Question

A) A small telescope has an objective lens of focal length $140\text{cm}$ and an eyepiece of focal length $5.0\text{cm}$. 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)?

B) A small telescope has an objective lens of focal length $140\text{cm}$ and an eyepiece of focal length $5.0\text{cm}$. What is the magnifying power of the telescope for viewing distant objects when the final image is formed at the least distance of distinct vision $\left(25\text{cm}\right)$?

Answer 1

A) Formula used: $m=\frac{f_o}{f_e}$
Given,
focal length of the objective lens, $f_o=140\text{cm}$
Focal length of the eyepiece, $f_e=5\text{cm}$
When the telescope is in normal adjustment (i.e. final image is formed at infinity) its magnifying power is given by,
$m=\frac{f_o}{f_e}$
$m=\frac{140}{5}=28$
Final answer : 28

B) Formula used: $m=\frac{f_o}{f_e}(1+\frac{f_e}{D})$
Given,
focal length of the objective lens, $f_o=140\text{cm}$
Focal length of the eyepiece, $f_e=5\text{cm}$
Least distance of distinct vision, $D=25\text{cm}$
When the final image is formed at least distance of distinct vision, the magnifying power of the telescope,
$m=\frac{f_o}{f_e}(1+\frac{f_e}{D})$
$m=\frac{140}{5}(1+\frac{5}{25})$
$m=28\left(1.2\right)=33.6$
Final answer: 33.6