Question

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.

Answer #1

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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