10 A red light is submerged 27 m beneath the surface of a li

10. A red light is submerged 2.7 m beneath the surface of a liquid with an index of refraction 1.37. What is the radius of the circle from which light escapes from the liquid into the air above the surface?

Solution

Depth h = 2.7 m

Index of refraction of liquid n = 1.37

Critical angle C = ?

At critical angle angle of refraction r = 90 o

From Snell\'s law , sin i / sin r = n \' / n     Where n \' = index of refraction of air = 1

                      sin C / sin 90 = 1 / 1.37

                         sin C = 0.7299

C = sin ^-1 0.7299

   = 46.88 ^o

We know tan C = radius of the circle / depth

                        = r / h

From this radius r = h tan C

                            = 2.7 tan 46.88

                            = 2.7(1.067)

                           = 2.883 m