Question

The length of the compound microscope is 14 cm. The magnifying power for relaxed eye is 25. If the focal length of eye lens is 5 cm, then the object distance for objective lens will be

• A. 1.8 cm
• B. 1.5 cm
• C. 2.1 cm
• D. 2.4 cm

Answer 1

Answer: (A)

1.8 cm

Answer 2

$\Rightarrow v _ { o } = 9 \mathrm {~cm}$

Magnifying power of microscope for relaxed eye

$m = \frac { v _ { o } } { u _ { o } } \cdot \frac { D } { f _ { e } }$ or $u _ { o } = \frac { 9 } { 5 } = 1.8 \mathrm {~cm}$