A converging lens has a focal length of `20 cm` in air. It is made of a material of refractive index `1.6`. If it is immersed in a liquid of refractive index `1.3`, what will be its new foacl length ?

Answer

51 cm

Explanation

`f_a = 20 cm, mu_g = 1.6, f_l = ?, mu_l = 1.3`
`(1)/(f_a) = ((mu_g)/(mu_a) - 1) ((1)/(R_1) - (1)/(R_2))`
`(1)/(20) = (1.6 - 1) ((1)/(R)-(1)/(R_2))` or `(1)/(R_1) - (1)/(R_2) = (1)/(12)`
Again, `(1)/(f_1) = ((mu_g)/(mu_l) - 1)((1)/(R_1) - (1)/(R_2))`
=`((1.6)/(1.3) - 1) xx (1)/(12) = (0.3)/(1.3 xx 12)`
`f_l = (1.3 xx 12)/(0.3) = 52 cm`.