The distance between the objective and eyepiece of an astronomical telescope in normal adjustment is $27\text{ cm}$ and its magnifying power is $8$. What is the focal length of the eyepiece?
Show Hint
You can isolate $f_e$ directly from any telescope system data using the short form: $f_e = \frac{L}{m + 1}$. Here, $f_e = \frac{27}{8 + 1} = \frac{27}{9} = 3\text{ cm}$. This completely removes the intermediate substitution step!
Understanding the Concept:
For an astronomical telescope under normal adjustment (where the final image forms at infinity):
The total length of the telescope tube ($L$) is the sum of the focal lengths: $L = f_o + f_e$
The magnifying power ($m$) is given by the ratio: $m = \frac{f_o}{f_e}$
Step 1: Set up the algebraic system from the given data.
We are given:
\[
L = 27\text{ cm} \implies f_o + f_e = 27 \quad \cdots (1)
\]
\[
m = 8 \implies \frac{f_o}{f_e} = 8 \implies f_o = 8f_e \quad \cdots (2)
\]
Step 2: Substitute equation (2) into equation (1) to find $f_e$.
\[
8f_e + f_e = 27
\]
\[
9f_e = 27 \implies f_e = \frac{27}{9} = 3\text{ cm}
\]