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

`"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) 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 hte 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 `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"))`
`or " "f_(B)=3.91f_("air")`
As the sign of `f_(B)` is same of that of `f_("air")` the lens will behave as a converging lens.