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 ?

Explanation

(a) Given refractive index of glass, `mu_(a)=1.5`
Refractive index of 1st medium, `mu_(1)=1.6`
Refractive index of Iind medium, `mu_(2)=1.3`
(i) For 1st medium `mu_(1) gt mu_(a) Rightarrow (mu_(a))/(mu_(1))lt 1`
Hence, `f gt 0` concave lens or diverging lens
(ii) For `II^(nd)` medium `mu_(2) gt mu_(a) Rightarrow (mu_(a))/(mu_(2))gt 1`
Hence, `f lt 0` convex lens or convergin lens.
(b) (i) For first medium, `(1)/(f_(1))=(1-(mu_(a))/(mu_(1)))((1)/(R_(1))-(1)/(R_(2)))_(-R_(2) gt0)^(R_(1)gt0)`
`=(1-(1.5)/(1.6))" "("Positive number) for convex lens")`
`" "(1-0.9)("Positive number")`
`" "(1.1)("Positive number)Original focal length"`
`" "(1)/(f)=(1-mu_(a))((1)/(R_(1))-(1)/(R_(2)))`
`" "(1)/(f)=-0.5" "["Negative number"]`
`Rightarrow (f_(1))/(f)=-(0.5)/(0.1)`
`Rightarrow " "f_(1)=-5f`

Hence, focal length will be 5 times the original focal length and its nature will become diverging.
(ii) For second medium
`(1)/(f_(2))=(1-(mu_(a))/(mu_(2)))((1)/(R_(1))-(1)/(R_(2)))=(1-(1.5)/(1.3))((1)/(R_(1))-(1)/(R_(2)))=(1-1.15)((1)/(R_(1))-(1)/(R_(2))=0.2((1)/(R_(1))-(1)/(R_(2))`
`Rightarrow " "(f_(2))/(f)=(0.2)/(0.1) Rightarrow f_(2)=2f`
Hence, focal length will be twice the original foal length and its nature (converging nature) will remain same.