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Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsRefraction at curved surface/LensEasy2 minPYQ_2023

PhysicsEasynumerical

The radius of curvature of each surface of a convex lens having refractive index $1.8$ is $20 cm$ . The lens is now immersed in a liquid of refractive index $1.5$ . The ratio of power of lens in air to its power in the liquid will be $x : 1$ . The value of $x$ is

Answer:

4.00

Solution:

The power of a lens is given by

$P_{1} = \frac{1}{f}$

Using Lens Maker's formula,

$P_{1} = (1 .8 - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$ and

$P_{2} = \frac{1}{f_{immersed}} = (\frac{1 .8}{1 .5} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

Hence, the ratio is,

$\frac{P_{1}}{P_{2}} = \frac{(1 .8 - 1)}{(\frac{1 .8}{1 .5} - 1)} = \frac{0 .8 \times 1 .5}{0 .3} = 4$

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Refraction at curved surface/Lens

⏱ 2mℹ️ Source: PYQ_2023

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