refractive index of themedium sin i sinr i 45 r 133 sin 45 sin ....

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refractive index of themediumÂ Â Â Â Â Â ? = sin i / sinr

Â Â Â Â Â Â Â Â Â Â Â Â Â Â i = 45 , r = ?

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1.33 = sin 45 / sin?

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â sin? = sin 45 /1.33

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ? =32.282^{0Â Â Â Â} is the angle ofrefraction .

b) from thefigureÂ Â Â Â Â Â Â Â Â Â Â tan 45 = Â Â Â 2.1 /Â Â xÂ Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â X = 2.1Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â from which the distance of the person 'x' can bedetermined.

c) if the pool isempty the object must be moved to the distance

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â d = tan(45 -32.282)* depth of the pool.

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