An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length \( f_o \) of the objective and the focal length \( f_e \) of the eyepiece are
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( f_o=30\,\mathrm{cm} \) and \( f_e=6\,\mathrm{cm} \)
For an astronomical telescope with the final image at infinity, the angular magnification is \( m=f_o/f_e \). The length of the tube (the separation of the two lenses) is \( f_o+f_e \).
The two conditions are:
\( f_o=5f_e \)
\( f_o+f_e=36\,\mathrm{cm} \)
Putting the first into the second:
\( 6f_e=36\,\mathrm{cm} \)
\( \Rightarrow\; f_e=6\,\mathrm{cm} \)
\( \begin{aligned} f_o &= 5f_e \\ &= 5\times6\,\mathrm{cm} \\ &= 30\,\mathrm{cm} \end{aligned} \)
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.”