Question

The length of the compound microscope is 15 cm. The magnifying power for relaxed eye is 25. If the focal length of eye lens is 6 cm then the object distance for objective lens will be ______.

Options

• 1.3 cm

• 1.5 cm

• 1.7 cm

• 1.9 cm

Answer 1

The length of the compound microscope is 15 cm. The magnifying power for relaxed eye is 25. If the focal length of eye lens is 6 cm then the object distance for objective lens will be 1.5 cm.

Explanation:

Let L be the length of compound microscope, v₀ be the image distance and f_e be focal length of the eye lens.

\[\therefore\] L = v₀ + f_e

\[\therefore\] 15 = v₀ + 6

\[\therefore\] v₀ = 9 cm

Magnifying power of microscope for relaxed eye,

m = \[\frac {\mathrm {v_0}}{\mathrm {u_0}}\] . \[\frac {\mathrm {D}}{\mathrm {f_e}}\]

25 = \[\frac {9}{\mathrm {u_0}}\] . \[\frac {25}{6}\]

\[\therefore\] u₀ = 1.5 cm