First slide

Ray optics

Question

In a compound microscope the objective lens and eye piece have focal length of 0.95 cm and 5 cm respectively, and are kept at a distance of 20 cm. The last image is formed at a distance of 25 cm from the eye piece calculate magnifying power:

Moderate

A

94
B

84
C

75
D

88

Solution

$f_{0} = 0 .95 m = \frac{V_{0}}{u_{0}} (1 + \frac{D}{f_{e}})$
$f_{e} = 5 cm u_{e} = \frac{25}{6}$
For eyepiece
$1 + \frac{D}{f_{e}} = \frac{V_{e}}{U_{e}} u = \frac{V_{e}}{6}$
$V_{0} = 20 - \frac{25}{6} = \frac{95}{6} m_{0} = \frac{\frac{95}{6} - 0 .95}{0 .95}$
$\approx \frac{94}{6}$
$m_{net} = \frac{94}{6} \times 6 = 94$

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