A convex lens of focal legnth `0.2 m` and made of glass `(mu = 1.50)` is immersed in water `(mu = 1.33)`. Find the change in the focal length of the lens.

Answer

`0.58 m`

Explanation

For glass lens in air, `.^(a)mu_(g) = 1.5` and `f_(a) = 0.2 m`
`(1)/(f_(a)) = (.^(a)mu_(g) - 1)((1)/(R_(1)) - (1)/(R_(2)))`
`(1)/(0.2) = (1.5 -1)((1)/(R_(1)) - (1)/(R_(2)))`
`:. (1)/(R_(1)) - (1)/(R_(2)) = 10 cm`
For the same lens in water,
`(1)/(f_(a)) = ((.^(a)mu_(g))/(.^(a)mu_(w)) - 1)((1)/(R_(1)) - (1)/(R_(2))) = ((1.5)/(1.33) - 1) xx 10`
`f_(w) = (1.33)/(1.71) = 0.78 m`
Change in focal length `= f_(w) - f_(a) = 0.78 - 0.20`
`= 0.58 m`