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 i

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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

TestHub · Accepted Answer

Using the lens maker's formula, the power of a lens is given by $P = \frac{1}{f} = (μ_{re l} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ .

Power of the lens in air is $P_{a} = (μ_{g} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ .

Power of the lens in liquid is $P_{l} = (\frac{μ _{g}}{μ _{l}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ .

Taking the ratio of the power in liquid to the power in air:

$\frac{P _{l}}{P _{a}} = \frac{\frac{μ _{g}}{μ _{l}} - 1}{μ _{g} - 1}$

Given $μ_{g} = 1.5$ and $μ_{l} = 1.25$ .

$\frac{P _{l}}{P _{a}} = \frac{\frac{1.5}{1.25} - 1}{1.5 - 1}$

$\frac{P _{l}}{P _{a}} = \frac{1.2 - 1}{0.5} = \frac{0.2}{0.5} = \frac{2}{5}$

The ratio is $2 : 5$ .

Answer: 2:5

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