Question

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?

• A. $24\text{ cm}$
• B. $12\text{ cm}$
• C. $6\text{ cm}$
• D. $3\text{ cm}$

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!

Answer 1

The Correct Option is D

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 $f_e$.
\[ 8f_e + f_e = 27 \] \[ 9f_e = 27 \implies f_e = \frac{27}{9} = 3\text{ cm} \]