Light enters a 45 45 90 right perpendicular to one face as

Light enters a 45 - 45 - 90 right perpendicular to one face as shown. a) If the prism is in air, what minimum index of refraction is required in order to achieve total internal reflection? b) If the prism is immersaed in water (n-1.33), at what does the light leave the prism?

Solution

a) light will enter straight through perpendicular face.

after that it will strike exit surface at 45 deg,

hencce @c < 45 deg

so taking @c = 45 deg

Applying snell\'s law.

ni sini = nr sinr

for @c , r = 90 deg

n * sin45 = 1 * sin90

n = 1.414 .............Ans

b) now nr is not 1 but 1.33

so 1.4141 * sin45 = 1.33 sin@r

@r = 48.75 deg with prism surafce

from initial direction,

delta@ = 48.75 -45 = 3.75 deg