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

Engineering

Physics

Curved Surface Refraction

Question

Two persons A and B wear glasses of optical powers (in air) P₁ = + 2 D and P₂ = + 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 (from A) to A. A appears to be present at distance 1m (from B) to B. Find out the refractive index of water in the swimming pool, in the form $\frac{X}{5}$ and fill value of X.

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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 µ. Then

$\frac{1}{f_{1}} = (\frac{3}{2} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \Rightarrow (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = \frac{2}{f_{1}}$

$\frac{1}{{f^{'}}_{1}} = (\frac{3}{2 μ} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\Rightarrow \frac{1}{{f^{'}}_{1}} = \frac{2}{f_{1}} (\frac{3 - 2 μ}{2 μ}) and \frac{1}{f_{2}^{'}} = \frac{2}{f_{2}} (\frac{3 - 2 μ}{2 μ})$

Now, for A we have

$- (\frac{1}{200}) - (\frac{1}{- x}) = \frac{2}{50} (\frac{3 - 2 μ}{2 μ})$

$\frac{1}{x} = \frac{1}{200} + \frac{2}{50} (\frac{3 - 2 μ}{2 μ})$ .....(1)

Also for B we have

$- \frac{1}{100} - (- \frac{1}{x}) = \frac{2}{100} (\frac{3 - 2 μ}{2 μ})$

So, $\frac{1}{x} = \frac{1}{100} + \frac{2}{100} (\frac{3 - 2 μ}{2 μ})$ ......(2)

from (1) and (2) we get

$µ = \frac{6}{5}$