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

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

Tardigrade Admin · Accepted Answer

L∞=vo+fe ⇒ 14=vo+5 ⇒ vo=9 cm Magnifying power of microscope for relaxed eye m=(vo/uo) ⋅ (D/fe) or 25=(9/uo) ⋅ (25/5) or uo=(9/5)=1.8 cm

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

1.8 cm

1.5 cm

2.1 cm

2.4 cm

$L_{\infty} = v_{o} + f_{e}$
$\Rightarrow 14 = v_{o} + 5$
$\Rightarrow v_{o} = 9 c m$
Magnifying power of microscope for relaxed eye
$m = \frac{v _{o}}{u _{o}} \cdot \frac{D}{f _{e}}$ or
$25 = \frac{9}{u _{o}} \cdot \frac{25}{5}$ or
$u_{o} = \frac{9}{5} = 1.8 c m$

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