The drawing shows a ray of light traveling through three materials whose surfaces are parallel to each other. The refracted rays (but not the reflected rays) are shown as the light passes through each material. A ray of light strikes the a–b interface at a 50.0? angle of incidence. The index of refraction of material a is na = 1.20. The angles of refraction in materials b and c are, respectively, 42.6? and 58.7?. Find the indices of refraction in these two media.
for interface a-b
Using snell's law:
n1*sin A1 = n2*sin A2
n1 = refractive index of medium a = 1.20
n2 = refractive index of medium b = ?
A1 = Angle of incidence = 50 deg
A2 = angle of refraction = 42.6 deg
So,
n2 = (n1*sin A1)/sin A2
n2 = (1.20*sin 50 deg)/(sin 42.6 deg)
n2 = refractive index of medium b = 1.358
Now for interface b-c
Using snell's law:
n2*sin A2 = n3*sin A3
n2 = refractive index of medium b = 1.358
n3 = refractive index of medium c = ?
A2 = Angle of incidence = 42.6 deg
A3 = angle of refraction = 58.7 deg
So,
n3 = (n2*sin A2)/sin A3
n3 = (1.358*sin 42.6 deg)/(sin 58.7 deg)
n3 = refractive index of medium c = 1.076
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