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Geometrical Optics Study Guide (page 2)

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Updated on Sep 27, 2011

Example

Two planes are set such that they form an angle of 100° with each other. A light is incident on the first surface at an angle of 60°. Find the angle of the reflected light from the second plane with respect to the normal to that plane.

Solution

First, we need to set our angle diagram and use the laws of reflection to determine the angle of the outgoing ray (see Figure 18.4).

Reflection of Light

In the ABC triangle, the angle at A can be found from the first law of reflection:

θi = θr = 60°

And since the triangle totals 180° with A at 30°, C is 100° and B is 50°.

θi = 90° – B = 90° – 50° = 40°

θ* i = 40°

Again, we use the law of reflection on the second surface:

θ* i = θ* r = 40°

Refraction of Light

The examples we studied previously are probably easy to understand if you have performed experiments such as shining a light on a mirror or a shiny metal surface. But what happens when light meets a transparent medium and not an opaque one? We all know that light is transmitted beyond the surface of separation. Light goes through the water in your glass and even through the glass itself. This phenomenon of light propagation from a transparent medium to another transparent medium is called refraction. But refraction is affected by the properties of the medium itself, and the direction of propagation is changed by it. The corresponding rays are called incident and refracted rays, as are the angles. Figure 18.6 shows an incident wave and the refraction on the separation surfaces.

Refraction of Light

On the far left, we have the incident ray refracting on the surface and propagating inside Medium 2 at an angle smaller than the incident angle:

θi < θr

This behavior happens when light comes from a medium where the speed of light is larger and propagates into a medium with reduced speed. The light bends toward the normal when vl > v2 where vl and v2 are the speed of light in Medium 1 and Medium 2.

On the right side of the picture, we see the light reflecting from the bottom of Medium 2 and refracting at the surface of separation a second time. This time, the angle of refraction is larger than the angle of incidence because the light returns to Medium 1 where the speed of propagation is larger.

θ* i > θ* r

The light bends away from the normal when vl < v2

Analyzing the geometry of the figure and relating to the speed in respective media, one can determine after some calculations a law of refraction relating the angles of incidence and refraction.

The laws of refraction are as follows:

  • The incident, refracted ray, and normal are in the same plane.
  • The angle of incidence and the angle of refraction are related by

In optics, we also work with a coefficient that relates the speed of light in the specific medium with the speed in vacuum, and it is called the index of refraction:

The index of refraction is a unitless quantity and it is material dependent. A few values for the most common substances are given in Table 18.l.

Refraction of Light

Because the index of refraction is so close to 1 for air, we will consider n = 1. One note on propagation in a different medium is that the frequency of the wave will remain constant while the wavelength changes with speed.

Returning to the laws of refraction, we will find a second version of the law of the angles by multiplying the right-hand-side fraction with the speed oflight in a vacuum:

and with the definition of the index of refraction:

or separating on the same side factors related to the incident and refracted rays, respectively:

n1 · sin θi = n2 · sin θr

This is also known as Snell's law of refraction (Willebroad van Roijen Snell, 1580-1626).

For the purpose of this lesson, we will consider the index of refraction constant, although in reality, n is wavelength dependent. This dependence is the explanation for phenomena such as the dispersion of light through a prism.

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