A light ray is incident at the interface between mediums a and b, with n a = 1.7
ID: 1376755 • Letter: A
Question
A light ray is incident at the interface between mediums a and b, with na = 1.72 and nb = 2.09. In medium a, the incident angle with respect to the normal is 61.5 degrees, and after it is refracted into medium b, it hits the interface between mediums b and c at exactly the critical angle needed for total internal reflection. What is the index of refraction of medium c, in order for this to occur?
A light ray is incident at the interface between mediums a and b, with na = 1.72 and nb = 2.09. In medium a, the incident angle with respect to the normal is 61.5 degrees, and after it is refracted into medium b, it hits the interface between mediums b and c at exactly the critical angle needed for total internal reflection. What is the index of refraction of medium c, in order for this to occur?Explanation / Answer
Note that from Snell's law,
n1sin(t1) = n2sin(t2)
where
n1 = index of refraction of first medium = 1.72
t1 = angle of incidence = 61.5 degrees
n2 = index of refraction of second medium = 2.09
t2 = angle of refraction
Thus,
t2 = 46.32238805 degrees
As this is a critical angle, then
nc = n2 sin t2
Thus,
nc = 1.51 [ANSWER]
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