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

Explanation

Here,`.^(a)mu_(g)=1.5`
Let `f_(air)` be the focal length of the lens in air, Then,
`(1)/(f_(air))(.^(a)mu_(g)-1)((1)/(R_(1))-(1)/(R_(2)))` or `((1)/(R_(1))-(1)/(R_(2)))=(1)/(f_(air)(.^(a)mu_(g)-1))`
`=(1)/(f_(air)(1.5-1))` or `((1)/(R_(1))-(1)/(R_(2)))=(2)/(f_(air)) .....(i)`
(i) When lens is dipped in medium `A`
Here, `.^(a)mu_(A)=1.65`
Let `f_(A)` be the focal length of the lens, when dipped in medium `A`. Then,
`(1)/(f_(A))(.^(A)mu_(g)-1)((1)/(R_(1))-(1)/(R_(2)))=((.^(a)mu_(g))/(.^(a)mu_(A))-1)((1)/(R_(1))-(1)/(R_(2)))`
Using the equation `(i)`, we have
`(1)/(f_(A))((1.5)/(1.65)-1)xx(1)/(f_(air))=-(1)/(5.5f_(air))`
or `f_(A)=-5.5f_(air)`
As the sign of `f_(A)` is opposite to that of `f_(air)` the lens will behave as a diverging lens.
(ii) When lens is dipped in medium `B` :
Here, `.^(a)mu_(B)=1.33`
Let `f_(B)` be the focal length of the lens, when dipped in medium `B`. Then,
`(1)/(f_(B))(.^(B)mu_(g)-1)((1)/(R_(1))-(1)/(R_(2)))=((.^(a)mu_(g))/(.^(a)mu_(B))-1)((1)/(R_(1))-(1)/(R_(2)))`
Using the equation `(i)`, we have
`(1)/(f_(B))((1.5)/(1.33)-1)xx(2)/(f_(air))=(0.34)/(1.33f_(air))`
As the sign of `f_(B)` is same as that of `f_(air)` the lens will behave as a converging lens.