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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsOptical instrumentsEasy2 minPYQ_2020

PhysicsEasysingle choice

Magnification of a compound microscope is $30$ . The focal length of eye-piece is $5 cm$ and the image is formed at a distance of ‘distinct vision of $25 cm$ . Magnification of the objective lens is

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Answer:

Solution:

$m = m_{1} \cdot m_{2} = 30$

for eye-piece $\frac{1}{v_{1}} - \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{- 1}{25} - \frac{1}{u} = \frac{1}{5}$

$∴ u = \frac{- 25}{6}$

$∴ m_{2} = \frac{- 25 \times 6}{- 25} = 6$

$∴ m_{1} = 5$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Geometrical OpticsSubtopic:Optical instruments

⏱ 2mℹ️ Source: PYQ_2020

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