A convex lens having refractive index 1.6 has focal length 12 cm, when it is in air. The focal length of that lens when placed in water is (refractive index of water = 1.28)

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Using the Lens Maker's formula for the lens in air:

$\frac{1}{f _{a}} = (μ_{g} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

$\frac{1}{12} = (1.6 - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

$\frac{1}{12} = 0.6 (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Using the Lens Maker's formula for the lens in water:

$\frac{1}{f _{w}} = (\frac{μ _{g}}{μ _{w}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

$\frac{1}{f _{w}} = (\frac{1.6}{1.28} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

$\frac{1}{f _{w}} = (1.25 - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

$\frac{1}{f _{w}} = 0.25 (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Dividing the two equations:

$\frac{f _{w}}{f _{a}} = \frac{0.6}{0.25}$

$f_{w} = 12 \times 2.4 = 28.8 cm$

$f_{w} = 288 mm$

Answer: $288 mm$

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