A convex lens of refractive index 3/2 has a power of `2.5^(@)`. If it is placed in a liqud of refractive index 2,the new power of the lens is

Options

A. 2.5 D
B. -2.5 D
C. 1.25 D
D. - 1.25 D

Answer

D

Explanation

We have, `P=(mu-1)[(1)/(R_(1))-(1)/(R_(2))]`
`2.5=((3)/(2)-1)((1)/(R_(1))-(1)/(R_(2)))`
`P=((3//2)/(2)-1)((1)/(R_(1))-(1)/(R_(2)))`
`therefore " " P=-1.25 D`