Define the magnifying power of a telescope. Write its expression. A small telescope has an objective lens of focal length $150\ \mathrm{cm}$ and an eyepiece of focal length $5\ \mathrm{cm}$. If this telescope is used to view a $100\ \mathrm{m}$ high tower $3\ \mathrm{km}$ away, find the height of the final image when it is formed $25\ \mathrm{cm}$ from the eyepiece.

Magnifying power $m=-\dfrac{f_o}{f_e}$ (normal adjustment). Height of the final image $=30\ \mathrm{cm}$.

CBSE Class 12 · 2012 · Geometrical OpticsLong Answer

Define the magnifying power of a telescope. Write its expression. A small telescope has an objective lens of focal length

150 cm

and an eyepiece of focal length

5 cm

. If this telescope is used to view a

100 m

high tower

3 km

away, find the height of the final image when it is formed

25 cm

from the eyepiece.

Answer: Magnifying power $m = - \frac{f _{o}}{f _{e}}$ (normal adjustment). Height of the final image $= 30 cm$ .

\textbf{Magnifying power} is the ratio of the angle subtended at the eye by the final image to that subtended by the object; in normal adjustment

m = - \frac{f _{o}}{f _{e}}

.
Angle subtended by the tower:

α = \frac{100}{3000} = \frac{1}{30} rad

. Image by objective at its focus:

h^{'} = f_{o} α = 150 \times \frac{1}{30} = 5 cm

.
Eyepiece (

f_{e} = 5

, image at

v_{e} = - 25 cm

\frac{1}{u _{e}} = \frac{1}{v _{e}} - \frac{1}{f _{e}} = - \frac{1}{25} - \frac{1}{5} = - \frac{6}{25} \Rightarrow u_{e} = - \frac{25}{6} cm

; linear magnification

m_{e} = \frac{v _{e}}{u _{e}} = 6

H = h^{'} \times m_{e} = 5 \times 6 = 30 cm

CBSE Class 12 Board · 2012 · Subjective