Download Geometric Optics: Reflection and Refraction of Electromagnetic Waves and more Study notes Physics in PDF only on Docsity!
Today we will begin our discussion of geometric optics by discussing reflection and refraction of electromagnetic waves.
We will begin by discussing the reflection of electromagnetic waves (usually light) from a surface. We will claim that the angle of incidence is equal to the angle of reflection and will provide an argument for this law for the case when the surface is a conductor.
We will then discuss the propagation of electromagnetic waves in matter and discover that the speed of the wave in matter is less than in vacuum, with the difference being specified in terms of the index of refraction of the material. We will introduce Snell’s law to determine the angle of refraction.
We will then discuss the intensities and polarization of the reflected wave. In particular, we will discuss the requirements on the incident angles to produce total internal reflection and linear polarization of the reflected wave.
To this point, we have been concerned with the propagation and polarization of electromagnetic waves in vacuum. Today we will focus on what happens to electromagnetic waves (usually light) in different materials.
To determine electric and magnetic fields in matter, we need to replace the vacuum constants ε 0 and μ 0 with the constants ε and μ that describe the material of interest. For the next few prelectures we will be studying geometric optics in which the wavelength of the radiation (usually light) is small compared to the objects with which it interacts. With this restriction, we can assume that the light travels in straight lines called rays.
Our primary focus will be on the reflection and refraction of these rays at the interface of two materials which are described in terms of their indices of refraction which are in turn determined by the values of ε for these materials..
We will now turn to the more general case in which there is also a refracted ray. The first order of business is to consider how Maxwell’s equations must be modified to hold in matter, rather than in vacuum. The modification consists simply in replacing the vacuum constants ε 0 and μ 0 with the constants ε and μ that describe the material of interest.
The main effect of this change is that the speed of electromagnetic waves in matter must be different from what it is in vacuum. Namely, the velocity of the wave in matter is equal to 1/sqrt(με) which is different from 1/sqrt(μ 0 ε 0 ), the speed of the wave in vacuum.
We introduce n , the index of refraction of the material, to specify this change in the speed of light. Namely, we define n as the ratio of the speed of light in vacuum to the speed of light in the material, Now, in our study of geometric optics, we will not be interested in magnetic materials, so that we can approximate μ by μ 0 and we see that n is approximately equal to the sqrt(ε/ε 0 ). Recalling that the dielectric constant κ of the material was defined as the ratio of ε to ε 0 , we see that n is approximately equal to the sqrt(κ).
The index of refraction is frequency dependent. In our study of geometric optics, we will be concerned with visible light. At these frequencies, for example, n ranges from a minimum of 1 for air and is about equal to 1.5 for glass and reaches a maximum of about 2.4 for diamond.
We now want to determine how the angle of refraction is related to the angle of incidence. This relationship is totally determined by the difference in the speeds of the waves in the materials at the interface.
We can derive this relationship by noting that the frequency of the incident wave must be equal to the frequency of the refracted wave; the oscillations must be in phase at the surface. Since the speed of light is equal to the product of the frequency and the wavelength, however, we see that this constancy of the frequency demands that the wavelength of the light must be different in the two materials if the speed is different in the two materials. In particular, we see that the ratio of the wavelengths of light ( λ 2 / λ 1 ) must be equal to the ratio of the velocities of light in the two materials ( v 2 /v 1 ). We know the ratio of the velocities, however, is just given by the ratio of the indices of refraction. i.e., v 2 /v 1 = n 1 /n 2. Consequently, we see that λ 2 / λ 1 is equal to n 1 /n 2..
Here we see the ray in material 1 is incident at an angle of θ 1 and is refracted in material 2 to an angle θ 2. We want to determine θ 2 in terms of θ 1. We start by drawing the wavefronts associated with each wave. In particular, we draw the wavefronts separated by one wavelength as shown. Here we have assumed that n 2 > n 1 , so that λ 2 is < λ 1.
