A compound microscope uses an objective lens of focal length $4\ \mathrm{cm}$ and an eyepiece lens of focal length $10\ \mathrm{cm}$. An object is placed at $6\ \mathrm{cm}$ from the objective lens. Calculate the magnifying power of the compound microscope when the final image is formed at the near point. Also calculate the length of the microscope.

CBSE Class 12 · 2011 · Geometrical OpticsLong Answer

A compound microscope uses an objective lens of focal length

4 cm

and an eyepiece lens of focal length

10 cm

. An object is placed at

6 cm

from the objective lens. Calculate the magnifying power of the compound microscope when the final image is formed at the near point. Also calculate the length of the microscope.

Answer: Magnifying power $= 7$ ; length of the microscope $\approx 19.14 cm$ .

Solution

\textbf{Objective:}

u_{o} = - 6 cm

f_{o} = 4 cm

\frac{1}{v _{o}} = \frac{1}{4} - \frac{1}{6} = \frac{1}{12} \Rightarrow v_{o} = 12 cm

;

m_{o} = \frac{v _{o}}{∣ u _{o} ∣} = 2

.
\textbf{Eyepiece} (near point,

m_{e} = 1 + \frac{D}{f _{e}} = 1 + \frac{25}{10} = 3.5

m = m_{o} m_{e} = 2 \times 3.5 = 7

.
For the eyepiece,

v_{e} = - 25 cm

\frac{1}{u _{e}} = \frac{1}{v _{e}} - \frac{1}{f _{e}} = - \frac{1}{25} - \frac{1}{10} = - \frac{7}{50} \Rightarrow ∣ u_{e} ∣ = 7.14 cm

L = v_{o} + ∣ u_{e} ∣ = 12 + 7.14 = 19.14 cm

CBSE Class 12 Board · 2011 · Subjective

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CBSE Class 12 · 2011 · Geometrical OpticsLong Answer

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