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 76.7. 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

from snell's law

n1sintheta1 = n2 sintheta2

1.4sin76.7 = 1.5 sintheta

theta = 65.27 degrees

When that ray hits the interface with material a, the equation becomes

1.5sin65.27 = 1.3 sintheta

sintheta = 1.05

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

1.5sin65.27 = 1.4sintheta

theta = 76.69 degrees

When this ray hits the air-c boundary

1.4sin76.7 = 1sintheta

sintheta>1 not possible

the ray originally in material b

1.5sin76.7 = 1.3 sintheta

sintheta = 1.12

total internal reflection

at the b-c boundary

1.5sin76.7 = 1.4 sintheta

sintheta = 1.04

total internal reflection

So the ray starting in b, remains in b

the ray starting in material a

1.3sin76.7 = 1sintheta

sintheta = 1.27 total internal reflection

when this ray hits the a-b boundary

1.3sin76.7 = 1.5sintheta

theta = 57.5 degrees

when the ray hits the b-c boundary

1.5sin57.5 = 1.4 sintehta

theta = 64.6 degrees

when the ray hits the c-air boundary on the bottom

1.4sin64.6 = 1sintheta

sin theta = 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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