Question

A prism made of saphire with a refractive index of 1.762 is immersed in oil with a refractive index of 1.390. Light initially traveling horizontally in the oil enters the prism as shown below. 60.0° a) What is the refracted angle inside the prism? b) What is the refracted angle when the light exits the prism? A. The refracted angle inside the prism is 43.17 B. The refracted angle inside the prism is 23.2 The refracted angle inside the prism is 47.4° D. The refracted angle inside the prism is 0.394" The refracted angle inside the prism is 39.3" F The refracted angle on leaving the prism is 30.00 DE

Answer 1

Use Snell’s law and the refractive indices of sapphire and oil to find the required refraction angle as shown below

a) is 90- 60 = 30° normal v from Snell's law at oil- prism interface B T 600 sini Norism 360° Sinr Noil sin 30° Sino 1:762 1.390 f 123.2 b 2 = 90-T = 90-23.2° = 66.8° o = 90-60 = 30 Now, B tota = 180° . = B + 30° + 66.8° 180° 6 > B = 83.2 So, angle of incidence at 2nd face is, a 90 B = 90-83:29 6.8° = Now, from Snell's law, Sini Sinol noil nprism sin 6.8° sinri 1:39 1.762 o 8.59