A ray of light is incident normally on the right face of a 30o-60o-90o cubic zir

Question

A ray of light is incident normally on the right face of a 30o-60o-90o cubic
zirconium prism (n2 = 2.193) as shown. The ray undergoes total internal
reflection at point P and then exits the prism into a glass stand (n3 = 1.662).
Finally, the ray exits the bottom of the glass stand at angle .. The cubic
zirconium prism and glass stand are immersed in ethyl alcohol (n1 = 1.361).

a) Determine the exit angle theta for the ray of light.

b) When the cubic zirconium prism and glass stand are removed from the
ethyl alcohol and placed into air, the ray undergoes total internal
reflection at the bottom of the glass base. (Previously the ray exited the
base at angle ..) Verify. Be sure to find the critical angle between the
base and air.

c) The ray now reflects up and down the inside of the glass base and exits
the left side just after the 7th reflection. If the glass base is 2.00 cm
thick, how far does the light ray travel horizontally from the point where it
enters the base to the point where it exits the base?

Explanation / Answer

a) sin x/siny=n2/n3 x=30 n2 = 2.193 n3=1.662 so siny = 0.2789 so y = 22.27 now siny/sin theta = n3/n1 sin theta = 0.31 so theta = 18 degrees = the exit angle b) angle on the glass stand and air = y = 22.27 critical angle = arcsin(n1/n3) = arcsin(1/1.662) = 36 degrees as 22.27 < 36 total internal reflection occurs c)distance it travelled till first reflection = 2/cos y = 2.1612cm total distance travelled = 7*2.1612 = 15.13cms

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