If FO and FE are the focal lengths of the objective and eyepiece of a telescope, respectively, then its magnifying power in normal adjustment will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → FO/FE
In normal adjustment the final image of a telescope is at infinity: the objective forms the image of the distant object at its focus, which is also the focus of the eyepiece. The magnifying power is the ratio of the angle subtended at the eye by the final image to the angle subtended by the object.
\( \displaystyle m = \frac{\beta}{\alpha} = \frac{F_O}{F_E} \)
Why the other options are incorrect
Option (1) FO + FE — Incorrect. \( F_O + F_E \) is the length of the telescope (distance between the lenses) in normal adjustment, not its magnifying power.
Option (2) FO × FE — Incorrect. The product of the focal lengths has the dimensions of area; a magnifying power has no dimensions.
Option (4) (FO + FE)/2 — Incorrect. Half the length of the telescope is not a magnifying power.
NCERT (2026-27):
Telescope – “… The magnifying power \( m \) is the ratio of the angle \( \beta \) subtended at the eye by the final image to the angle \( \alpha \) which the object subtends at the lens or the eye. Hence
\( \displaystyle m \approx \frac{\beta}{\alpha} \approx \frac{h}{f_e}\cdot\frac{f_o}{h} = \frac{f_o}{f_e} \qquad (9.46) \)
In this case, the length of the telescope tube is \( f_o + f_e \).”