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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsOptical instrumentsMedium2 minQB

PhysicsMediumsingle choice

In a compound microscope, the final image is formed at a distance 'd' from the eyepiece, whose focal length is $f_{e}$ , where $d > D =$ LDDV . The object is placed at a distance 'u' from the objective lens, whose focal length is $f_{0}$ . The angular magnification of the microscope is

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Solution:

Objective: $m_{o} = f_{0} / (u - f_{0})$ . Eyepiece, final image at distance $d : m_{e} = D (1/ d + 1/ f_{e})$ .

$m = m_{o} \times m_{e} = [D f_{0} / (u - f_{0})] \times [1/ d + 1/ f_{e}] .$

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Optical instruments

⏱ 2mℹ️ Source: QB

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