Match List-I with List-II (Symbols have their usual meaning)List-ILis…

The correct answer is Option (1) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)

(A) No chromatic aberration — this matches (III) Reflecting type telescope. Its objective is a mirror, and reflection does not depend on wavelength.

(B) Magnifying power M = (1 + D/f_e) — this matches (I) Simple microscope, with the final image at the near point D.

(C) Magnifying power M = f_o/f_e — this matches (IV) Refracting type telescope in normal adjustment (final image at infinity).

(D) v_o + u_e — this matches (II) Length of compound microscope: the image distance of the objective plus the object distance of the eyepiece, \( L = v_o + |u_e| \).

Note: In the question, (C) \( M = f_o/f_e \) holds for both (III) Reflecting type telescope and (IV) Refracting type telescope. The question can still be solved, because (A) No chromatic aberration fits only (III), and Option (1) is the only option with (A) - (III), (B) - (I) and (D) - (II).

NCERT (2026-27):
Telescope – “… Telescopes with mirror objectives are called reflecting telescopes. There is no chromatic aberration in a mirror.”

Points to Ponder – “… 8. Magnifying power m of a simple microscope is given by m = 1 + (D/f), where D = 25 cm is the least distance of distinct vision and f is the focal length of the convex lens.”

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

    \( \displaystyle m \approx \frac{\beta}{\alpha} \approx \frac{h}{f_e}\cdot\frac{f_o}{h} = \frac{f_o}{f_e} \qquad (9.46) \)”