Question:
The magnifying power of a telescope of length $76$ cm in the normal adjustment is $75$. Then the focal lengths of the objective and eyepiece lenses are, respectively,
The magnifying power of a telescope of length $76$ cm in the normal adjustment is $75$. Then the focal lengths of the objective and eyepiece lenses are, respectively,
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For telescopes:
- $M = \frac{f_o}{f_e}$
- Length = sum of focal lengths
Updated On: Apr 30, 2026
- $50$ cm, $25$ cm
- $70$ cm, $5$ cm
- $75$ cm, $1$ cm
- $73$ cm, $2$ cm
- $60$ cm, $15$ cm
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The Correct Option is C
Solution and Explanation
Concept:
For telescope in normal adjustment:
\[
M = \frac{f_o}{f_e}, \quad L = f_o + f_e
\]
Step 1: Write given equations.
\[ \frac{f_o}{f_e} = 75 \quad \cdots (1) \] \[ f_o + f_e = 76 \quad \cdots (2) \]
Step 2: Substitute from (1).
\[ f_o = 75 f_e \]
Step 3: Use in (2).
\[ 75 f_e + f_e = 76 \Rightarrow 76 f_e = 76 \Rightarrow f_e = 1\ \text{cm} \]
Step 4: Find $f_o$.
\[ f_o = 75 \times 1 = 75\ \text{cm} \]
Step 1: Write given equations.
\[ \frac{f_o}{f_e} = 75 \quad \cdots (1) \] \[ f_o + f_e = 76 \quad \cdots (2) \]
Step 2: Substitute from (1).
\[ f_o = 75 f_e \]
Step 3: Use in (2).
\[ 75 f_e + f_e = 76 \Rightarrow 76 f_e = 76 \Rightarrow f_e = 1\ \text{cm} \]
Step 4: Find $f_o$.
\[ f_o = 75 \times 1 = 75\ \text{cm} \]
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