A planet is observed by an astronomical refracting telescope in norma…

The correct answer is Option (2) → (A), (B) and (C) only

Convert the powers into focal lengths with f = 1/P (P in dioptres, f in metres).

    \( \displaystyle f_o = \frac{1}{2}\,\mathrm{m} = 50\,\mathrm{cm} \)

    \( \displaystyle f_e = \frac{1}{20}\,\mathrm{m} = 5\,\mathrm{cm} \)

In normal adjustment the image formed by the objective lies at the focus of the eyepiece, so the final image is at infinity.

(A) The distance between the objective and the eyepiece is 55 cm. — Correct. The tube length is \( f_o + f_e = 50 + 5 = 55\,\mathrm{cm} \).

(B) The effective object distance for the eye piece is 5 cm. — Correct. The intermediate image lies at the focus of the eyepiece, 5 cm from it.

(C) The final image formed is inverted. — Correct. The objective forms a real, inverted image and the eyepiece magnifies it without inverting it again.

(D) The angular magnification of the telescope is 20. — Incorrect. In normal adjustment

    \( \displaystyle m = \frac{f_o}{f_e} = \frac{50\,\mathrm{cm}}{5\,\mathrm{cm}} = 10 \)

20 is the power of the eyepiece, not the magnification.

NCERT (2026-27):
Telescope – “… Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image. 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 \( f_o + f_e \).”