If focal length of the lens when immersed in water be $\mathrm{f}_{\mathrm{W}}$ .

Focal length in air can be calculated by the formula,

$\frac{1}{f_a}=\left(\frac{\mu_1}{\mu_2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \Rightarrow \frac{1}{f_a}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{R}-\frac{1}{-R}\right) \\ \Rightarrow \frac{1}{f_a}=(0.5) \times \frac{2}{\mathrm{R}} \\$

$\Rightarrow f_a=R$ - (i)

But given the focal length of lens in air , $\mathrm{f}_a=10 \mathrm{~cm}$

Hence , $true$

Focal length in water can be calculated by the formula,

$\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{\mu_1}{\mu_2}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ \Rightarrow \frac{1}{\mathrm{f}_{\mathrm{n}}}=\left(\frac{1.5}{\frac{4}{3}}-1\right)\left(\frac{1}{\mathrm{R}}-\frac{1}{-\mathrm{R}}\right) \\ \Rightarrow \frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{4.5}{4}-1\right)\left(\frac{2}{\mathrm{R}}\right) \\ \Rightarrow \frac{1}{\mathrm{f}_{\mathrm{w}}}=0.125 \times \frac{2}{10} \\ \Rightarrow \mathrm{f}_{\mathrm{w}}=40 \mathrm{~cm}$

will be the focal length of the convex lens when immersed in water.

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Reflection & Refraction

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