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. `-1.25D`
B. `-1.5D`
C. `1.25D`
D. `1.5D`

Answer

A

Explanation

Focal length of a convex lens having power
`2.5D,=(1)/(2.5)m`
Also focal length of a lens in a medium of refractive index `mu` is given by
`(1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))`
`rArr 2.5=(1)/(f)=((3)/(2)-1)((1)/(R_(1))-(1)/(R_(2)))" ...(i) (in air)"`
`rArr (1)/(f')=((3)/(4)-1)((1)/(R_(1))-(1)/(R_(2)))" ...(ii) "[because .^(t)mu_(s)=(3)/(4)]" in liquid" `
Dividing the two, `2.5f'=(0.5)/(-0.25)`
`rArr" "(1)/(f')=(-5)/(25xx0.25)=-1.25D`