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Light is traveling in glass, and hits a glass/unknown surface. In the glass the light beam is making an angle of 45.0 o with the normal to the surface. The glass has an index of refraction of 1.52. (A) If the refracted light ray leaves the glass at an 55o from the normal, what is the index of refraction for the unknown surface

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cjrodriguez89

Answer:

n = 1.31

Explanation:

  • When a ray of light crosses the separation surface between two transparent media, there exists a fixed relationship between the indexes of refraction of both media, related with the angles of incidence and refraction, which is known as Snell's Law.
  • The Snell's Law can be written as follows:

        [tex]n_{i} * sin( \theta_{i}) = n_{r} * sin( \theta_{r}) (1)[/tex]

  • In our case the ray is incident from the glass, so ni = n glass = 1.52
  • The angle of incidence is the angle that the ray makes with the normal to the separation surface, so θi=45º.
  • The angle of refraction is the angle that the refracted ray makes with the normal, so θr= 55º
  • Replacing by the values in (1), and solving for nr, we have:

       [tex]n_{r} =\frac{n_{i} * sin \theta_{i} }{sin \theta_{r} } = \frac{sin (45)*1.52}{sin (55)} = 1.31 (2)[/tex]

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