Question

P[roble# 1. A light ray is incident normally (.e. perpendicularly on the interface) from WATER on one face of a 30-60-90-degree prism of dense flint glass. The index of refraction of the prism is n2= 1.66. The index of refraction of the WATER is n = 1.33. A. Calculate the angle Bo and show that the ray will Have total internal reflection. B. What are the value of the angles C. If we added solution to the WATER to increase its index of refraction bad O4a To 1.5 to eliminate the total internal reflection at P. What is the refracted angle at P then? normal ac n = 1.33 normal to ab water 10.3 12 Water and case llow Post n=15 normal on bo tin2=1.66

Answer 1

normal A) from sade, La=60, Ld=90, Le=30=2P dh 300 GO :: 0, -60° 0,2633 Golo coinical angle c is given by, 02 -1.66 300 b f 1024 nz Sine normal ? 1.66 ㅗ Sinc 1.33 Sinc 1.33 1,66 =0.8 30 6- sinh' (0.8) = 53.13. As O,>c, total internal sefterson will occur. B) From sefe £6:30°, Lf=90° - Leef (becan ause normal I makes ao with surface). Lefe = 180-60=120°. = 120-90=300 At f 27 sinos Sir Oy 1.33 = Sin 30 1.66 Sinda sindu = 1.66 x sin 30 = 0.624. 1.33 ;. 04 = sin(0.624) = 38.6° c) At P, (e), - siol sin O. 10 = 1,5 Ocnew 1.66 Sin da sin 60 - Sind! = 1.66 xsin 60-0.958 02-Sihl (0.958) = 73.34 St Scho, angle of refraction = 0 libb 60 P=15 n2=1.66 1.5