Question

The tube length of an astronomical telescope is 105 cm and for normal setting magnifying power is 20. The focal length of the objective is :

• A. 10 cm
• B. 20 cm
• C. 25 cm
• D. 100 cm

Answer 1

The correct option is D 100 cm

In an astronomical telescope with objective lens with focal length $f_o$ and eye piece lens of focal length $f_e$ in the normal mode (where final image is formed at infinite distance for relaxed vision) magnifying power is given by $M$ $=$ $\frac{f_o}{f_e}$ and the tube length (distance between objective and eye piece) is given by $L$ $=$ $f_o$ $+$ $f_e$.

Given that magnifying power is 20.

Therefore, $\frac{f_o}{f_e}$ $=$ 20.

$f_o$ $=$ $20f_e$

$20f_e$ $+$ $f_e$ $=$ 105

So, $f_e$ $=$ $\frac{105}{21}$ $=$ 5 $cm$.

Therefore, the focal length of eye piece is 5 $cm$.

Focal length of objective $f_o$ $=$ 20$f_e$

$=20\times 5=100cm$.

Hence, the Correct Option is ${}^{\prime }{D}^{\prime }$.