A diverging lens of refractive index `1.5` and focal length `15 cm` in air has same radii of curvature for both sides. If it is immersed in a liquid of refractive index `1.7`, calculate focal length of the lens in liquid.

Explanation

`mu = 1.5, f = -15 cm`
Let `R_1 = -R, R_2 = + R`
`(1)/(f) = (mu - 1)((1)/(R_1)-(1)/(R_2))`
`(1)/(-15) = (1.5 - 1)(-(1)/( R) - (1)/(R )) = -(1)/( R)`
`R = 15 cm`
Let `f'` be focal length of lens in liquid
`(1)/(f')=((mu_g)/(mu_l) -1)((1)/(R_1) - (1)/(R_2))`
=`((1.5)/(1.7) - 1)((1)/(-15) - (1)/(15))`
=`-(0.2)/(1.7) ((-2)/(15)) = (0.4)/(1.7 xx 15)`
`f' = (1.7 xx 15)/(0.4) = 63.75 cm`
The diverging lens will behave as converging lens in the liquid.