Explain what happens when a convex lens of refractive index `1.2` is immersed in a liquid of refractive index `1.3`.

Explanation

If `f` is focal length of the lens of
`mu = 1.2`, then from `(1)/(f) = (mu - 1)((1)/(R_1) - (1)/(R_2))`
=`(1.2 - 1) ((1)/(R_1) - (1)/(R_2))`
:. `(1)/(R_1) - (1)/(R_2) = (1)/((0.2) f)` …(i)
Let `f'` be focal length of this lens, when immersed in a liquid of refractive index `mu' = 1.3`.
`:. (1)/(f') = ((mu)/(mu') - 1) ((1)/(R_1) - (1)/(R_2))`
using (i).
`(1)/(f') = ((1.2)/(1.3) - 1) xx (1)/((0.2) f) = -(1)/(13 xx 0.2 f)`
`f' = -2.6 f`
The lens will behave as concave lens and its focal length will be `2.6` times the initial focal length.