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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsOptical instrumentsMedium2 minPYQ_2012

PhysicsMediumsingle choice

The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is $20 cm$ . The focal length of lenses are

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Solution:

Given, $M = \frac{f _{0}}{f _{e}} = 9$ and $f_{0} + f_{e} = 20$ $f_{o} = 9 f_{e}$ So, $9 f_{e} + f_{e} = 20$ $∴ f_{e} = 2 cm f_{o} = 9 \times 2 f_{0} = 18 cm$

Stream:NEETSubject:PhysicsTopic:Geometrical OpticsSubtopic:Optical instruments

⏱ 2mℹ️ Source: PYQ_2012

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