Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsRefraction at curved surface/LensEasy2 minPYQ_2020

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A thin convex lens of refractive index $1.5$ has $20 c m$ focal length in air. If the lens is completely immersed in a liquid of refractive index $1.6,$ its focal length will be

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Solution:

From lens formula

$\frac{1}{f} = ({}_{a}{μ_{g}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

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

Also, ${}_{l}{μ_{g}} = \frac{μ_{g}}{μ_{l}} = \frac{1.5}{1.6}$

$∴ \frac{1}{f^{'}} = ({}_{l}{μ_{g} - 1}) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{1}{f'} = (\frac{15}{16} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$ ...(ii)

Dividing Equation (i) by Equation (ii), we get

$\frac{f}{f^{'}} = \frac{(\frac{15}{16} - 1)}{(1.5 - 1)} = - \frac{1}{8}$

$f' = - 8 f$

$= - 160 c m$

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

⏱ 2mℹ️ Source: PYQ_2020

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