Question

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 1

Correct Answer - (i) diverging
(ii) converging
Here, `mu_(g) = 1.5`.
The focal length of the lens in air is
`(1)/(f_(a)) = ((mu_(g))/(mu_(a)) - 1)((1)/(R_(1)) - (1)/(R_(2)))`
`= ((1.5)/(1) - 1)((1)/(R_(1)) - (1)/(R_(2)))`
`(1)/(R_(1)) - (1)/(R_(2)) = (2)/(f_(a))`
(i) When lens is dipped in medium `A` of `mu_(A) = 1.65`,
`(1)/(f_(A)) = ((mu_(g))/(mu_(A)) - 1)((1)/(R_(1)) - (1)/(R_(2)))`
`= ((1.5)/(1.65) -1) xx (2)/(f_(a)) = (-0.15 xx 2)/(1.65 f_(a))`
`f_(A) = (1.65 f_(a))/(0.15 xx 2) = -5.5 f_(a)`
`:.` In meduim `A`, the lens will behave as a diverging lens of `f_(A) = - 5.5 f_(a)`
(ii) When is dipped in medium `B` of `mu_(B) = 1.33`.
`(1)/(f_(B)) = ((mu_(g))/(mu_(B))-1)((1)/(R_(1)) - (1)/(R_(2)))`
`(1)/(f_(B)) - ((1.55)/(1.33) -1) xx (2)/(f_(a)) = (0.17 xx 2)/(1.33 f_(a))`
`f_(B) = (1.33 f_(a))/(0.34) = 3.91 f_(a)`
`:.` In medium `B`, the lens behave as a converging lens of `f_(B) = 3.91 f_(a)`.