Question

Describe terrestrial telescope on the basis of the following points:
(i) A clear labelled diagram;
(ii) Expression for magnifying power when the final image is at least distance of distinct vision D.

Answer 1

Hint: A terrestrial telescope is a refracting telescope that works on the principle of refraction. It uses three convex lenses called the objective lens, the erecting lens and the eye lens. The combination of these lenses forms an enlarged image of the object.

Formula used:
$m=\dfrac{\beta }{\alpha }$

Complete step by step answer:
(i) A telescope is an optical instrument that helps to see far away objects clearly. A telescope is used to see celestial bodies like stars, moons and other planets.
A terrestrial telescope is a refracting telescope that works on the principle of refraction. For this, it uses a lens. A terrestrial telescope uses three convex (converging) lenses as you can see in the ray diagram below.

The three lenses are called the objective lens, the erecting lens and the eye lens.
The rays of light from a very far object are incident on the objective lens. They refract through the objective lens and intersect at a point on the other side of the lens. Due to this an intermediate image (A’B’) of the object is formed between the erecting lens and the objective lens.
The image A’B; acts as an object for the erecting lens and one more intermediate image (A’’B’’) is formed on the other side of this lens. However, this image formed is inverted as shown.
Lastly, the rays of light refract through the eye lens and a final image of the object is formed. As you can see the final image is formed, is erect and enlarged and this can be seen through the eye lens.
In this way, a terrestrial telescope formed an enlarged image of a very far away object.
(ii) Magnifying power (m) of a telescope is defined as the ratio of the angle subtended by the final image ($\beta $) at the eye to the angle subtended by the object ($\alpha $)
This means that $m=\dfrac{\beta }{\alpha }$ … (i)
The erecting lens is placed in such a way that the image AB’ is formed at a distance of 2f (f is focal length of the lens) from it. When an object is placed at a distance of 2f from the lens, the image formed is real and formed at a distance of 2f form the lens on the other side. In addition, the size of the image is equal to the size of the object. This can be seen in the ray diagram.
Therefore, $A'B'=A''B''$ …. (ii).
From the diagram we get that $\tan \alpha =\dfrac{A'B''}{{{f}_{o}}}$ and $\tan \beta =\dfrac{A''B''}{{{f}_{e}}}$.
$\Rightarrow {{f}_{0}}\tan \alpha =A'B'$ …. (iii)
And ${{f}_{e}}\tan \beta =A''B''$ ….. (iv).
From (ii), (iii) and (iv) we get that ${{f}_{0}}\tan \alpha ={{f}_{e}}\tan \beta $ ….. (v).
Since, the object and the image formed are very far, $\alpha $ and $\beta $ are small angles.
Therefore, $\tan \alpha \approx \alpha $ and $\tan \beta \approx \beta $.
Substitute these values in (v).
$\Rightarrow {{f}_{0}}\alpha ={{f}_{e}}\beta $
$\Rightarrow \dfrac{{{f}_{0}}}{{{f}_{e}}}=\dfrac{\beta }{\alpha }$.
But $m=\dfrac{\beta }{\alpha }$
$\Rightarrow m=\dfrac{{{f}_{0}}}{{{f}_{e}}}$
If the final image is formed at least distance of distinct vision D, the magnifying power is equal to $\Rightarrow m=\dfrac{{{f}_{0}}}{{{f}_{e}}}\left( 1+\dfrac{D}{{{f}_{e}}} \right)$ .

Note:
The other type of refracting telescope is an astronomical telescope. It is used to enlarge the celestial bodies.
However, it uses only two convex lenses. The erecting lens is absent. As a result, the final image formed is inverted.
The rays of light refract through the eye lens and a final image of the object is formed.