Geometric Optics

13.1.1  Reflection and Refraction

Light in any medium travels along a straight line from the source to any point in that medium. Since this is true, we can make a huge simplification of the picture of light's properties and propagation. This is responsible for the fact that the properties of light, as determined by experiment, were well-understood long before it was known that light is an electromagnetic wave. Explaining the experimentally observed properties in terms of electromagnetic theory is quite complex (although Maxwell did succeed in explaining some things) due to the interaction of light with matter. Propagation of the EM wave through a vacuum is rather simple and easily described with Maxwell's equations. Getting the detailed interaction with matter requires understanding the interaction of light with atoms. As such, it requires quantum mechanics at some level but certainly one needs to consider the detailed properties of electrons in atomic and molecular structure to make considerable progress.

Luckily, it had already been shown, by Christian Huygens and others, that most of the gross properties of the interaction of light and matter are understandable in terms of wave mechanics alone if one assumes that the primary difference between light propagation in a vacuum and light propagation in matter is one of a change of wave speed.

Although Huygen's model will not be covered in detail, it makes sense to include some discussion of it here since it explains the basic properties of reflection and refraction. Huygen's model starts by assuming that light is a wave phenomenon - a novel idea in Huygen's day as most scientists believed light to be of a particle nature. What the wave is, what medium it travels in when it is not in air or water, etc. (e.g. light coming from the Sun to earth was assumed to traverse a vacuum), is not important. Maxwell would later explain that no medium was necessary: electric and magnetic fields oscillate and essentially continually recreate each other to produce the traveling wave. The model is that light travels as a wave by having each point on the wavefront, the set of planes which contains the very nearly straight electric and magnetic field lines, be a point source of spherical wavelets. For our purposes, we merely need to say that the wavefronts move, as a whole, along a straight line in any uniform medium.

The propagation of light in any medium has a characteristic speed of travel in that medium. The index of refraction, n, is defined as being the ratio of the speed of light in a medium (call it v) to the speed of light in vacuum (usually referred to as c). Thus,

n = c v

(13.1.1.1)

We can easily show graphically that media with different indices of refraction will have rays of light ``bend'' at the interface of those media. The bending is due to a change of speed as the wavefront in one media crosses the interface since the part of the wavefront that enters the new media goes faster or slower than the part of the wavefront still in the original media. The Java applet below shows the effect.

The laws for reflection and refraction are therefore easy to derive, but we state them as just empirical rules the way they were originally discovered. We talk about the incident ray as the one coming toward the interface. The reflected ray stays in the original medium but moves away from the interface. The refracted ray moves away from the interface but in the new medium.

qincident = qreflected nincidentsinqincident = nrefractedsinqrefracted (13.1.1.2)

The second relationship is called the law of refraction or Snell's Law.

Total Internal Reflection

One application of the law of refraction is to note that if light is traveling in a medium with an index of refraction that is higher than the index of the medium outside the medium, then it is possible that no refraction can take place. In this case, the angle of incidence is such that the equation

sinqrefraction = nincident nrefraction sinqincident

(13.1.1.3)

cannot be solved for any value of sinq_refraction since the ratio of indices is greater than 1. It's clear that this occurs for any angle q_incident which is greater than the critical angle

qcrit. = sin-1 nrefraction nincident .

(13.1.1.4)

Let's try some example problems.

Problem 1:

During the nightime hours, you shine a flashlight on a pool which has a coin at the bottom. The pool depth is 4.0 m and the flashlight beam strikes the pool at 1.5 m from the edge starting a distance 1.2 m above the pool (see figure ). What is the distance of the coin from the same pool edge?

Figure 13.2: A flashlight beam shining into a pool.

Solution:

We first seek to find the appropriate angles for determining the position of the coin. In this case, we need the angle f as shown in figure 13.3. We can calculate this from the law of refraction once we know the angle q. We assume the flashlight is incident from air ( n_air = 1.0) into water ( n_water = 1.33).

nairsinq = nwatersinfÞ f = sin-1 é ê ë nair nwater sinq ù ú û Þ = sin-1 é ê ë nair nwater sin é ê ë tan-1 1.5 m 1.2 m ù ú û ù ú û = sin-1 é ê ë 1 1.33 sin é ê ë tan-1 1.5 m 1.2 m ù ú û ù ú û f = 36° (13.1.1.5)

We can calculate x once we know d in figure 13.3.

tanf = d 4.0 m Þd = (4.0 m)tan36° = 2.9 m

(13.1.1.6)

Finally,

x = d + 1.5 m = 4.4 m

(13.1.1.7)

Figure 13.3: Determining the angles needed to solve the problem of locating the coin at the bottom of the pool.

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Problem 2:

A corner reflector consists of glass with corners set at 90^° so that incoming light tends to be reflected directly back to the source. If the incoming light is in air, what is the minimum index of refraction of the glass so that all the light is reflected for an incident angle of 90^°?

Figure 13.4: Two pieces of glass set so as to reflect back any light which enters along the horizontal.

