Question

Draw a labelled ray diagram of an astronomical telescope and derive the formula of its magnifying power.

Hint:

For an astronomical telescope, the longer the focal length of the objective lens and the shorter the focal length of the eyepiece, the greater the magnifying power.

Answer 1

An astronomical telescope is designed to view distant objects in the sky, such as stars and planets, at a magnified scale. It consists of two lenses: the objective lens and the eyepiece.
- The objective lens (\(O\)) has a large focal length and is placed at a distance from the object. It forms a real, inverted image of the object at its focal plane.
- The eyepiece (\(E\)) is used to magnify the image formed by the objective lens. It is placed close to the focal plane of the objective lens and is used as a magnifying glass.
In the diagram: - \(O\) is the objective lens.
- \(E\) is the eyepiece.
- \(F_o\) and \(F_e\) are the focal points of the objective and eyepiece, respectively.
- The image formed by the objective lens is at the focal plane \(F_o\), and it is magnified by the eyepiece.
The magnifying power \(M\) of the telescope is the ratio of the angular size of the image seen through the telescope (\(\theta'\)) to the angular size of the object when viewed with the unaided eye (\(\theta\)). The formula for magnifying power of the telescope is:
\[ M = \frac{\theta'}{\theta} = \frac{f_o}{f_e} \] Where: - \(f_o\) is the focal length of the objective lens,
- \(f_e\) is the focal length of the eyepiece.
The magnifying power is directly proportional to the focal length of the objective lens and inversely proportional to the focal length of the eyepiece.