Refractive index of a glass convex lens is 1.5 . The radius of curvature of each of the two surfaces of the lens is 40 cm. The ratio of the power of the lens when immersed in a liquid of refractive in

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Refractive index of a glass convex lens is 1.5 . The radius of curvature of each of the two surfaces of the lens is 40 cm. The ratio of the power of the lens when immersed in a liquid of refractive index 1.25 to that when placed in air is

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The power of a lens is given by the lens maker's formula:

$P = \frac{1}{f} = (μ_{re l} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

When the lens is placed in air, the relative refractive index is $μ_{re l} = \frac{μ _{g}}{μ _{a}} = \frac{1.5}{1} = 1.5$ .

The power in air is:

$P_{a} = (1.5 - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = 0.5 (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

When the lens is immersed in the liquid, the relative refractive index is $μ_{re l} = \frac{μ _{g}}{μ _{l}} = \frac{1.5}{1.25} = 1.2$ .

The power in the liquid is:

$P_{l} = (1.2 - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = 0.2 (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Taking the ratio of the power of the lens in the liquid to that in air, we get:

$\frac{P _{l}}{P _{a}} = \frac{0.2}{0.5} = \frac{2}{5}$

Thus, the required ratio is $2 : 5$ .

Answer: $2 : 5$

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