Question

In an astronomical telescope of $135\text{ cm}$ length kept in normal adjustment, if the difference between the focal lengths of the objective and eyepiece is $125\text{ cm}$, then the magnification of the telescope is:

• A. $13.5$
• B. $12.5$
• C. $26$
• D. $28$

Hint:

For an astronomical telescope in normal adjustment, \[ L=f_o+f_e \] and \[ m=\frac{f_o}{f_e}. \] If the sum $S$ and difference $D$ of two quantities are known, then \[ \text{Larger quantity}=\frac{S+D}{2} \] and \[ \text{Smaller quantity}=\frac{S-D}{2}. \] Using this shortcut, the focal lengths can be obtained immediately without lengthy substitution.

Answer 1

The Correct Option is C

Step 1: Recall the normal-adjustment relations.
For an astronomical telescope in normal adjustment, the tube length equals the sum of the focal lengths, $L = f_o + f_e$, and the magnification is $m = \dfrac{f_o}{f_e}$.
Step 2: Write the two given conditions.
The length gives $f_o + f_e = 135\,cm$, and the difference gives $f_o - f_e = 125\,cm$ (objective is the larger).
Step 3: Solve for the objective focal length.
Adding the two equations, $2f_o = 260$, so $f_o = 130\,cm$.
Step 4: Solve for the eyepiece focal length.
Subtracting, $2f_e = 135 - 125 = 10$, so $f_e = 5\,cm$.
Step 5: Compute the magnification.
$m = \dfrac{f_o}{f_e} = \dfrac{130}{5} = 26$.
Step 6: Conclude.
The magnifying power of the telescope is $26$, which is option (3).
\[ \boxed{m = 26} \]