Geometric Optics: Reflection and Refraction of Electromagnetic Waves, Study notes of Physics

Download Geometric Optics: Reflection and Refraction of Electromagnetic Waves and more Study notes Physics in PDF only on Docsity! Prelecture 24: Slide 2 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. Prelecture 24: Slide 3 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 ε0 and µ0 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.. Prelecture 24: Slide 6 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 (λ2/λ1) must be equal to the ratio of the velocities of light in the two materials (v2/v1). We know the ratio of the velocities, however, is just given by the ratio of the indices of refraction. i.e., v2/v1 = n1/n2. Consequently, we see that λ2/λ1 is equal to n1/n2. . Here we see the ray in material 1 is incident at an angle of θ1 and is refracted in material 2 to an angle θ2. We want to determine θ2 in terms of θ1. 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 n2 > n1, so that λ2 is < λ1. 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 θ1 is equal to λ1/L, while in medium 2, sin θ2 is equal to λ2/L. Consequently, we can combine these equations to eliminate L to obtain, λ2/sinθ2 = λ1/sinθ1. We can rewrite this expression as λ2/λ1 = sinθ2/sinθ1. Recall that we earlier derived that the ratio λ2/λ1 was just given by the ratio n1/n2. Combining these results, we arrive at Snell’s law. Namely, that n1sin θ1 = n2sinθ2. 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 nsinθ in each medium at the interface is the same. Prelecture 24: Slide 7 We now know the relationships between all of angles involved in the reflection and refraction of light. Namely, the angle of incidence equals the angle of reflection and n1sin θ1 = n2sinθ2. What about the intensities of these waves? How much is reflected and how much is refracted? While the answers to these questions can be obtained from Maxwell’s equations, the derivations are beyond the scope of this course. We will be able to determine, however, intensities for a couple of special cases in the following slides. We simply note here the results for two extreme cases. Namely, for glancing incidence (θ1 ~ 90o), we have complete reflection (R ~ 1), while for normal incidence, it can be shown that R = ((n2-n1)/(n2+n1)) 2. Note that for an air-glass interface, for example, only 4% of the incident light is reflected at normal incidence. Prelecture 24: Slide 8 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 sinθ2/sinθ1 is equal to n1/n2. Now if the light is incident at angle θ1 from the medium with the larger index of refraction, we see that the light will be refracted at a larger angle θ2 ( i.e., sinθ2/sinθ1 = n1/n2 > 1). Consequently, as θ1 increases, θ2 also increases and is in fact always larger than θ1. However, there is a limit to this process. Namely, θ2 can never be larger than 90o. Therefore, for all angles θ1 such that sin θ1 > n2/n1, 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., θc = sin-1(n2/n1). For example, light in glass which is incident on a glass-air interface has a critical angle θc = sin-1(1.0/1.5) = 41.8o. Light which is incident at any angle θ1 > θc will be totally reflected. This property of total internal reflection is the basis for optical fiber communication.