Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsRefraction at curved surface/LensEasy2 minPYQ_2020

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A thin glass (refractive index $1.5$ ) lens has an optical power of $- 5 D$ in the air. Its optical power in a liquid medium with refractive index $1.6$ will be

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Answer:

Solution:

$\frac{1}{f_{a}} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$= (1.5 - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \dots \dots \dots \dots . . (i)$

$a n d \frac{1}{f_{m}} = \frac{μ_{g} - μ_{m}}{μ_{m}} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{1}{f_{m}} = (\frac{1.5}{1.6} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \dots \dots \dots . . (i i)$

Thus,

$\frac{f_{m}}{f_{a}} = \frac{(1.5 - 1)}{(\frac{1.5}{1.6} - 1)} = - 8$

$f_{m} = - 8 \times f ₐ$

$= - 8 \times \frac{- 1}{5} (∵ f_{a} = \frac{1}{p} = - \frac{1}{5} m)$

$= 1.6 m$

$∴ P_{m} = \frac{μ}{f_{m}}$

$= \frac{1 .6}{1 .6} = 1 D$

Subject:PhysicsTopic:Geometrical OpticsSubtopic:Refraction at curved surface/Lens

⏱ 2mℹ️ Source: PYQ_2020

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