Explain the construction of a refracting telescope with a figure and derive the equation for its magnification.

(N/A) An astronomical telescope is used to observe very large celestial bodies. Its ray diagram is shown in the figure.
In this telescope, two convex lenses are placed such that their principal axes coincide.
The lens facing the object is called the objective, and the lens near the eye is known as the eyepiece.
The diameter and focal length of the objective are greater than those of the eyepiece.
When the telescope is focused on a distant object, parallel rays from the object form a real, inverted, and small image $A'B'$ at the second principal focus of the objective. This image acts as the object for the eyepiece.
The eyepiece is moved to and fro to obtain the final, magnified, and inverted image at a certain distance.
In such a telescope, rays from the object are refracted by the objective to form an image. Thus, it is called a refracting telescope.
Magnification $(m)$ of the telescope is defined as the ratio of the angle subtended by the final image at the eye $(\beta)$ to the angle subtended by the object at the objective $(\alpha)$:
$m = \frac{\beta}{\alpha}$
From the geometry of the ray diagram:
For the objective, $\tan \alpha \approx \alpha = \frac{h}{f_0}$
For the eyepiece, $\tan \beta \approx \beta = \frac{h}{f_e}$
Therefore, the magnification is:
$m = \frac{h/f_e}{h/f_0} = \frac{f_0}{f_e}$