Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsOptical instrumentsEasy2 minPYQ_2020

PhysicsEasysingle choice

The magnifying power of a telescope is $10$ and length of telescope is $1.1 m$ for normal adjustment. The magnification when image is formed at least distance of distinct vision is

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Answer:

Solution:

Magnifying power of a telescope is $\frac{f_{0}}{f_{e}} = 10$
Length of telescope is $f_{0} + f_{e} = 1.1 m$
$∴$ $f_{0} = 1 m$ & $f_{e} = 0.1 m$
When final image is at D; then magnification is
$m = f_{0} (\frac{1}{D} + \frac{1}{f_{e}}) = 100 (\frac{1}{25} + \frac{1}{10})$
$m = 14$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Geometrical OpticsSubtopic:Optical instruments

⏱ 2mℹ️ Source: PYQ_2020

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