Im still lost on how to do this one 5 15 pts A light ray ent

I\'m still lost on how to do this one

5. (15 pts) A light ray enters a glass enclosed fish tank From ar it eries the gKatr ,- with respect to the plane of the surface and then emerges into èhe watert. The index far the glass is 1.50 and for water 1.33 (a) What is the angle of refraction in the glass? (b) What is the angle of refraction in the water? (c) Is there any incident angle in air for which the ray will not enter the water due to total internal reflection? If so, find it

Solution

Air - glass interface:

Angle of incidence i = 20 ^o

Index of refraction of air n = 1

Index of rrefraction of glass n \' = 1.5

Index of refraction of water n \" = 1.33

From snell\'s law, sin i / sin r = n \' / n

sin 20 / sin r = 1.5 / 1= 1.5

             sin r = sin 20 / 1.5

                     = 0.228

                  r = sin ^-1 (0.228)

                    = 13.18 ^o

(b).Glass -water interface:

Angle of incidence i \' = r = 13.18 o

Angle of refraction r \' = ?

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

sin 13.18 / sin r \' = 1.33/1.5

                          = 0.8866

                sin r \' = sin 13.18 / 0.8866

                         = 0.2571

                    r \' = 14.9 ^o

(c).If the light passes from denser medium to rarer medium then only total internal reflection occur.

In our problem , light enter from rarer to denser.So, total internal reflection doesnot occured.