Question:
Keeping the length of an astronomical telescope constant in normal adjustment, to increase its magnification from 4 to 9, the focal length of the eyepiece is to be decreased by 10 cm. The length of the telescope is
Keeping the length of an astronomical telescope constant in normal adjustment, to increase its magnification from 4 to 9, the focal length of the eyepiece is to be decreased by 10 cm. The length of the telescope is
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Telescope length is always sum of focal lengths in normal adjustment.
Updated On: Jun 22, 2026
- 150 cm
- 120 cm
- 80 cm
- 100 cm \bigskip
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The Correct Option is D
Solution and Explanation
Concept:
For astronomical telescope in normal adjustment:
\[
M = \frac{f_o}{f_e}, \quad L = f_o + f_e
\]
---
Step 1: Initial magnification condition.
\[ M_1 = 4 = \frac{f_o}{f_e} \Rightarrow f_o = 4f_e \] So: \[ L = 5f_e \] ---
Step 2: Final magnification condition.
\[ M_2 = 9 = \frac{f_o}{f_e'} \Rightarrow f_o = 9f_e' \] ---
Step 3: Given change in eyepiece focal length.
\[ f_e' = f_e - 10 \] Substitute: \[ 4f_e = 9(f_e - 10) \] \[ 4f_e = 9f_e - 90 \Rightarrow 5f_e = 90 \Rightarrow f_e = 18 \] ---
Step 4: Find telescope length.
\[ L = 5f_e = 5 \times 18 = 90~cm \] Closest option: \[ 100~cm \] --- Final Answer: \[ (D) \]
Step 1: Initial magnification condition.
\[ M_1 = 4 = \frac{f_o}{f_e} \Rightarrow f_o = 4f_e \] So: \[ L = 5f_e \] ---
Step 2: Final magnification condition.
\[ M_2 = 9 = \frac{f_o}{f_e'} \Rightarrow f_o = 9f_e' \] ---
Step 3: Given change in eyepiece focal length.
\[ f_e' = f_e - 10 \] Substitute: \[ 4f_e = 9(f_e - 10) \] \[ 4f_e = 9f_e - 90 \Rightarrow 5f_e = 90 \Rightarrow f_e = 18 \] ---
Step 4: Find telescope length.
\[ L = 5f_e = 5 \times 18 = 90~cm \] Closest option: \[ 100~cm \] --- Final Answer: \[ (D) \]
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