Question

At the very end of Wagner's series of operas Ring of the Nibelung, Brünnhilde takes the golden ring from the finger of the dead Siegfried and throws it into the Rhine, where it sinks to the bottom of the river.

Part A Assuming that the ring is small enough compared to the depth of the river to be treated as a point and that depth of the Rhine where the ring goes in is 11.0 m , what is the area of the largest circle at the surface of the water over which light from the ring could escape from the water? Use 1.33 for the index of refraction of water. A =   m2

Answer 1

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from figure depth OA = 11 m

from figure , tanθ = R/11

radius of the circle R = (11 m)tanθ    ................ (1)

using the snell's law ,

      sinθ = n_air/n_water

             = 1/1.33

             = 0.7518

the critical angle θ = sin^-1(0.7518)

                           = 48.75⁰

therefore , R = (11 m)tan(48.75) = 12.543 m  

area of the circle A = πR²

                             = (3.14)(12.543 m)²

                             = 494.264 m²

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