The magnifying power of a telescope is 6 in normal adjustment. The distance between its two lenses is 35 cm. The focal lengths of objective and eye-piece are, respectively:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 30\,\mathrm{cm},\ 5\,\mathrm{cm} \)
For an astronomical telescope in normal adjustment the magnifying power is the ratio of the focal lengths, and the distance between the lenses is their sum.
\( \displaystyle m = \frac{f_{o}}{f_{e}} = 6 \;\Rightarrow\; f_{o} = 6f_{e} \)
\( \displaystyle f_{o} + f_{e} = 35\,\mathrm{cm} \;\Rightarrow\; 7f_{e} = 35\,\mathrm{cm} \;\Rightarrow\; f_{e} = 5\,\mathrm{cm} \)
\( \displaystyle f_{o} = 6 \times 5 = 30\,\mathrm{cm} \)