Two persons A and B wear glasses of optical powers (in air) P₁=+2 D and P₂=+1 D 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 2 ~m ( from A ) to A . 'A' appears to be present at distance 1 m (from B ) to B . Find out the refractive index of water in the swimming pool, in the from X 10 and fill value of X .

Step-by-step solution

We have f₁=50 ~cm and f₂=100 ~cm Let the real distance between A and B be x . Also let refractive index of liquid be . Then 1 f₁ = ( 3 2 -1 ) ( 1 R₁ - 1 R₂ ) ( 1 R₁ - 1 R₂ )= 2 f₁ 1 f₁ = ( 3 2 -1 ) ( 1 R ₁ - 1 R ₂ ) 1 f₁ = 2 f₁ ( 3-2 2 ) And 1 f ₂^ = 2 f ₂ ( 3-2 2 ) Now, for A we have - ( 1 200 )- ( 1 -x )= 2 50 ( 3-2 2 ) 1 x = 1 200 + 2 50 ( 3-2 2 ) ...(1) Also for B we have - 1 100 - (- 1 x )= 2 100 ( 3-2 2 ) so 1 x = 1 100 + 2 100 ( 3-2 2 ) ...(2) from (1) and (2) we get 2(3-2 ) 100(2 ) + 1 100 = 1 200 + 2(3-2 )