Match List-I with List-II,
(Symbols have their usual meaning, D is the least distance of the distinct vision)
| List-I | List-II | ||
|---|---|---|---|
| Optical Instrument | Magnifying Power | ||
| (A) | For a simple microscope the magnifying power when the final image is at near point | (I) | \( m = (f_o/f_e) \) |
| (B) | For a compound microscope, the magnifying power is | (II) | \( m = (1 + D/f) \) |
| (C) | For an astronomical telescope to see distant objects, the magnifying power is | (III) | \( m = D/f \) |
| (D) | For a simple microscope, when the final image is at infinity, the magnifying power is | (IV) | \( m = (v/u)(1 + D/f_e) \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
The magnifying power compares the angle subtended at the eye by the image with that subtended by the object at the least distance of distinct vision D (for a telescope, by the distant object).
(A) For a simple microscope the magnifying power when the final image is at near point — this matches (II) \( m = (1 + D/f) \). With the image at D, the lens equation gives \( u = -Df/(D+f) \), so \( m = D/|u| = 1 + D/f \).
(B) For a compound microscope, the magnifying power is — this matches (IV) \( m = (v/u)(1 + D/f_e) \). The objective magnification \( v/u \) is multiplied by the eyepiece acting as a simple microscope with the final image at D.
(C) For an astronomical telescope to see distant objects, the magnifying power is — this matches (I) \( m = (f_o/f_e) \). In normal adjustment the final image is at infinity and \( m = f_o/f_e \).
(D) For a simple microscope, when the final image is at infinity, the magnifying power is — this matches (III) \( m = D/f \). The object is at the focus, so the angle it subtends is \( h/f \) against \( h/D \) at the near point.
NCERT (2026-27):
Summary – …
