A thin biconvex lens of refractive index 3 2 and radius of curvature 30 cm is put in water (refractive index = 4 3 ). Its focal length is

Using lens-makers formula 1 f = μ 1 μ 2 - 1 1 R 1 - 1 R 2 1 f = 3 / 2 4 / 3 - 1 1 0 .3 + 1 0 .3 Or 1 f = 9 8 - 1 2 0 .3 Or 1 f = 1 8 × 2 0 .3 or f = 1 .20 m

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PhysicsGeometrical OpticsRefraction at curved surface/LensEasy2 minPYQ_2020

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A thin biconvex lens of refractive index $\frac{3}{2}$ and radius of curvature $30 cm$ is put in water (refractive index = $\frac{4}{3}$ ). Its focal length is

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Answer:

Solution:

Using lens-makers formula

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

$\frac{1}{f} = (\frac{3 / 2}{4 / 3} - 1) (\frac{1}{0 .3} + \frac{1}{0 .3})$

Or $\frac{1}{f} = (\frac{9}{8} - 1) (\frac{2}{0 .3})$

Or $\frac{1}{f} = \frac{1}{8} \times \frac{2}{0 .3}$ or $f = 1 .20 m$

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

⏱ 2mℹ️ Source: PYQ_2020

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