In a compound microscope, the focal length of the objective lens and the eye lens are 4 mm and 25 mm and the length of the tube is 16 cm . For relaxed eye position, find its magnifying power.

Question

In a compound microscope, the focal length of the objective lens and the eye lens are 4 mm and 25 mm and the length of the tube is 16 cm . For relaxed eye position, find its magnifying power. (A) 32.75 (B) 327.5 (C) 0.3275 (D) None of the above

Tardigrade Admin · Accepted Answer

fo=4 mm=0.4 cm fe=25 mm=2.5 cm Length of the tube, Lt = 16 cm Magnifying power for relaxed eye position is given by: m=((Lt - fo - fe) D/fo fe) m=((. 16 - 0 . 4 - 2 . 5 .) × 25/0 . 4 × 2 . 5) m=327.5

Q. In a compound microscope, the focal length of the objective lens and the eye lens are $4 mm$ and $25 mm$ and the length of the tube is $16 c m$ . For relaxed eye position, find its magnifying power.

$32.75$

$327.5$

$0.3275$

None of the above

$f_{o} = 4 mm = 0.4 c m$
$f_{e} = 25 mm = 2.5 c m$
Length of the tube, $L_{t}$ = $16 c m$
Magnifying power for relaxed eye position is given by:
$m = \frac{( L _{t} - f _{o} - f _{e} ) D}{f _{o} f _{e}}$
$m = \frac{( 16 - 0.4 - 2.5 ) \times 25}{0.4 \times 2.5}$
$m = 327.5$

Q. In a compound microscope, the focal length of the objective lens and the eye lens are 4mm and 25mm and the length of the tube is 16cm . For relaxed eye position, find its magnifying power.

32.75

327.5

0.3275