  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  B
Increases to a factor of 2.5  C
Increases to a factor of 1.2  D
Decreases to a factor of 1.2  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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