Question

Describe labelled diagram of Terrestrial telescope and derive expression for its magnifying power.

Answer 1

Where, $O\to$ Objective, $E\to$ Eye lens, $L\to$ Inverter, $PQ\to$ Image.
Derivation of magnifying power: As the object is situated far away, hence visual and formed on the objective is equal to the visula angle formed on the eyepiece.
As the angles $\alpha$ and $\beta$ are very small. Hence $\tan \alpha \approx \alpha$ and $\tan \beta \approx \beta$.
$\therefore m=\frac{\beta }{\alpha }=\frac{\tan \beta }{\tan \alpha }$ ......(i)
In figure, $\tan \alpha =\frac{PQ}{OQ}$
and $\tan \beta =\frac{P^{\prime }{Q}^{\prime }}{Q^{\prime }{O}^{\prime }}$
But $PQ=-P^{\prime }{Q}^{\prime }$, $OQ=f_0$ and $O^{\prime }{Q}^{\prime }=-u_e$
Then $\tan \alpha =\frac{-P^{\prime }{Q}^{\prime }}{f_0}$
and $\tan \beta =\frac{P^{\prime }{Q}^{\prime }}{-u_e}$
Putting the values of $\tan \alpha$ and $\tan \beta$ in equation (i), we get
$m=\frac{\frac{P^{\prime }{Q}^{\prime }}{-u_e}}{\frac{-P^{\prime }{Q}^{\prime }}{f_0}}$
$=\frac{P^{\prime }{Q}^{\prime }}{-u_e}\times \frac{f_0}{-P^{\prime }{Q}^{\prime }}$
$m=\frac{f_0}{u_e}$