A convex lens made up of glass of refractive index `1.5` is dippedin turn
(i) in a medium of refractive index `1.65`
(ii) in a medium of refractive index `1.33`
(a) Will it behave as converging or diverging lens in the two cases ?
(b) How will its focal length changes in the two media ?

Answer

(a) diverging lens in medium of ref. index `1.65` ; converging lens in medium of ref. index `1.33`
(b) `f_1 = -5.5 f, f_2 = 3.9 f`.

Explanation

(i) When the lens is in air, its focal length is
`(1)/(f) = (1.5 - 1) ((1)/(R_1) - (1)/(R_2)) = (1)/(2) ((1)/(R_1) - (1)/(R_2))`
(ii) When the lens is dipped in medium of `mu = 1.65`,
`(1)/(f_1) = ((1.5)/(1.65) -1)((1)/(R_1) - (1)/(R_2)) = (0.15)/(1.65) ((1)/(R_1) -(1)/(R_2))`
=`-(1)/(11) ((1)/(R_1)-(1)/(R_2))`
`(f_1)/(f) = (1//2)/(-1//11) = -(11)/(2) = -5.5` or `f_1 = -5.5 f`
`:.` The lens behaves as concave lens of focal length `-5.5 f`
(iii) When the lens is dipped in medium of `u = 1.33`
`(1)/(f_2) = ((1.5)/(1.33) -1) ((1)/(R) - (1)/(R_2)) = (0.17)/(1.33) ((1)/(R_1) - (1)/(R_2))`
`(f_2)/(f) = (1//2)/(17//133) = (133)/(2 xx 17) = 3.9` or `f_2 = 3.9 f`
`:.` The lens behaves as convex lens of focal length `3.0 f`.