Question

Describe the astronomical 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

(i) Where, $O\to$ Objective lens, $O^{\prime }\to$ Eye lens, $PQ\to$ Image.
(ii) Expression for magnifying power: If angle made by an object at least distance of distinct vision is $\alpha$ and angle made by an image is $\beta$. Then magnification.
$m=\frac{\beta }{\alpha }$ .......... (i)
Object is situated at infinity, therefore angle subtended by an object on objective lens is said to be angle subtended by an object on eye.
Now, $tan\beta =\frac{P^{\prime }{Q}^{\prime }}{O^{\prime }{Q}^{\prime }}$ and $tan\alpha =\frac{P^{\prime }{Q}^{\prime }}{O^{\prime }{Q}^{\prime }}$
As $\alpha$ and $\beta$ are very small hence,
$tan\beta \sim \beta$ and $tan\alpha \sim \alpha$
$\therefore \beta =\frac{P^{\prime }{Q}^{\prime }}{O^{\prime }{Q}^{\prime }}$ and $\alpha =\frac{P^{\prime }{Q}^{\prime }}{O{Q}^{\prime }}$
Putting the values of $\beta$ and $\infty$ in eqn. (i), we get.
$m=\frac{P^{\prime }{Q}^{\prime }/O^{\prime }{Q}^{\prime }}{P^{\prime }{Q}^{\prime }/O{Q}^{\prime }}=\frac{O{Q}^{\prime }}{O^{\prime }{Q}^{\prime }}$ ......... (ii)
By sign convention distance of P'Q' from $L_e=u_e(-v_e)=O^{\prime }{Q}^{\prime }$ and focal length of objective lens $L_o$ is
$f_o=O{Q}^{\prime }(+ve)$
$\therefore m=\frac{f_o}{-u_e}$ .......... (iii)
When image is formed at least distance of distinct vision.
From formula : $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
or $\frac{1}{f_e}=\frac{1}{-v_e}-\frac{1}{-u_e}$
or $\frac{1}{f_e}=-\frac{1}{v_e}+\frac{1}{u_e}$
But, $v_e=D$
$\frac{1}{f_e}=-\frac{1}{D}+\frac{1}{u_e}$
or $\frac{1}{u_e}=\frac{1}{f_e}+\frac{1}{D}$
Substituting above value in eqn. (iii),
$m=-f_o(\frac{1}{f_e}+\frac{1}{D})$
$m=-\frac{f_o}{f_e}(1+\frac{f_e}{D})$