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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsOptical instrumentsMedium2 minQB

PhysicsMediumsingle choice

The focal lengths of objective lens and eye lens of a Galilean Telescope are respectively 30 cm and 3.0 cm. Telescope produces virtual, erect image of an object situated far away from it at least distance of distinct vision from the eye lens. In this condition, the Magnifying Power of the Galilean Telescope should be:

Options:

Answer:

Solution:

$M = \frac{f _{o}}{f _{e}} (1 + \frac{f _{e}}{D}) = \frac{30}{3} (1 + \frac{3}{25}) = 10 \times 1.12 = 11.2$ . Positive, because the image is erect. (Printed solution's stray minus sign corrected.)

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Optical instruments

⏱ 2mℹ️ Source: QB

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