Two persons A and B wear glasses of optical powers (in air) `P_1 = + 2 D` and `P_2 = + 1D` respectively. The glasses have refractive index 1.5. Now they jump into a swimming pool and look at each other. B appears to be present at distance 2m (f rom A) to A. A appears to be present at distance 1m (from B) to B. The refractive index of water in the swimming pool, in the form `(X)/(10)` and find X.

Explanation

We have `f_(1)=50` cm and `f_(2)=100cm`
let the real distance between A and B be x also let refractive index of liquid be `mu` then
`(1)/(f_(1))=((3)/(2)-1)((1)/(R_(1))-(1)/(R_(2)))implies((1)/(R_(1))-(1)/(R_(2)))=(2)/(f_(1))`
`(1)/(f'_(1))=((3)/(2mu)-1)((1)/(R_(1))-(1)/(R_(2)))implies(1)/(f'_(1))=(2)/(f_(1))((3-2mu)/(2mu))` M
Now for A we have
`-((1)/(200))-((1)/(-x))=(2)/(50)((3-2mu))/(2mu))`
Also for B we have
`-(1)/(100)-(-(1)/(x))=(2)/(100)((3-2mu)/(2mu))`
`(1)/(x)=(1)/(100)+(2)/(100)((3-2mu)/(2mu))` .(ii)
`implies(2(3-2mu))/(100(2mu))+(1)/(100)=(1)/(200)+(2(3-2mu))/(50(2mu))`
`implies(2(3-2mu))/((2mu))[(1)/(50)-(1)/(100)]=(1)/(100)-(1)/(200)=(1)/(200)`
`implies((3-2mu))/(2mu)=(1)/(2)implies6-4mu=mu`
so `mu=(6)/(5)=(12)/(10)`