The magnification of a refracting type telescope is 4. The separation between the objective and the eyepiece is 35 cm and the final image is formed at infinity. The focal lengths of the objective and eyepiece are, respectively.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 28 cm and 7 cm
For a refracting telescope in normal adjustment (final image at infinity), the magnifying power is \( f_o/f_e \) and the tube length is \( f_o + f_e \).
\( \displaystyle m = \frac{f_o}{f_e} = 4 \;\Rightarrow\; f_o = 4f_e \)
\( f_o + f_e = 35\,\mathrm{cm} \)
\( 4f_e + f_e = 35\,\mathrm{cm} \;\Rightarrow\; f_e = 7\,\mathrm{cm} \)
\( f_o = 4 \times 7\,\mathrm{cm} = 28\,\mathrm{cm} \)