Question

Describe terrestrial telescope on the basis of the following points:
(i) A clear labelled diagram;
(ii) Expression for magnifying power when the final image is at least distance of distinct vision D.

Answer 1

(a) Construction: The construction of terrestrial telescope is similar to the astronomical telescope. The only difference is that a third convex lens $\left(L_r\right)$ called field lens or erecting lens is 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 is equal to the visual angle formed on the eyepiece.
As the angles $\alpha$ and $\beta$ are very small, hence $tan\alpha \approx \alpha$ and $tan\beta \approx \beta$.
$m=\frac{\beta }{\alpha }=\frac{tan\beta }{tan\alpha }\alpha \approx \alpha$ and
In fig., $tan\alpha =\frac{PQ}{OQ}=\frac{P^{\prime }{Q}^{\prime }}{O{Q}^{\prime }}$ ........9i) tan
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 $m=-\frac{P^{\prime }{Q}^{\prime }}{Q^{\prime }O}\times \frac{OQ}{P^{\prime }{Q}^{\prime }}=-\frac{OQ}{O^{\prime }Q}$
But, $OQ=f_o$ and $O^{\prime }{Q}^{\prime }=-u_e^{\prime }$
(By sign convention)
$m=\frac{-f_e}{u_e}=\frac{f_e}{u_e}$ ......(ii)
When image is formed infinity: $u_e=f_e$
$\therefore m=\frac{f_o}{f_e}$.