Question

Problems 30 Air n=1 . 1. Light is incident on a right triangular polystyrene (n = 1.59) prism at an angle of 6 as shown on the right. A. If , is 5º what is 0.? B. What is the minimum value for 0; in order to get total internal reflection at the right face (hypotenuse) of the the prism? 0, n=1.59 60

Answer 1

R
n=1 re 600 120° 60-01 Di للا On R21 Qz=130°0, 17 g 1.59 600/ Snell's law at the hot Using interface. in sini = n₂ sino, - Sino = ㅗ 물 Sini : singi - 0.054 => 0, = sin?0.054) Tisg sing' ni » O- 3.142° O2 = 7200 +01= 123:1420 Urring to shell's law at the right interface n₂ Sino2 = nisinor.

02= 30+0 = 304 3.14° = 3314 Applying Snell's law at the right interfau h₂ Sino2 = n, sinor.. & Sinar na Sinaz = 15.9% sint23142 » Sinar -- 0.869 - Or = 60.374° (B) For minimum on that causes total internal reolection, teefraction angle will be go° = n₂ Sino2 = n, singoo » Sinoz DO2 = sin ► 03- 38.97 nina sint (Mike)

re ut > Q1 = 02-30' = 38.970-30 ° > Q. = 8.97 Snell's law at the Again wring left boundary na sino = n, sind: » Sindi = n2 sino = 1.59 Sinsay ni s) Sindi = 0.247 => O 14.355 Am so Minimum 14.355 should be the oi in order to cause total interenal revention at the right boundary. ol is
Just go through the steps and notice in the figure how to find the angles by simple geometry