Question

The magnification produced by an astronomical telescope in normal adjustment is 10, and the length of the telescope is 1.1m. The magnification, when the image is formed at the least distance of distinct vision is
A. 6
B. 18
C. 16
D. 14

Answer 1

Hint: In the problem, the magnification power in the normal adjust of the telescope is given. In the normal adjustment of the telescope, the final image is formed at infinity. The length of the telescope is also given, which is the sum of the focal length of the objective lens and the focal length of the eyepiece. From these two relations, we can determine the focal length of the two lenses and find the magnification of the image formed at infinity.

Complete step by step answer:
We know that the telescope consists of two lengths, one is the objective, and the other is the eyepiece. In the normal adjustment of the telescope, the final image is formed at infinity and the magnification, in this case, is given by,
$M={\displaystyle \frac{f_o}{f_e}}$
$f_0$ is the focal length of the objective.
$f_e$ is the focal length of the eyepiece.
It is given that the magnification, in this case, is 10. So, we can write,
$10={\displaystyle \frac{f_o}{f_e}}$
$f_o=10f_e$ … equation (1)
The length of the telescope can be expressed as the sum of the focal length of the objective lens and the focal length of the eyepiece.
$L=f_o+f_e$
It is given that the length of the telescope is said to 1.1 m and using the relation in equation (1), we can write,
$1.1m=10f_e+f_e$
$\therefore f_e=0.1m$
$\therefore f_0=1m$
So, for the object is formed at the distance of distinct vision, we have to find the corresponding focal length of the eyepiece, which is given by,
${\displaystyle \frac{1}{{f_e}^{\prime }}}=-[{\displaystyle \frac{1}{D}}+{\displaystyle \frac{1}{f_e}}]$ ..equation (2)
So, the magnification of the telescope in this adjustment is,
$M={\displaystyle \frac{f_o}{{f_e}^{\prime }}}$
Substituting equation (2) in the above equation we get,
$M=-f_o[{\displaystyle \frac{1}{D}}+{\displaystyle \frac{1}{f_e}}]$
We know that the value of the least distinct vision is 0.25m. So, substituting the value in the above equation, we get,
$M=-(1m)[{\displaystyle \frac{1}{0.25m}}+{\displaystyle \frac{1}{0.1m}}]$
$\therefore M=-14$
So, the magnification will be 14 in this case.
So, the answer to the question is option (D).
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.