State how focal length of a glass (Refractive index 1.5) changes when it is completely immersed in:
(i) Water (Refractive index 1.33)
(ii) A liquid (Refractive index 1.65)

Explanation

Given : Refractive index of glass lens, `mu_g = 1.5` 
(i) When it is dipped in water of refractive index, `mu_w = 1.33`
 Refractive index of glass w.r.t. Water,
`""_wmu_g = (mu_g)/(mu_w) = (1.5)/(1.33)`
By lens formula,
`1/(f_a) = (""_amu_g - 1)(1/(R_1) - 1/(R_2))` (in air) ….(1)
`1/(f_w) = (""_wmu_g - 1)(1/(R_1) - 1/(R_2))`(in water)....(2)
By equations (1) and (2),
`(f_w)/(f_a)  = (""_amu_g - 1)/(""_wmu_g - 1)`
`= (1.5 -  1)/((1.5)/(1.33) - 1)`
`= (0.5 xx 1.33)/(0.17)`
`= 3.91`
`implies f_w = 3.91 f_a`
It means that when lens is dipped in water its focal length increases.
(ii) For liquid of refractive index 1.65.
`1/(f_l) = (""_lmu_g - 1)(1/(R_1) - 1/(R_2)) "     " ….(3)` 
 By equations (1) and (3)
`(f_t)/(f_a) = (""_amug - 1)/(""_lmu_g - 1)`
`=( 1.5 - 1)/((1.5)/(1.65) -1) = (-0.5 xx 1.65)/(0.15)`
`f_l = -5.5 f_a`
  Since, `f_l` and `f_a` are of opposite sign, so behaviour of lens will change. Converging lens will behave as diverging lens and vice versa.