Question

If the length of an astronomical telescope is 80 cm and its magnification is 15, then find the focal length of the objective lens and eyepiece.

Answer 1

Hint: A telescope is used to observe astronomical objects. It is constructed in such a way that the final image is formed at infinity. The telescope consists of two convex lenses called the objective and the eyepiece. The ratio of the focal length of the objective to the focal length of the eyepiece gives the magnification of the telescope. The length of the telescope cab is represented as the sum of the focal lengths of the objective and the eyepiece.

Formula Used:
The length of the telescope in its normal adjustment (adjustment where the final image is formed at infinity) can be represented as the sum of the focal length of the objective and the focal length of the eyepiece. So, we can write,
$L={{f}_{o}}+{{f}_{e}}$
Where
L is the length of the telescope.
${{f}_{0}}$ is the focal length of the objective.
${{f}_{e}}$ is the focal length of the eyepiece.
The magnification of a telescope in normal adjustment is the ratio of the focal length of the objective to the focal length of the eyepiece gives the magnification of the telescope. So, we can write,
$M=\dfrac{{{f}_{o}}}{{{f}_{e}}}$

Complete step by step answer:
In the problem, it is given that the length of the telescope is 80 cm. The length of the telescope in its normal adjustment (adjustment where the final image is formed at infinity) can be represented as the sum of the focal length of the objective and the focal length of the eyepiece. So, we can write,
$L={{f}_{o}}+{{f}_{e}}$
Where
L is the length of the telescope.
${{f}_{0}}$ is the focal length of the objective.
${{f}_{e}}$ is the focal length of the eyepiece.
It is also mentioned that the magnification produced by the telescope is 15. The magnification of a telescope in normal adjustment is the ratio of the focal length of the objective to the focal length of the eyepiece gives the magnification of the telescope. So, we can write,
$M=\dfrac{{{f}_{o}}}{{{f}_{e}}}$
Where M is the magnification produced by the telescope.
$15=\dfrac{{{f}_{0}}}{{{f}_{e}}}$
$\therefore {{f}_{o}}=15{{f}_{e}}$ … equation (2)
Substituting equation (2) in equation (1), we get,
$80cm=15{{f}_{e}}+{{f}_{e}}$
$\therefore {{f}_{e}}=\dfrac{80}{16}$
$\therefore {{f}_{e}}=5cm$
$\therefore {{f}_{o}}=75cm$
So, the focal length of the eyepiece is 5 cm, and the focal length of the objective is 75 cm.

Note: Convex lenses are used in the making of a telescope. Since the convex lens is a converging lens, it will have as much light as possible and converge all the light to a focal point, so the faraway astronomical object becomes more clear. There are mainly two types of telescope that use visible light for its observation, one is the refracting type telescope, and the other is the reflecting type telescope.