A convex lens is made of glass of refractive index 1.5.If radius of curvature of the each of its twosurfaces is 20 cm,find the ratio of the power of the lens,when placed in air to its power ,when immersed inside a liquid of refractive index 1.25.

Explanation

`R_(1)=+20cm` and `R_(2)=-20cm`
`""^(a)mu_(g)=1.5` and `""^(a)mu_(1)=1.25cm`
`1/(f_("air"))=(""^(a)mu_(g)-1)(1/(R_(1))-1/(R_(2)))`
`=(1.5-1)(1/20-1/(-20))`
`1/(f_("air"))=0.5(1/20+1/30)`
`=0./5((1+1)/20)=1/20`
`impliesf_("air")=20cm=0.20m`
Power of lens in air `=P_(1)=1/(f_("air")("inm"))`
`P_(a)=1/0.20=+5D`
Now let `F_(e)=` Focal lenght of the lens in a given liquid
`1/(f_(1))=(""^(1)mu_(g)-1)(1/(R_(1))-1/(R_(2)))`
`=((""^(a)mu_(g))/(""^(a)mu_(1))-1)(1/(R_(1))-1/(R_(2)))`
`1/(f_(1))=(1.5/1.25-1)(1/20-1/(-20))`
`1/(f_(1))=((1.5-1.25)/1.25)(1/20+1/20)`
`=0.25/1.25 xx2/20`
`1/(f_(1))=1/50`
`impliesf_(1)=50cm=0.5m`
Power of lens in liquid `=P_(1)=1/(f_(1)("in m"))=1/0.5`
`(P_(a))/(P_(1))=1/5=5:2`