Question

What happens to the focal length of a convex lens when it is immersed in water? Refractive index of the material of the lens is greater than that of water.

Answer 1

Hint : The refractive indices of the material of the lens and that of water are different. So by application of the lens maker’s formula we can find the focal length of the lens when the refractive index of the medium changes.

Formula Used: The formulae used in the solution are given here.
 $\Rightarrow \dfrac{1}{f} = \left( {{}^1{\mu _2} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) $ where $ f $ is the focal length of the mirror, $ {}^1{\mu _2} $ is the refractive index of water with respect to the lens, $ {R_1} $ is the radius of curvature of the lens surface closer to the light source, $ {R_2} $ is the radius of curvature of the lens surface farther from the light source.

Complete step by step answer
The refractive index of a lens is the relative measure, that is, it is the ratio of velocities of light in two mediums. It is dependent upon the relative refractive index of the surrounding medium where it is measured.
Given that, the refractive index of the material of the lens is greater than that of water.
More the refractive index, the more the light beam bends away from normal while travelling from high density medium to low density medium.
 $\Rightarrow {}^w{\eta _g} < {}^a{\eta _g} $ , where $ {}^w{\eta _g} $ is the refractive index of the material of glass in water and $ {}^a{\eta _g} $ is the refractive index of the material of glass in air.
According to the lens makers’ formula,
 $\Rightarrow \dfrac{1}{f} = \left( {\eta - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) $ where $ f $ is the focal length of the mirror, $ \eta $ is the refractive index of lens with respect to water, $ {R_1} $ is the radius of curvature of the lens surface closer to the light source, $ {R_2} $ is the radius of curvature of the lens surface farther from the light source
From this formula, we come to know that, $ f\alpha \dfrac{1}{{\eta - 1}} $ .
Thus, the focal length is inversely proportional to the refractive index.
 $ \therefore {f_w} > {f_a} $ where $ {f_w} $ is the focal length of the lens in water and $ {f_a} $ is the focal length of the lens in air.
Thus, the focal length of the lens will increase.

Note
When any lens or glass slab is immersed in a liquid whose refractive index is equal to that of lens, refractive index of lens will be equal to 1 relative to that liquid.
According to the lens makers’ formula,
 $\Rightarrow \dfrac{1}{f} = \left( {1 - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) $ where $ \eta = 1 $
 $ \Rightarrow \dfrac{1}{f} = 0 $
The focal length of the lens in this case, will tend to infinity.
 $\Rightarrow f = \infty $ .
But the extent of the change in focal length depends on the curvature of the lens surface that is in contact with the water.