Question

Light is incident along the normal on face AB of a glass prismwith refractive index 1.52.

A. Find the largest value the angle α can have withoutany light refracted out of the prism at face AC if the prism isimmersed in air.

B. Find the largest value the angle α can have withoutany light refracted out of the prism at face AC if the prism isimmersed in water.

air refactive index = 1.00029

water refractive index = 1.33

Answer 1

air refactive index n _a = 1.00029

water refractive index n _w = 1.33

refractive index  of a glass prism n =1.52

incident angle on AC surface i = 90 - α

for without any light refracted out of the prism at face AC ,refracted angle r = 90 ^o

from snell's law ( sin i / sin r ) = ( n 2 / n 1 )

here i = 90 - α

       r = 90

     n 2 = n a = 1.00029

     n 1 = n = 1.52

substitute these values in above, we get

( sin ( 90 - α ) / sin 90 ) = (1.00029 / 1.52 )

                                       = 0.658

                sin ( 90 - α )    = 0.658

                       90 - α = sin ^-1 ( 0.658 )

                                   = 41.15

                               α = 90 - 41.15

                                   = 48.85 ^o

( b ) . when prism immerssed in water

  

           

from snell's law ( sin i / sin r ) = ( n 2 / n 1 )

here i = 90 - α

       r = 90

     n 2 = n w = 1.33

     n 1 = n = 1.52

substitute these values in above, we get

( sin ( 90 - α ) / sin 90 ) = (1.33 / 1.52 )

                                       = 0.875

                sin ( 90 - α )    = 0.875

                       90 - α = sin ^-1 ( 0.875 )

                                   = 61

                               α = 90 - 61

                                   = 29 ^o