The focal length of a convex lens of `R.I. 1.5` is `f` when it is placed in air. When it is immersed in a liquid is behaves as a converging lens its focal length becomes `xf(x gt 1)`. The refractive index of the liquid

Options

A. `gt (3)/(2)`
B. `lt (3)/(2)` and `gt 1`
C. `lt (3)/(2)`
D. All of these

Answer

B

Explanation

`(1)/(f_("air")) = (1)/(f)`
= `(1.5 -1) ((1)/(R_(1)) - (1)/(R_(2)))`
= `(1)/(2) ((1)/(R_(1)) - (1)/(R_(2)))`
`(1)/(f_(m)) = ((1.5)/(mu) - 1) ((1)/(R_(1)) - (1)/(R_(2)))`
`(1)/(xf) = ((1.5)/(mu)-1) xx (2)/(f)`
`because x gt 1 therefore ((1.5)/(mu) - 1) xx 2 lt 1`
`(3)/(mu) lt 3`
`mu gt 1`
Again for converging nature `mu lt 1.5`
`mu lt (3)/(2)`