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.

Answer

`-63.75 cm`

Explanation

Here, `mu_g = 1.5, f_a = 15 cm, f_l = ?, mu_l = 1.7`
If `R_1 = R, R_2 = -R`
From `(1)/(f) = (mu - 1) ((1)/(R_1) - (1)/(R_2))`
=`(mu - 1)((1)/(R) + (1)/(R)) = (2 (mu - 1))/( R)`
`(1)/(f_a) = (2)/( R) ((mu_g)/(mu_a) -)` …(i)
`(1)/(f_1) = (2)/(R)((mu_g)/(mu_l)-1)` ...(ii)
Divide (i) by (ii),
`(f_l)/(f_a)=(((mu_g)/(mu_a)-1 ))/(((mu_g)/(mu_l) - 1)) = (((1.5)/(1) -1))/(((1.5)/(1.7) - 1)) = (0.5)/(-0.2//1.7)`
`f_l = (-0.5 xx 1.7)/(0.2) xx f_a = -4.25 xx 15 = -63.75 cm`.