Question

An astronomical telescope has objective and eyepiece focal lengths $40cm$ and $4cm$ respectively. To view an object $200cm$ away from the objective, the lenses must be separated by a distance:
$A)\text{ }54.0cm$
$B)\text{ 37}.3cm$
$C)\text{ 46}.0cm$
$D)\text{ }50.0cm$

Answer 1

Hint: The total tube length will be the distance between the two lenses. It will be the sum of the image distance of the image formed by the objective and the focal length of the eyepiece. This is because, for the final image to be formed at infinity, the object for the eyepiece (which is the image formed by the objective) must be situated at its focus.

Formula used:
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Complete step by step answer:
The final image formed is at infinity for which the object for the eyepiece must be situated at its focus. Now, the object for the eyepiece is nothing but the image formed by the objective. Therefore, it can be said that the distance between the lenses will be the sum of the image distance of the image formed by the objective and the focal length of the eyepiece.
Now, the focal length $f$, image distance $v$ and object distance $u$ for a lens is given by
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ --(1) [Lens formula]
Now, let us analyze the question.
Let the separation between the lenses, that is, the tube length of the telescope be $L$.
Let the image distance of the image formed by the objective be ${{v}_{0}}$.
The focal length of the objective is ${{f}_{0}}=40cm$.
The focal length of the eyepiece is ${{f}_{e}}=4cm$.
The object distance for the objective is ${{u}_{0}}=-200cm$.
(The negative sign is due to sign convention).
Now, as explained above, the tube length will be the sum of the image distance of the image formed by the objective and the focal length of the eyepiece.
$\therefore L={{v}_{0}}+{{f}_{e}}$ --(2)
Now, using (2), we get
$\dfrac{1}{{{f}_{o}}}=\dfrac{1}{{{v}_{o}}}-\dfrac{1}{{{u}_{o}}}$
$\therefore \dfrac{1}{{{v}_{o}}}=\dfrac{1}{{{f}_{o}}}+\dfrac{1}{{{u}_{o}}}$
Putting the values of the variables in the above equation, we get
$\dfrac{1}{{{v}_{o}}}=\dfrac{1}{40}+\dfrac{1}{-200}=\dfrac{1}{40}-\dfrac{1}{200}=\dfrac{5-1}{200}=\dfrac{4}{200}=\dfrac{1}{50}$
$\therefore {{v}_{o}}=50cm$ --(3)
Now, using (2) and (3), we get
$L=50+4=54.0cm$
Hence, the required separation between the lenses in the telescope is $54.0cm$.

So, the correct answer is “Option A”.

Note:
Students may make the mistake of thinking that the tube length of the telescope will be the sum of the focal length of the objective and the eyepiece. However, they must understand that this is valid only in cases of special astronomical telescopes where the object is at infinity and the image is also produced at infinity. In the general case, the tube length of the telescope is the sum of the image distance of the objective and the object distance of the eyepiece.