A light ray in air strikes the right angle prism shown in Fig. P33.40. The prism angle at Bis 30.0° . This ray consists of two different wavelengths. When it emerges at face AB, it has been split into two different rays that diverge from each other by. Find the index of refraction of the prism for each of the two wavelengths

The index of refraction, n, is the ratio of the speed of light in a vacuum, c, to the speed of light in a medium, c': One consequence of this difference in speed is that when light goes from one medium to another at an angle, the propagation vector in the new medium has a different angle with respect to the normal.

I-or the first ray, since the angle IS 60 at point B, pnsm angle at point B Will be 60 _ I he angle of refraction tor the upper ray

would be 72° (60° 42° )- Now apply the Snell's Law to get the index of refraction-

Hence, the index of refraction for the upper ray Would be 1.098

For the second ray, the angle of refraction for the upper ray Would be 805° (60° +12°+8.5° )- Now apply the Snell's Law to

n - sin(60°) = 1 - sin(80.5°XIndex of refraction: Prism =n, Air 2 1)

For the second ray, the angle of refraction for the upper ray would be 80.5° (60° +12°+8.5° ). Now apply the Snell's Law to

get the index of refraction-

Hence, the index of refraction for the lower ray Would be 1.139

Result 4 0f 4

1-098, 1-139