Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsOptical instrumentsMedium2 minPYQ_2014

PhysicsMediumsingle choice

The magnifying power of a telescope is $9.$ When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be $20 cm$ . The focal length of lenses are

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Solution:

$\frac{f _{0}}{f _{e}} = 9, ∴ f_{0} = 9 f_{e}$ Also $f_{0} + f_{e} = 20 (∵$ final image is at infinity $)$ $9 f_{e} + f_{e} = 20, f_{e} = 2 cm, ∴ f_{0} = 18 cm$

Stream:BITSATSubject:PhysicsTopic:Geometrical OpticsSubtopic:Optical instruments

⏱ 2mℹ️ Source: PYQ_2014

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