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 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