Question

A planet is observed by an astronomical reflecting telescope having an objective of focal length $16m$ and an eye-piece of focal length $2cm$. Then :

• A. the distance between the objective and the eye - piece is $16.02m$
• B. the angular magnification of the planet is $800$
• C. the image of planet is inverted
• D. the objective is larger than eye - piece

Answer 1

Answer: (A), (B), (D)

the distance between the objective and the eye - piece is $16.02m$
the angular magnification of the planet is $800$
the objective is larger than eye - piece

A telescope uses two co-axially placed convex lenses in such a way that the focus of objective lens is past the focus of the eye piece as evident here from the focal lengths of the objective lens and the eye piece. $\therefore$ The objective lens is larger than eye piece.
Length of the tube is given as
$L=f_o+f_e=16m+2cm=16.02m$
Angular magnification is given as:
$m=\frac{f_o}{f_e}=\frac{1600}{2}=800$

Since, the first lens or the objective lens produces a real and inverted image of the object to be observed and this real image formed acts as an object for the eye piece convex lens and is between the focus and the optical center, so, an inverted, virtual and enlarged image of the object is formed. $\therefore$ The final image is inverted.

Hence, the correct answers are OPTIONS A,B and D.