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# Geometrical Optics Study Guide (page 3)

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Updated on Sep 27, 2011

#### Example

Light travels from a vacuum through the medium of an index of refraction of 1.55. Find the speed of light in the second medium and the wavelength in each medium if the frequency is 6 · 1014 Hz. If the light comes at an incident angle of 30°, what is the refraction angle?

#### Solution

We want to set all known data and then determine what expressions are useful to determine the unknowns. We know that the first medium is a vacuum and the second is a material that has an index of refraction of 1.55.

n1 = 1

n1 = 1.55

θ1 = 30°

f = 6 · 1014 Hz

v = ?

λ1 = ?

λ2 = ?

θ2 = ?

From the definition of the index of refraction, we can determine v2:

We remember that v = λ · f, and that frequency stays the same with refraction. Then:

v1 = λ · f

And since the first medium is a vacuum, the speed v1 equals c:

λ1 = 600 nm

And in the second medium,

λ2 = 323 nm

Then to determine the refraction angle, we apply Snell's law:

n1 · sin θi = n2 · sin θr

1 · sin 30° = 1.55 · n1 · sin θi

sin θr = 0.32

## Internal Reflection

Let's summarize with a specific case: light rays going from a medium with a large optical density (large n) to a smaller n. In this case, the angle of refraction is larger than the angle of incidence, and so the light will move away from the normal. What happens when the light is at 90° from the normal? In this case, the light is no longer transmitted into the second medium but is returned to the first medium. This is the case of total internal reflection, and the limit angle of internal reflection can be determined by applying Snell's law (see Figure 18.12).

n1 · sin θi = n2 · sin θr

n1 · sin θi = n2 · sin 90°

n1 · sin θi = n2

We will call the angle of incidence for which total internal reflection happens a critical angle θc.

n1 · sin θi = n2

#### Example

Can a light ray propagating from air to water suffer total internal reflection at any angle?

#### Solution

This is a general question that addresses the expression of the critical angle of total internal reflection:

and n > 1 for any of the two media in the experiment.

If we have Medium 1 to be air, then nl = 1 and the equation becomes

But n2 > 1 and because the limits for the sine function are between –1 and + 1, the above equation has no solution. Hence, no critical angle exists for total internal reflection to happen.

Practice problems of this concept can be found at: Geometrical Optics Practice Questions

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