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:

30 cm, 5 cm

The correct answer is Option (1) → 30 cm, 5 cm

Given:

Magnifying power, $M = 6$

Distance between lenses (length of telescope), $L = 35\ \text{cm}$

For telescope in normal adjustment:

$M = \frac{f_o}{f_e}$, where $f_o$ = focal length of objective, $f_e$ = focal length of eye-piece

Length of telescope: $L = f_o + f_e = 35\ \text{cm}$

From magnifying power: $f_o = M f_e = 6 f_e$

Substitute in telescope length:

$L = f_o + f_e = 6 f_e + f_e = 7 f_e = 35\ \text{cm}$

$f_e = \frac{35}{7} = 5\ \text{cm}$

$f_o = 6 f_e = 6 \cdot 5 = 30\ \text{cm}$

∴ Focal lengths: Objective = 30 cm, Eye-piece = 5 cm