Question

A simple telescope consisting of an objective of focal length $60cm$ and a single eye lens of the focal length $5cm$ is focused on a distant object in such a way that parallel rays come out from the eye lens. If the object subtends an angle ${2^ \circ }$ at the objective, the angular width of the image:
$\left( A \right)$ ${10^ \circ }$
$\left( B \right)$ ${24^ \circ }$
$\left( C \right)$ ${50^ \circ }$
$\left( D \right)$ ${\dfrac{1}{6}^ \circ }$

Answer 1

This question is based on the magnification of the telescope. In the telescope, the magnification of the telescope is defined as the focal length of the objective by the focal length of the eyepiece. Also, the angle subtended by both of them is the image and the object.
Formula used:
Magnification of telescope is given as,
$\Rightarrow M = \dfrac{{Focal{\text{ length of objective}}}}{{Focal{\text{ length of eyepiece}}}} = \dfrac{{angle{\text{ subtended by image }}}}{{angle{\text{ subtended by object}}}}$

Complete step by step solution: Telescopes are used to view distant objects clearly with larger magnification and that’s why when we have to see the stars or the moon we have seen them with the help of different types of telescope. Since these objects are not possible to see with the naked eyes, our great scientists developed this and made life more beautiful by giving us the chance to see the world.
In the question, it is given that the focal length of $60cm$ and a single eye lens having the focal length $5cm$ which is being focused on a distant object.
Telescopes are used to view distant objects clearly with larger magnification and that’s why when we have to see the stars or the moon we have seen them with the help of different types of telescope. Since these objects are not possible to see with the naked eyes, our great scientists developed this and made life more beautiful by giving us the chance to see the world.
So now we have to find the angle subtended by the image.
Therefore to calculate this we will first know the formula to find it.
Magnification of telescope is given as,
$\Rightarrow M = \dfrac{{Focal{\text{ length of objective}}}}{{Focal{\text{ length of eyepiece}}}} = \dfrac{{angle{\text{ subtended by image }}}}{{angle{\text{ subtended by object}}}}$
Now we will put the values and will be able to get the required angle subtended by an image.
$\Rightarrow \dfrac{{60}}{5} = \dfrac{{{\text{angle subtended by image}}}}{2}$
Now we will solve this and will get the following result,
$\Rightarrow angle{\text{ subtended by image = 2}}{{\text{4}}^ \circ }$

Therefore the angle subtended by an image will be ${24^ \circ }$.

Note: Telescopes are meant for viewing distant objects, producing a picture that's larger than the image that may be seen with the unaided eye. Telescopes gather much more light than the eye, permitting dim objects to be observed with larger magnification and higher resolution.