Draw a ray diagram to show the working of a compound microscope. Deduce an expression for the total magnification when the final image is formed at the near point. In a compound microscope, an object is placed at a distance of $1.5\ \mathrm{cm}$ from the objective of focal length $1.25\ \mathrm{cm}$. If the eyepiece has a focal length of $5\ \mathrm{cm}$ and the final image is formed at the near point, estimate the magnifying power of the microscope.

CBSE Class 12 · 2010 · Geometrical OpticsLong Answer

Draw a ray diagram to show the working of a compound microscope. Deduce an expression for the total magnification when the final image is formed at the near point. In a compound microscope, an object is placed at a distance of

1.5 cm

from the objective of focal length

1.25 cm

. If the eyepiece has a focal length of

5 cm

and the final image is formed at the near point, estimate the magnifying power of the microscope.

Answer: Magnifying power $= 30$ .

Solution

Total magnification at near point:

m = \frac{v _{o}}{∣ u _{o} ∣} (1 + \frac{D}{f _{e}})

.
\textbf{Objective:}

u_{o} = - 1.5 cm

f_{o} = 1.25 cm

\frac{1}{v _{o}} = \frac{1}{f _{o}} + \frac{1}{u _{o}} = \frac{1}{1.25} - \frac{1}{1.5} \Rightarrow v_{o} = 7.5 cm

m = \frac{v _{o}}{∣ u _{o} ∣} (1 + \frac{D}{f _{e}}) = \frac{7.5}{1.5} (1 + \frac{25}{5}) = 5 \times 6 = 30

CBSE Class 12 Board · 2010 · Subjective

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CBSE Class 12 · 2010 · Geometrical OpticsLong Answer

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