# Q 9.28  A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. What is the magnifying power of the telescope for viewing distant objects when  (b) the final image is formed at the least distance of distinct vision (25cm)?

Given,

the focal length of the objective lens $f_{objective}=140cm$

the focal length of the eyepiece lens $f_{eyepiece}=5cm$

normally, least distance of vision = 25cm

Now,

as we know magnifying power when the image is at d = 25 cm is

$m=\frac{f_{objective}}{f_{eyepiece}}(1+\frac{f_{eyepiece}}{d})=\frac{140}{5}(1+\frac{5}{25}) = 33.6$

Hence magnification, in this case, is 33.6.

Exams
Articles
Questions