An astronomical telescope of ten fold angular magnification has a length of 44 cm. The focal length of the objective is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 40\,\mathrm{cm} \)
For an astronomical telescope in normal adjustment, the angular magnification is the ratio of the focal lengths, and the length of the tube is their sum.
\( \displaystyle m = \frac{f_o}{f_e} = 10 \;\Rightarrow\; f_o = 10 f_e \)
\( \displaystyle L = f_o + f_e = 10 f_e + f_e = 11 f_e = 44\,\mathrm{cm} \)
\( f_e = 4\,\mathrm{cm} \)
\( f_o = 10 \times 4 = 40\,\mathrm{cm} \)
Note: The same question was also asked in CUET (UG) 2023 on 27-05-2023 (Slot 3).