The powers of the objective and the eyepiece of an astronomical telescope are 0.5 D and 20 D respectively. Its magnifying power in normal adjustment is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 40
The focal length of a lens in metres is the reciprocal of its power in dioptres, \( f=1/P \). In normal adjustment the final image is at infinity, and the magnifying power of an astronomical telescope is \( m=f_o/f_e \).
\( \begin{aligned} f_o &= \frac{1}{P_o} \\ &= \frac{1}{0.5\,\mathrm{D}} \\ &= 2\,\mathrm{m} \end{aligned} \)
\( \begin{aligned} f_e &= \frac{1}{P_e} \\ &= \frac{1}{20\,\mathrm{D}} \\ &= 0.05\,\mathrm{m} \end{aligned} \)
\( \begin{aligned} m &= \frac{f_o}{f_e} \\ &= \frac{2\,\mathrm{m}}{0.05\,\mathrm{m}} \\ &= 40 \end{aligned} \)
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
\( \displaystyle m\approx\frac{\beta}{\alpha}\approx\frac{h}{f_e}\cdot\frac{f_o}{h}=\frac{f_o}{f_e} \)\( \qquad (9.46) \)”