Question

For a medium composed of upper, middle, and lower layers with velocities of 6, 8, and 10 km/s, calculate the angle of incidence in the 8 and 10 km/s layers for a ray with an incidence angle of 10 degrees in the 6 km/s layer. What is the smallest angle of incidence in the 6 km/s layer that causes total internal reflection at the 8 km/s - 10 km/s interface?

Answer 1

Answer: Po 6kmis - 12 8km/s n3 lokms We know n=0.750 Again ng = 8 = 45 →ng=.802 n = 4 m2 = 4 x 8 m ng zoon, 2 -0.756, en 3=0.60, Given 0,= 100 from Snell's Law n, sino an, Sino, → Simo, = minsino, =ų, sino, - sino₂ = y simio > O2 = Sinil(.231)- 13.35 Agan sing2 = n₃ Sinez Sino 3 = 12 sine, = 5 sin 13.3 - 03 - Sin'' (1288) - 1673 Let o be the angle for which total internal reflection skmls - 10 km/s inter face than Simo = n3 singoo 7 sino = n3 = 0.8 -> o - Sin-1(0.6) - 53.13° in3

Let the connesponding angle of incidence on 6 kms medium is o for the may to be totally internal reflected on 8 km/s - 10km/s then n, sino an, sino -> Sinol = ng sino -0.75X0.8 - o - Sint(0.6) – 36.86°