u _{o}} = - 5

.
Objective (

f_{o} = 1.25

\frac{1}{f _{o}} = \frac{1}{v _{o}} - \frac{1}{u _{o}} = \frac{1}{5∣ u _{o} ∣} + \frac{1}{∣ u _{o} ∣} = \frac{6}{5∣ u _{o} ∣} \Rightarrow ∣ u_{o} ∣ = \frac{6}{5} \times 1.25 = 1.5 cm

CBSE Class 12 Board · 2012 · Subjective

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CBSE Class 12 · 2012 · Geometrical OpticsLong Answer

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