Question: Red light is incident in air on a 30 degree - 60 degree - 90 degree pr

Question

Question:

Red light is incident in air on a 30 degree - 60 degree - 90 degree prism as shown. The incident beam is directed at an angle of phi 1 = 40.9 degree with respect to the horizontal and enters the prism at a height h = 29 cm above the base. The beam leaves the prism to the air at a distance d = 77.9 along the base as shown. What is phi 1, max, the maximum value of phi 1 for which the incident beam experiences total internal reflection at the horizontal face of the prism? Suppose now that the violet beam is incident at height h, but makes an angle phi 1,v = 60 degree with the horizontal. What is phi 3,v, the angle the transmitted beam makes with the horizontal axis?

Explanation / Answer

Draw an altitude from the incident point to the base. Then the given angle (30 deg) is part of a right triangle. Note that in the 30-60-90 triangle, the length along the base from the 30-deg angle to the altitude is h*sqrt(3).
Then you see the triangle next to that, inclusive of the angle phi-2. Quick inspection yields:
tan(phi-2)=h/(d-h*sqrt(3))

I'm not sure if this is strictly trigonometry or physics. It would seem to me that you must then use Snell's Law to determine the index of refraction of the prism.
n1*sin(phi-1)=n2*sin(90-phi2) If you follow snell's law, it is necessary that the second angle be 90-phi2 to remain consistent. n1=n_air=1.
n2=sin(phi1)/cos(phi2)

Apply snell's law as the ray leaves the prism:
n2*sin(90-phi2)=n1*sin(90-phi3), again, geometry dictates this application.
cos(phi3)=[sin(phi1)/cos(phi2)]*cos(phi2)
phi3=90-phi1 from the above relation.

Subbing all this in:

phi2=arctan(0.29/(0.779-0.29*sqrt(3)))= 46.34 degrees

phi3=90-46.34

=43.66 degrees

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