Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsOptical instrumentsEasy2 minPYQ_2020

PhysicsEasysingle choice

The magnifying power of a telescope with tube length $60 c m$ is $5 .$ What is the focal length of its eye piece?

Answer:

Solution:

$m = \frac{f_{0}}{f_{e}}$
$5 = \frac{f_{0}}{f_{e}}$
$f_{o} = 5 f_{e}$
$f_{o} + f_{e} = 60$
$6 f_{e} = 60$
$f_{e} = 10$

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Optical instruments

⏱ 2mℹ️ Source: PYQ_2020

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