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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsOptical instrumentsHard2 minPYQ_2015

PhysicsHardsingle choice

In an astronomical telescope in normal adjustment a straight black line of the length $L$ is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is $l$ . The magnification of the telescope is:

Options:

Answer:

Solution:

Magnification by eyepiece
$m = \frac{f}{f + u}$
$- \frac{l}{L} = \frac{f_{e}}{f_{e} + (- (f_{0} + f_{e})}$
$\Rightarrow \frac{l}{L} = \frac{f_{e}}{f_{0}}$
$m . p . = \frac{f_{0}}{f_{e}} = \frac{L}{l}$

Stream:NEETSubject:PhysicsTopic:Geometrical OpticsSubtopic:Optical instruments

⏱ 2mℹ️ Source: PYQ_2015

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