Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsOptical instrumentsEasy2 minPYQ_2020

PhysicsEasysingle choice

A microscope is having objective of focal length $1 cm$ and eyepiece 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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Solution:

For a compound microscope,
$m = \frac{L}{f_{0}} (1 + \frac{D}{f_{e}})$
(Where length of tube $L = 30 cm$ , focal length of objective lens $f_{0} = 1 cm$ , focal length of eye-piece $f_{e} = 6 cm, D = 25 cm$ )

$�� m = \frac{30}{1} (1 + \frac{25}{6}) = 30 \times \frac{(6 + 25)}{6}$
$\Rightarrow m = 5 \times 31$
$\Rightarrow m = 155$

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Optical instruments

⏱ 2mℹ️ Source: PYQ_2020

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