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Question

A lens is made of flint glass (refractive index $$= 1.5)$$. When the lens is immersed in a liquid of refractive index $$1.25$$, the focal length :


A
Increases to a factor of 1.25
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B
Increases to a factor of 2.5
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C
Increases to a factor of 1.2
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D
Decreases to a factor of 1.2
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Solution

The correct option is C Increases to a factor of $$2.5$$
From lens maker's formula,

$$\dfrac { 1 }{ f } =\left( \dfrac { { \mu  }_{ 2 } }{ { \mu  }_{ 1 } } -1 \right) \left( \dfrac { 1 }{ { R }_{ 1 } } -\dfrac { 1 }{ { R }_{ 2 } }  \right) $$

When lens is in air,

$$\dfrac { 1 }{ f_{air} } =\left( 1.5-1 \right) \left( \dfrac { 1 }{ { R }_{ 1 } } -\dfrac { 1 }{ { R }_{ 2 } }  \right) \quad \longrightarrow \left( 1 \right) $$

When lens is in air of R.I.$$\mu $$

$$\dfrac { 1 }{ f } =\left( \dfrac { 1.5 }{ 1.25 } -1 \right) \left( \dfrac { 1 }{ { R }_{ 1 } } -\dfrac { 1 }{ { R }_{ 2 } }  \right) \quad \longrightarrow \left( II \right) $$

(I)/(II) = $$\dfrac { f }{ fair } =\dfrac { 1/2 }{ \dfrac { 1}{5}  } =\dfrac { 5 }{ 2 } $$

Hence focal length of lens increases by a factor of 2.5.

Physics

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