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