The eyepiece of an astronomical telescope has a focal length of 5 cm. If the angular magnification in normal adjustment is 10, the distance between the eyepiece and the objective should be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 55 cm
In normal adjustment the final image is at infinity. The angular magnification of an astronomical telescope is then \( m=f_o/f_e \), and the objective and the eyepiece are a distance \( f_o+f_e \) apart.
The focal length of the objective is:
\( \begin{aligned} f_o &= m\,f_e \\ &= 10\times5\,\mathrm{cm} \\ &= 50\,\mathrm{cm} \end{aligned} \)
The distance between the objective and the eyepiece is:
\( \begin{aligned} L &= f_o+f_e \\ &= 50\,\mathrm{cm}+5\,\mathrm{cm} \\ &= 55\,\mathrm{cm} \end{aligned} \)
NCERT (2026-27):
Telescope – “… The magnifying power m is the ratio of the angle β subtended at the eye by the final image to the angle α which the object subtends at the lens or the eye. Hence
\( \displaystyle m\approx\frac{\beta}{\alpha}\approx\frac{h}{f_e}\cdot\frac{f_o}{h}= \)\( \displaystyle \frac{f_o}{f_e} \)\( \qquad (9.46) \)
In this case, the length of the telescope tube is fo + fe.”