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`,
`R_1 = R, R_2 = -R`
`mu_l = 1.7, f_l = ?`
As `(1)/(f_a) = ((mu_g)/(mu_a) - 1) ((1)/(R_1) - (1)/(R_2))`
`:. (1)/(15) = ((1.5)/(1) - 1) ((1)/( R) + (1)/( R) = (1)/( R), R = 15 cm`
Again `(1)/(f_l) = ((mu_g)/(mu_l) - 1) ((1)/( R) + (1)/( R))`
=`((1.5)/(1.7) - 1) xx (2)/(15) = (-0.4)/(1.7 xx 15)`
`:. f_l = (-1.7 xx 15)/(0.4) = -63.75 cm`.