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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsOptical instrumentsMedium2 minQB

PhysicsMediumsingle choice

A microscope is having objective of focal length 1 cm and eye-piece of focal length 6 cm. If tube length is 30 cm and image is formed at the least distance of distinct vision, what is the magnification produced by the microscope. Take $D = 25 cm$ .

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Answer:

Solution:

$m = \frac{L}{f _{o}} (1 + \frac{D}{f _{e}})$ with $L = 30 cm, f_{o} = 1 cm, f_{e} = 6 cm, D = 25 cm$ .

$= \frac{30}{1} (1 + \frac{25}{6}) = 30 \times \frac{31}{6} = 155 ≅ 150.$

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Optical instruments

⏱ 2mℹ️ Source: QB

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