Two thin symmetrical lenses of different nature and of different material have equal radii of curvature `R=15cm`. The lenses are put close together and immersed in water `(mu_(w)=4//3)` . The focal length of the system in water is 30cm. The difference between refractive indices of the two lenses is

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. `(1)/(3)`
D. `(3)/(4)`

Answer

C

Explanation

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))` ....`(1)`
`(1)/(f_(1))=((mu_(2))/(mu_(w))-1)(-(1)/(R)+(1)/(R))=((mu_(2))/(mu_(w))-1)(-(2)/(R))` ....`(2)`
Adding `(1)` and `(2)` 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)`