Question

Describe the terrestrial telescope on the basis of the following points.
(i) Labelled Ray diagram.
(ii) Derivation of formula for magnifying power, when final image is formed at the least distance of distinct vision.

Answer 1

Construction: The construction of terrestrial telescope is similar to the astronomical telescope. The only difference is that a third convex lens$\left(L_f\right)$ called field lens or erecting lens if kept in between the objective and erecting lens is kept in between. The objective and erecting lens is $f_o+zf$.
Where $f_o$ is the focal length of objective and $f_u$ the focal length of erecting lens. The eye lens can be shifted used the rack and pinion arrangement.
(b) Ray diagram:
(c) Derivation of magnifying power: As the object is situated far away, hence visual and formed on the objective equal to the visual angle formed on the eyepiece.
As the angles $\alpha$ and $\beta$ are very small. Hence tan$\alpha \approx \alpha$ and $tan\beta \approx \beta$.
$\therefore m=\frac{\beta }{\alpha }=\frac{tan\beta }{tan\alpha }\alpha \approx \alpha$ and
In Fig., $tan\alpha =\frac{PQ}{OQ}=\frac{P^{\prime }{Q}^{\prime }}{O{Q}^{\prime }}$ ..........(1)
and $tan\beta =\frac{P^{\prime }{Q}^{\prime }}{Q^{\prime }O}$
Putting the values of tan $\alpha$ and tan $\beta$ in eqn. (i), we get
$m=\frac{P^{\prime }{Q}^{\prime }/Q^{\prime }{O}^{\prime }}{-P^{\prime }{Q}^{\prime }/OQ}$
or $m=\frac{P^{\prime }{Q}^{\prime }}{Q^{\prime }O}\times \frac{OQ}{P^{\prime }}{Q}^{\prime }=-\frac{OQ}{O^{\prime }{Q}^{\prime }}$
But, $OQ=f_o$ and $O^{\prime }{Q}^{\prime }=-u_o$,
$\therefore m=\frac{-f_o}{-u_e}=\frac{f_o}{u_e}$ (By sign convention) ........(ii)
When image is formed infinity: $u_e=f_e$
$\therefore m=\frac{f_o}{f_e}$.