Focal length of a converging lens air in $R$ . If it is dipped in water of refractive index $1.33$ , then its focal length will be around (Refractive index of lens material is $1.5$ )

R

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2 R

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4 R

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R / 2

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Solution

The correct option is C $4 R$
As we know,
$\frac{f_{l}}{f_{a}} = \frac{_{a} μ_{1}}{_{l} μ - 1} \Rightarrow f_{1} = 4 R$

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Focal length of a converging lens air in R. If it is dipped in water of refractive index 1.33, then its focal length will be around (Refractive index of lens material is 1.5)

The correct option is C 4RAs we know,flfa=aμ1lμ−1⇒f1=4R

Focal length of a converging lens air in $R$ . If it is dipped in water of refractive index $1.33$ , then its focal length will be around (Refractive index of lens material is $1.5$ )

The correct option is C $4 R$
As we know,
$\frac{f_{l}}{f_{a}} = \frac{_{a} μ_{1}}{_{l} μ - 1} \Rightarrow f_{1} = 4 R$