An astronomical telescope has a converging eye-piece of focal length 5 cm and objective of focal length 80 cm. When the final image is formed at the least distance of distinct vision (25 cm), the separation between the two lenses is
A. 75 cm
B. 80 cm
C. 84.2 cm
D. 85.0 cm

Hint: An astronomical telescope is an optical instrument used to see the magnified image of distant heavenly bodies. There are two main lenses in an astronomical telescope namely, objective and eye-piece. To solve this question, use the formula for separation between two lenses in an astronomical telescope. Substitute the given values and find the separation between the lenses.
.
Formula used:
$L= {f}_{o}+ \dfrac {{f}_{e}D}{{f}_{e}+D}$

Complete solution:

Given:
Focal length of eye-piece, ${f}_{e}$= 5 cm
Focal length of objective, ${f}_{o}$= 80 cm
Distance of image, v= 25 cm=D
For an astronomical telescope, separation between two lenses is given by,
$L= {f}_{o}+ \dfrac {{f}_{e}D}{{f}_{e}+D}$
Substituting values in above equation we get,
$L= 80+ \dfrac {5 \times 25}{5+25}$
$\Rightarrow L= 80 + \dfrac {125}{30}$
$\Rightarrow L= 80 +4.17$
$\Rightarrow L= 84.17$
$\Rightarrow L\approx 84.2 cm$
Thus, the separation between two lenses is 84.2 cm.

So, the correct answer is option C i.e. 84.2 cm.

Note:
Astronomical telescopes are used to see heavenly bodies. Students must remember that the objective is a convex lens of large aperture and an eye-piece is a convex lens of small aperture. Students must also remember that for an astronomical telescope the focal length of an object is always greater than the focal length of the eye-piece or eye lens. Intermediate image formed due to this telescope is real, inverted and small. Whereas the final image formed is virtual, inverted and small.