Can someone please help me with this problem. I am getting 52.4 degrees for thet
ID: 1474771 • Letter: C
Question
Can someone please help me with this problem. I am getting 52.4 degrees for theta4 and 32.7 degrees for theta5. I was told it was wrong and I want to make sure I dont miss this problem. Can you please show work.
In the figure, light is incident at angle 1 = 37.0 ° on a boundary between two transparent materials. Some of the light travels down through the next three layers of transparent materials, while some of it reflects upward and then escapes into the air. If n1 = 1.32 , n2 1.40, n3 = 1.30 and n-1.47 , what is the value of (a) 4 and (b) 5? Air e, ni n2 ng n4Explanation / Answer
use Snell's law at n1-air interface
theta1 = 37 degrees
Apply, sin(theta1)/sin(theta5) = n5/n1
sin(37)/sin(theta5) = 1/1.32
==> sin(theta5) sin(37)*1.32
sin(theta5) = 0.794
theta5 = sin^-1(0.794)
= 52.6 degrees <<<<<<<<<<<<<-------------------Answer
Use Snell's law at n1-n2 interface
sin(theta1)/sin(theta2) = n2/n1 -----(1)
Use Snell's law at n2-n3 interface
sin(theta2)/sin(theta3) = n3/n2 -----(2)
Use Snell's law at n3-n4 interface
sin(theta3)/sin(theta4) = n4/n3 -----(3)
multiply equations 1,2 and 3
we get
sin(theta1)/sin(theta4) = n4/n1
sin(37)/sin(theta4) = 1.47/1.32
sin(theta4) = (1.32/1.47)*sin(37)
sin(theta4) = 0.54
theta4 = sin^-1(0.54)
= 32.7 degrees <<<<<<<<<<<<<<---------------Answer