We can derive the necessary relationship by focusing on the two triangles that share a hypotenuse L as shown. In medium 1 we can see that the sin θ 1 is equal to λ 1 /L, while in medium 2, sin θ 2 is equal to λ 2 /L. Consequently, we can combine these equations to eliminate L to obtain, λ 2 /sin θ 2 = λ 1 /sin θ 1. We can rewrite this expression as λ 2 / λ 1 = sin θ 2 /sin θ 1.
Recall that we earlier derived that the ratio λ 2 / λ 1 was just given by the ratio n 1 /n 2. Combining these results, we arrive at Snell’s law. Namely, that n 1 sin θ 1 = n 2 sin θ 2. Note that while we drew the incident ray in medium 1 and the refracted ray in medium 2, the mathematics of this derivation is unchanged if ray 2 were actually the incident ray and ray 1 were the refracted ray. Snell’s law simply states that the product of n sin θ in each medium at the interface is the same.
We have just noted that only 4% of the incident light is reflected at an air-glass interface at normal incidence. We will now see, that is possible, however, for some angles of light incident from glass to air, to get all of the light reflected.
We start from Snell’s law and note that sin θ 2 /sin θ 1 is equal to n 1 /n 2. Now if the light is incident at angle θ 1 from the medium with the larger index of refraction, we see that the light will be refracted at a larger angle θ 2 ( i.e., sin θ 2 /sin θ 1 = n 1 /n 2 > 1).
Consequently, as θ 1 increases, θ 2 also increases and is in fact always larger than θ 1. However, there is a limit to this process. Namely, θ 2 can never be larger than 90o. Therefore, for all angles θ 1 such that sin θ 1 > n 2 /n 1 , there will be no refracted ray; all of the incident light will be reflected.
We call this maximum angle of incidence the critical angle. i.e., θ c = sin-1( n 2 /n 1 ). For example, light in glass which is incident on a glass-air interface has a critical angle θ c = sin-1(1.0/1.5) = 41.8o. Light which is incident at any angle θ 1 > θ c will be totally reflected. This property of total internal reflection is the basis for optical fiber communication.
We now know through what angles light rays are reflected and refracted at an interface and something about the relative intensities of the beams. What about the polarization of these beams? Can an unpolarized incident beam produce a polarized reflected beam? The answer to this question is yes, if the unpolarized beam is incident a particular angle, called Brewster’s angle. Here’s how it works.
Here we see incident and reflected rays each making an angle θ 1 with the normal and a refracted ray making an angle θ 2 with the normal. We know the E field for each ray must oscillate in a plane that is perpendicular to that ray. We indicate the components of this oscillation with circles and double headed arrows. The E field component indicated by circles is the component perpendicular to the plane of incidence; this component is parallel to the surface and is therefore perpendicular to all rays. This component of the polarization will always be present.
The E field component indicated by double headed arrows is the component in the plane of incidence. This component for the reflected ray would be generated by electrons in the surface moving perpendicular to the direction of the refracted ray. Now, if the angle between the refracted ray and the reflected ray happens to be equal to 90o, that motion would be parallel to the direction of the reflected ray and therefore, could not give rise to any polarization in the plane of incidence for the reflected ray.
We can determine this special angle, Brewster’s angle, by requiring that θ 1 + θ 2 = 90 o. We can therefore write that sin θ 2 = sin(π/2 – θ 1 ) = cos θ 1. Applying Snell’s law for this angle of incidence, we obtain n 1 sin θ 1 = n 2 cos θ 1. Therefore, Brewster’s angle is determined from the relation: tan θ 1 = n 2 /n 1. If light is incident from the air at an air- glass interface at an angle of 33.7o^ (i.e., tan(33.7o) = 1/1.5), for example, the reflected ray will be linearly polarized perpendicular to the plane of incidence. For other angles of incidence, the reflected ray will be partially polarized in this direction. This result explains why Polaroid sunglasses have a vertical transmission axis in order to block the dominantly horizontally polarized reflected light from horizontal surfaces.
