Question

In: Physics

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

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

Solutions

Expert Solution

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

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