Correct Answer - c.
Let `f_(1)` and `f_(2)` be the focal length in water.
Then
`(1)/(f_(1))=((mu_(1))/(mu_(w))-1)((1)/(R)+(1)/(R))=((mu_(1))/(mu_(w))-1)((2)/(R))` (i)
`(1)/(f_(2))=((mu_(2))/(mu_(w))-1)(-(1)/(R)-(1)/(R))=((mu_(2))/(mu_(w))-1)((-2)/(R))` (ii)
Adding (i) and (ii), we get `(1)/(f_(1))+(1)/(f_(2))=(2(mu_(1)-mu_(2)))/(mu_(w)R)`
or `(1)/(30)=(2(mu_(1)-mu_(2)))/(mu_(W)R)`
`:. (mu_(1)-mu_(2))=(mu_(w)R)/(60)`
Substituting the values,
`(mu_(1)-mu_(2))=(4xx15)/(3xx60)=(1)/(3)`