In figure 2, light is incident at angle theta 1 = 40.1 degree on a boundary betw
ID: 1901410 • Letter: I
Question
In figure 2, light is incident at angle theta 1 = 40.1 degree on a boundary between two transparent materials. Some of the light then travels down through the next three layers of transparent materials, while some of it reflects upward and then escapes into the air. What are the values of theta 4 and theta 5? As shown in the figure, the indices of refraction for the 4 transparent layers are: 1.28, 1.42, 1.32, and 1.47 respectively.Explanation / Answer
You should apply Snell law at every interface between n1 and n2 n1sinO1=n2sinO2 ==> sinO2=n1 sinO1/n2= 1.28* sin(40.1)/1.42=0.5806==> O2=35.5 between n2 and n3 n2sinO2=n3sinO3 ==> sinO3=n2 sinO2/n3= 1.42* 0.5806/1.32=0.6246==> O3=38.65 between n3 and n4 n3sinO3=n4sinO4 ==> sinO4=n3 sinO3/n4= 1.32* 0.6246/1.47=0.5609 ==>O4=34.11 also u can use directely snell law between n1 and n4 sinO4=1.28*sind(40.1)/1.47=0.5609 ==> O4=34.11 between n1 and air to find O5 n1sinO1=sinX ==> sinX=n1 sinO1= > X=55.53 O5=90-X=90-55.53=34.47