The drawing shows three layers of different materials, with air above and below

Question

The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers are parallel. The index of refraction of each layer is given in the drawing. Identical rays of light are sent into the layers, and light zigzags through each layer, reflecting from the top and bottom surfaces. The index of refraction for air is nair = 1.00. For each layer, the ray of light has an angle of incidence of 82.2?. For the cases in which total internal refection is possible from either the top or bottom surface of a layer, determine the amount by which the angle of incidence exceeds the critical angle.

Explanation / Answer

Using the equation
nsin = nsin
take the ray in material c as it hits the interface with material b
1.40sin82.2 = 1.50sin
= 67.07°
So the portion of the light ray which enters b will be only 67.07° from the perpendicular to the surface.
When that ray hits the interface with material a, the equation becomes
1.5sin67.07 = 1.3sin
sin = 1.06
as you cannot have a value of sine larger than 1.00, there will be total internal reflection of this ray off the a-b boundry.
When the ray again hits the b-c boundry, the equation is
1.5sin67.07 = 1.40sin
= 80.06°

When this ray hits the air-c boundary (assuming the materials are long enough), the equation is
1.4sin80.06 = 1.00sin
sin = 1.37 oops, not possible
again there is total internal reflection, so this ray is trapped in the materials between the air on the bottom and the a-b boundary
Now lets look at the ray originally in material b
1.5sin80.06 = 1.3sin
sin = 1.13 ---total internal reflection
at the b-c boundary
1.5sin80.06 = 1.4sin
sin = 1.05--- total internal reflection
So the ray starting in b, remains in b.
Now lets look at the ray starting in material a
1.3sin80.06= 1.00sin
sin = 1.28 ---total internal reflection.
when this ray hits the a-b boundary
1.3sin80.06 = 1.5sin
= 58.53°
when the ray hits the b-c boundary
1.5sin58.53 = 1.4sin
= 65.50
when the ray hits the c-air boundary on the bottom
1.4sin65.50 = 1.00sin
sin = 1.27 ---total internal reflection
so this ray will travel in all three layers a,b,and c and be reflected by the boundary with air on both top and bottom.

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