Question

An astronomical telescope in normal adjustment receives light from a distant source S the tube length is now decreased slightly, then
A. No image will be formed
B. A virtual image of S will be formed at a finite distance
C. A large, real image of S will be formed behind the eyepiece far away from it
D. A small, teal image of S will be formed behind the eyepiece does to it

Answer 1

Hint: In this question use the direct relationship between the focal length of objective, focal length of eyepiece and tube length that is $L = {f_o} + {f_e}$. Use the concept that if the tube length is decreased then the position at which the image will be formed changes as now the image formed will lie between the principal focus and the optical center. This will help approaching the answer.

Complete step-by-step answer:
We all know in an astronomical telescope the tube length is given as the sum of the focal length of the objective and the focal length of the eyepiece.
$L = {f_o} + {f_e}$
Where, L = length of the tube
              ${f_o}$ = focal length of the objective
              ${f_e}$ = focal length of the eye piece
Now it is given that in normal adjustment it receives light from a distant source S.
Now if the tube length is slightly decreased then the image formed by the objective lens will lie between the principal focus and the optical center of the eye lens instead of lying at the focus of the eye lens.
So as the image does not form at the focus of the eye lens, so the image formed is a virtual (i.e. it is not a real image) and erect image.
And the position of this image is at a finite distance from the eye lens.
So a virtual image of S will be formed at a finite distance.
So this is the required answer.
Hence option (B) is the correct answer.

Note – Astronomical telescopes are generally used to observe celestial bodies and are far more long ranged as to the terrestrial telescope. Moreover, an advanced astronomical telescope does not require an image-erecting system for inverted images.