Which of the following are true for optical instruments?(A) A simple…

The correct answer is Option (3) → (B), (D) and (E) only

(A) A simple microscope is a converging lens of large focal length. — Incorrect. The magnifying power of a simple microscope is \( 1 + D/f \) (image at the near point), which is large only when the focal length is small.

(B) A telescope is used to provide angular magnification of distant objects. — Correct. A telescope does not enlarge the object itself; it makes the angle subtended by a distant object at the eye larger.

(C) The objective lens of a telescope has a smaller focal length than the eyepiece. — Incorrect. The magnifying power is \( f_{o}/f_{e} \), which has to be large, so the objective has the larger focal length.

(D) The magnifying power of a telescope is the ratio of the focal length of the objective lens to the focal length of the eyepiece. — Correct. For a telescope in normal adjustment, \( m = f_{o}/f_{e} \).

(E) A simple microscope is a converging lens of small focal length. — Correct. A small \( f \) gives a large \( 1 + D/f \).

NCERT (2026-27):
The Microscope – “A simple magnifier or microscope is a converging lens of small focal length (Fig. 9.23).”

Telescope – “The telescope is used to provide angular magnification of distant objects (Fig. 9.25). It also has an objective and an eyepiece. But here, the objective has a large focal length and a much larger aperture than the eyepiece.”

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

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