Question

A small telescope has an objective lens of focal length $144\text{cm}$ and an eyepiece of focal length $6.0\text{cm}$. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Answer 1

Step 1: Draw the diagram.

Step 2: Find the magnifying power.

Given,
Focal length of the objective lens, $f_o=144\text{cm}$
Focal length of the eyepiece, $f_e=6.0\text{cm}$
The magnifying power of the telescope can be calculated as,
$m=\frac{f_o}{f_e}$
$m=\frac{144}{6}$
$m=24$

Step 3: Find the separation between objective lens and eyepiece.

As we know that,
Focal length of the objective lens,
$f_o=144\text{cm}$
Focal length of the eyepiece, $f_e=6.0\text{cm}$
Therefore,
The separation between the objective lens and the eyepiece is, $L=f_o+f_e$
$L=144+6$
$L=150\text{cm}$

Final Answer: $m=24,L=150\text{cm}$.