A convex lens has a focal length of 0.1 m in air. Calculate its power. If the lens is completely dipped in `CS_(2)` of refractive index 1.66, then what will be the change in power of the lens? Given R.I of the convex lens = 1.50.

Explanation

`f=0.1m`
`therefore " Power"=(1)/(f)=(1)/(0.1m)`
`P=+10D`
w.k.t `(1)/(f)=P=(n-1)(k)`
`therefore k=(P)/(n-1)=(10)/(1.5-1)=(10)/(0.5)=20`
Also, `" "_(1)n_(2)=(n_(2))/(n_(1))=(n_(g))/(n_(cs_(2))) =(1.5)/(1.66)=0.9036`
`therefore (1)/(f_(e))=k(""_(cs_(2))n_(g)-1) =20(0.9036-1)`
`therefore P_(e) = 20xx(-0.0964)= -1.928D`
Lens behaves as a diverging lens of power = -1.93 D
Change in power `DeltaP=P_(f)-P_(1)= -1.93-(+10)= -11.93D`