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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsOptical instrumentsEasy2 minPYQ_2020

PhysicsEasysingle 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 lengths of the lenses are

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Answer:

Solution:

$M = \frac{f_{0}}{f_{e}}$

$9 = \frac{f_{0}}{f_{e}}$ or $f_{0} = 9 f_{e}$

Also, $L = f_{o} + f_{e}$ or $20 = f_{o} + f_{e}$

Or $20 = 9 f_{e} + f_{e}$ or $20 = 10 f_{e}$

Or $f_{e} = 2 c m$

$∴ f_{o} = 9 \times 2 c m = 18 c m$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Geometrical OpticsSubtopic:Optical instruments

⏱ 2mℹ️ Source: PYQ_2020

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