Question

A small telescope has an objective lens of focal length 140 cm and an eye piece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects when the telescope is in normal adjustment (i.e. when the final image is at infinity)

• A. 13
• B. 7
• C. 6
• D. 28

Answer 1

Answer: (D)

28

Answer 2

The magnifying power ($m$) of a telescope in normal adjustment (final image at infinity) is given by the ratio of the focal length of the objective lens ($f_o$) to the focal length of the eyepiece ($f_e$). $m = \frac{f_o}{f_e}$ Given $f_o = 140$ cm and $f_e = 5$ cm. $m = \frac{140 \text{ cm}}{5 \text{ cm}} = 28$