Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsOptical instrumentsEasy2 minPYQ_2020

PhysicsEasysingle choice

In a compound microscope, the objective lens and eyepiece have a focal length of $0.95 c m$ and $5 c m$ respectively and are kept at a distance of $20 c m$ . If the last image is formed at a distance of $25 c m$ from the eyepiece, calculate magnifying power.

Options:

Answer:

Solution:

$f_{0} = 0.95 m = \frac{V_{0}}{u_{0}} (1 + \frac{D}{f_{e}})$
$f_{e} = 5 c m u_{e} = \frac{25}{6}$
For eyepiece
$1 + \frac{D}{f_{e}} = \frac{V_{e}}{U_{e}} u = \frac{V_{e}}{6}$
$V_{0} = 20 - \frac{25}{6} = \frac{95}{6} m_{0} = \frac{\frac{95}{6} - 0.95}{0.95} \approx \frac{94}{6}$
$m_{n e t} = \frac{94}{6} \times 6 = 94$

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Optical instruments

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...