Question

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

Answer 1

Magnifying power of a telescope is defined as the ratio of the angle subtended at the eye by the image formed at the least distance of distinct vision to the angle subtended at the eye by the object lying in infinity.

Magnifying power $M=\frac{f_0}{f_e}(1+\frac{f_e}{D})$

$f_0=$ focal length of the object =150 cm; $f_e$= focal length of the eye piece= 5cm ; D= least distance of the distinct vision= 25cm

$M=\frac{-150}{5}(1+\frac{5}{25})=-36$

$M=\frac{\beta }{\alpha }=\frac{tab\beta }{tan\alpha }$ ( As angles are small)

$tan\alpha =\frac{H}{u}=\frac{100}{3000}=\frac{1}{30}$ where H= height of the object and u= distance of the object from the objective.

$M=\frac{tan\beta }{\left(1/30\right)}=\frac{-36}{30}=\frac{H^{\prime }}{D}$; H'= height of the image and D= distance of the image formation

Thus, $H^{\prime }=\frac{-36\times 25}{30}=-30cm$

Negative sign indicate that we get an inverted image.