Draw a labelled ray diagram of a compound microscope and write an expression for its magnifying power. The focal lengths of the objective and eye-lens of a compound microscope are $2\ \mathrm{cm}$ and $6.25\ \mathrm{cm}$ respectively. The distance between the lenses is $15\ \mathrm{cm}$. (i) How far from the objective lens will the object be kept so as to obtain the final image at the near point of the eye? (ii) Also calculate its magnifying power.

(i) Object distance $=2.5\ \mathrm{cm}$ from the objective. (ii) Magnifying power $=20$.

CBSE Class 12 · 2008 · Geometrical OpticsLong Answer

Draw a labelled ray diagram of a compound microscope and write an expression for its magnifying power. The focal lengths of the objective and eye-lens of a compound microscope are

2 cm

and

6.25 cm

respectively. The distance between the lenses is

15 cm

.
(i) How far from the objective lens will the object be kept so as to obtain the final image at the near point of the eye?
(ii) Also calculate its magnifying power.

Answer: (i) Object distance $= 2.5 cm$ from the objective.
(ii) Magnifying power $= 20$ .

Solution

Magnifying power (image at near point):

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

.
\textbf{Eyepiece:} final image at

v_{e} = - 25 cm

f_{e} = 6.25 cm

\frac{1}{u _{e}} = \frac{1}{v _{e}} - \frac{1}{f _{e}} = - \frac{1}{25} - \frac{1}{6.25} = - \frac{1}{5} \Rightarrow u_{e} = - 5 cm

. So the objective's image is

5 cm

before the eyepiece, i.e.

v_{o} = 15 - 5 = 10 cm

.
\textbf{Objective:}

f_{o} = 2 cm

v_{o} = 10 cm

\frac{1}{u _{o}} = \frac{1}{v _{o}} - \frac{1}{f _{o}} = \frac{1}{10} - \frac{1}{2} = - \frac{2}{5} \Rightarrow u_{o} = - 2.5 cm

.
(i)

u_{o} = 2.5 cm

.
(ii)

m = \frac{v _{o}}{∣ u _{o} ∣} (1 + \frac{D}{f _{e}}) = \frac{10}{2.5} (1 + \frac{25}{6.25}) = 4 \times 5 = 20

CBSE Class 12 Board · 2008 · Subjective

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CBSE Class 12 · 2008 · Geometrical OpticsLong Answer

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