Solution:

The condition of reflecting all the light out is satisfied if the incident angle at the back of the reflector is 45^° and this angle equals or exceeds the critical angle since then the incident angle on the other surface is also 45^° and total reflection again occurs. We need to find the minimum index of refraction for the glass so that the critical angle occurs at 45^°.

sinqcrit. = nair nglass, min. Þ nglass, min. = nair sinqcrit. = 1.0 sin45° nglass, min. = Ö2 = 1.41 (13.1.1.8)

So the index of refraction must be greater than 1.41.

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Problem 3:

A ray of light traveling in a block of glass with index of refraction 1.52 is incident on the top surface at an angle of 57.2^° with respect to the normal. If a layer of oil is on the top surface of the glass, the incident ray is totally reflected. What is the maximum possible index of refraction of the oil?

Figure 13.5: A layer of oil on top of glass leads to total internal reflection for a ray of light in the glass.

Solution:

The index of refraction is set by the same condition as in the previous problem.

sinqcrit. = noil nglass Þ noil = nglasssinqcrit.Þ noil £ nglasssin57.2° £ 1.52sin57.2° noil, min. £ 1.28 (13.1.1.9)

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Problem 4:

A light ray enters a glass slab at point A and then undergoes total internal reflection at point B. What minimum value for the index of refraction of the glass can be inferred from this information?

Figure 13.6: A ray of light is incident on a slab of glass and total internal reflection happens when the ray is incident on another surface of the glass.

Solution:

We first work out the angle of incidence at point B by determining the angle of refraction at point A.

nairsin45° = nglasssinqAÞ qA = sin-1 é ê ë nair nglass sin45° ù ú û (13.1.1.10)

The angle of incidence at point B must then be

qB = 90° - qA

(13.1.1.11)

The minimum value for q_B is the critical angle for total internal reflection, so

qB ³ sin-1 é ê ë nair nglass ù ú û 90° - sin-1 é ê ë nair nglass sin45° ù ú û ³ sin-1 é ê ë nair nglass ù ú û Þ cos é ê ë sin-1 é ê ë nair nglass sin45° ù ú û ù ú û ³ nair nglass   æ  ú Ö 1 - nair2 2nglass2   ³ nair nglass Þ 1 ³ 3nair2 2nglass2 Þ   æ  ú Ö 2 3   ³ nair nglass Þ nglass ³   æ  ú Ö 3 2   = 1.22 (13.1.1.12)

13.1.2  Polarization

Polarization is another of the several properties shared by all transverse waves, hence is is applicable to electromagnetic waves. We define light to be linearly polarized if the direction of the electric field vector is aligned along a single axis. Most natural light sources emit unpolarized light in that the electromagnetic waves are emitted with a random mixture of linear polarizations, i.e. any given wave can be polarized to any direction transverse to the direction of propagation.

Polarization by Reflection

One important aspect of polarization is that many dielectric materials can partially and fully polarize light that reflects off of it. In figure 13.7 we see that incident light which arrives unpolarized, i.e. has a mixture of all possible polarization angles with respect to the incident direction can be resolved into just two components. That is to say, we take any individual polarization direction and resolve it into two components, then add together the components for all polarization directions. We choose the component directions to be parallel and perpendicular to the page, defined to be the plane of incidence. For any incident angle, we find that the reflected light is weaker than the incident light or the refracted light since the reflected light is partially polarized, i.e. it carries light with more polarization along one component than another. For a particular incident angle, called Brewster's angle, the polarization of the reflected light is only perpendicular to the plane of incidence. The light which has polarization components parallel to the plane of incidence wind up in the refracted ray so the refracted light is weaker than the incident light, but stronger than the reflected light. We also find that light incident at the Brewster angle has reflected and refracted rays which are perpendicular to each other, hence

qB + f = 90°

(13.1.2.13)

Combining this fact with the law of refraction yields

nisinqB = nrsinf = nrsin(90° -qB) Þ nisinqB = nrcosqB Þ qB = tan-1 nr ni (13.1.2.14)

Figure 13.7: Light incident on a reflecting surface of dielectric material is partially polarized. Light incident at the Brewster angle is fully polarized.

If the incident and reflected rays travel in air, then we have a simplified form for this relation called Brewster's Law, i.e.

qB = tan-1nr

(13.1.2.15)

Polarization by reflection is the mechanism by which many sunglasses work. Sunlight reflecting off of a surface gives partially polarized light. Sunglass lenses with a polarization filter allow only light of a particular polarization to enter, thereby reducing all the light whose polarization does not line up with the filter. The intensity of light transmitted by a polarization filter is

I = I0cos2q

(13.1.2.16)

where I₀ is the original intensity of light and q is the angle between the polarization direction of the light incident on the filter and the polarization direction of the filter. By reducing the reflection-polarized light, sunglasses with polarization filters can significantly reduce the amount of light we perceive as glare.

Circular Polarization

As we stated in the material on mechanical waves, any wave you see can be the resultant of a number of waves superposed upon each other. If two waves with equal amplitude and equal phases but perpendicular polarizations are superposed, then the resultant wave turns out to be polarized at 45^° to the polarization axes of either of the two waves. If the two waves are not in phase but out of phase by a quarter-cycle, then circular polarization results. This is shown in figure 13.8.

Figure 13.8: Two waves with electric field vectors aligned along orthogonal axes are out of phase by a quarter-cycle. If these waves are superposed, the resultant wave is circularly polarized.

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