Match List I with List II
| LIST I | LIST II |
| A. Magnifying power of a simple microscope | I. \( \displaystyle\frac{L}{f_o} \times \frac{D}{f_e} \) |
| B. Magnifying power of a compound microscope | II. \( \displaystyle\frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right) \) |
| C. Magnifying power of a telescope (normal adjustment) | III. \( \displaystyle\left(1 + \frac{D}{f_e}\right) \) |
| D. Magnifying power of a telescope when final image is formed at least distance of distinct vision | IV. \( f_o f_e^{-1} \) |
(Symbols have their usual meanings)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-III, B-I, C-IV, D-II
The magnifying power of an optical instrument is the ratio of the angle subtended at the eye by the final image to the angle subtended by the object seen directly (for a microscope, with the object at the near point \( D \)).
(A) Magnifying power of a simple microscope — this matches (III) \( \left(1 + \frac{D}{f_e}\right) \). A single convex lens of focal length \( f_e \) with the image at the near point \( D \) gives \( m = 1 + D/f \).
(B) Magnifying power of a compound microscope — this matches (I) \( \frac{L}{f_o} \times \frac{D}{f_e} \). The objective magnifies by about \( L/f_o \) (\( L \) is the tube length) and the eyepiece by \( D/f_e \), and the total is the product of the two.
(C) Magnifying power of a telescope (normal adjustment) — this matches (IV) \( f_o f_e^{-1} \). With the final image at infinity, the magnifying power is the ratio of the focal lengths, \( f_o/f_e \).
(D) Magnifying power of a telescope when final image is formed at least distance of distinct vision — this matches (II) \( \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right) \). Bringing the final image to \( D \) makes the eyepiece magnify a little more, which multiplies \( f_o/f_e \) by the factor \( 1 + f_e/D \).
NCERT (2026-27):
Summary – “… 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. … For a compound microscope, the magnifying power is given by m = me × m0 … Approximately,
\( \displaystyle m = \frac{L}{f_o} \times \frac{D}{f_e} \)
where fo and fe are the focal lengths of the objective and eyepiece, respectively, and L is the distance between their focal points.
9. Magnifying power m of a telescope is the ratio of the angle β subtended at the eye by the image to the angle α subtended at the eye by the object.
\( \displaystyle m = \frac{\beta}{\alpha} = \frac{f_o}{f_e} \)”