Question

Define magnifying power of a telescope. Write its expression. A small telescope has an objective lens of focal length $150\,cm$ and an eye piece of focal length $5\,cm$ . If this telescope is used to view a $100\,m$ high tower $3\,km$ away, find the height of the final image when it is formed $25\,cm$ away from the eye piece.

Answer 1

In order to answer this question, first we will define the magnifying power of a telescope and then we will write the expression for the magnifying power of a telescope. And after that, we will first rewrite the given facts from the question with their respective terms and units, and then we will apply the formula of magnifying power of the telescope and we will do further calculations.

Complete step by step answer:
The angle subtended at the eye by the picture created at the least distance of distinct vision to the angle subtended at the eye by the object lying in infinity is defined as the magnifying power of a telescope.Expression of the magnifying power of a telescope:-
$M = \dfrac{{{f_0}}}{{{f_e}}}(1 + \dfrac{{{f_e}}}{D})$

Given that: Focal length of an object, ${f_o} = 150\,cm$.
Focal length of the eye piece, ${f_e} = 5\,cm$.
Least distance of the distinct vision, $D = 25\,cm$.
So, we will apply the formula of magnifying power of a telescope:
$M = \dfrac{{ - 150}}{5}(1 + \dfrac{5}{{25}}) = - 36$
$\Rightarrow M = \dfrac{\beta }{\alpha } \\\ \Rightarrow M= \dfrac{{\tan \beta }}{{\tan \alpha }}$................(As angles are small)
Since,
$\tan \alpha = \dfrac{H}{u} \\\ \Rightarrow \tan \alpha = \dfrac{{100}}{{3000}} \\\ \Rightarrow \tan \alpha = \dfrac{1}{{30}}$

Here, $H$ is the height of the object and $u$ is the distance of the object from the objective.
$M = \dfrac{{\tan \beta }}{{\dfrac{1}{{30}}}} \\\ \Rightarrow M = \dfrac{{ - 36}}{{30}} = \dfrac{{{H^`}}}{D}$
Here, ${H^`}$ is the height of the image and $D$ is the distance of the image formation.Therefore,
${H^`} = \dfrac{{ - 36 \times 25}}{{30}} \\\ \therefore {H^`}= - 30\,cm$
Here, negative signs indicate that we get an inverted image.

Note: The magnification of telescopes has practical limits. The laws of optics and the nature of the human eye determine these. Under ideal conditions, the maximum useable power is equal to 50-60 times the aperture of the telescope (in inches). Higher power typically results in a weak, low-contrast image.