Question:
An astronomical telescope in normal adjustment has magnifying power \(5\) for distant objects. The separation between the objective and the eyepiece is \(36\) cm. The focal lengths of the objective and the eyepiece are respectively
An astronomical telescope in normal adjustment has magnifying power \(5\) for distant objects. The separation between the objective and the eyepiece is \(36\) cm. The focal lengths of the objective and the eyepiece are respectively
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For an astronomical telescope in normal adjustment:
\[
M=\frac{f_o}{f_e}
\]
\[
L=f_o+f_e
\]
Solve the two equations simultaneously.
Updated On: Jun 16, 2026
- \(30\) cm, \(30\) cm
- \(30\) cm, \(6\) cm
- \(6\) cm, \(30\) cm
- \(6\) cm, \(6\) cm
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The Correct Option is B
Solution and Explanation
Concept:
For an astronomical telescope in normal adjustment,
\[
M=\frac{f_o}{f_e}
\]
and
\[
L=f_o+f_e
\]
where \(L\) is the tube length.
Step 1: Use magnifying power. \[ \frac{f_o}{f_e}=5 \] \[ f_o=5f_e \]
Step 2: Use the telescope length. \[ f_o+f_e=36 \] \[ 5f_e+f_e=36 \] \[ 6f_e=36 \] \[ f_e=6\text{ cm} \] \[ f_o=30\text{ cm} \] \[\begin{aligned} \boxed{f_o=30\text{ cm},\quad f_e=6\text{ cm}} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.
Step 1: Use magnifying power. \[ \frac{f_o}{f_e}=5 \] \[ f_o=5f_e \]
Step 2: Use the telescope length. \[ f_o+f_e=36 \] \[ 5f_e+f_e=36 \] \[ 6f_e=36 \] \[ f_e=6\text{ cm} \] \[ f_o=30\text{ cm} \] \[\begin{aligned} \boxed{f_o=30\text{ cm},\quad f_e=6\text{ cm}} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.
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