In Fig. 33-71, two light rays pass from air through five layers of transparent p
ID: 2069695 • Letter: I
Question
In Fig. 33-71, two light rays pass from air through five layers of transparent plastic and then back into air. The layers have parallel interfaces and unknown thicknesses; their indexes of refraction are n1 = 1.6, n2 = 1.8, n3 = 1.5, n4 = 1.5, and n5 = 1.6. Ray b is incident at angle ?b = 26 degree. Relative to a normal at the last interface, at what angle do (a) ray a and (b) ray b emerge? (Hint: Solving the problem algebraically can save time.) If the air at the left and right sides in the figure were, instead, glass with index of refraction 1.5, at what angle would (c) ray a and (d) ray b emerge?Explanation / Answer
Snell's law states n1sin(theta1)=n2sin(theta2)=...n7sin(theta7) accounting for both sides of the layers surrounded by air.
a) Since ray a is perpendicular to the surface, there will be no refraction so theta1=theta7=0 degrees
b) Since the index of n1=n7 for ray b, then theta1=theta7=26 degrees
c) Even when changed with glass, the angle of ray a is still perpendicular so theta1=theta7=0 degrees
d) Same as part b because n1 still equals n7.