As shown in the figure Figure 1 a layer of water covers a sl

As shown in the figure (Figure 1), a layer of water covers a slab of material X in a beaker. A ray of light traveling upwards follows the path indicated. Using the information on the figure, find the index of refraction of material X. Find the angle the light makes with the normal in the air.

Solution

From figure ,angle of incidence i = 65 o

Angle of refraction r = 48 o

Index of refraction of material X is n = n

Index of refraction of water n\' = (4/3)

From Snell\'s law sin i /sin r = n \' / n

                   sin 65 / sin 48 = (4/3) / n

           1.219 = (4/3) / n

From this n = (4/3) /1.219

                  = 1.3333/1.219

                  = 1.093

(b).water -air interface:

Angle of incidence i \' = r = 48 o

Angle of refraction r \' = ?

Index of refraction of water n \' = 4/3 = 1.3333

Index of refraction of air n \" = 1

From Snell\'s law sin i \' / sin r \' = n \" / n \'

                         sin 48 / sin r \' = 1 / 1.3333

                                 sin r \' = 1.333(sin 48 )

                                         = 0.9908

                                      r \' = sin ^-1(0.9908)

                                          = 82.24 ^o