Reflection, Refraction and Polarization
Introduction When you solved Problem 5.2 using the standing-wave approach, you found a rather curious behavior as the wave propagates and meets the boundary. A new wave emerges from the boundary, traveling in the opposite direction and inverted, as seen in Fig. 1 t 0 Before reflection After reflection Figure 1 The shape of a traveling pulse before and after reflection. The wave emerging from the boundary is called a reflected wave. The existence of a reflected wave and its inverted nature is ÒautomaticÓ when you solve the problem using the standing-wave approach of Problem 5.2, because the solutions you found there are valid for all times t > 0. On the other hand, we saw in Chapter 6 that the solution to this problem for t > 0 can be written as ζ ( x, t ) 1 a ( x ct) + 1 2 2 a ( x + ct) (1) In the figure above, a(x) is a square shape. From the compuphy 241 MenŽndez 168
tational point of view, this solution is much easier to implement than the standing-wave approach we used in Problem 5.2. However, while the standing wave solution to Problem 5.2 is valid for all times, it is apparent that Eq. 1 is not valid when the pulse reaches the boundary. In fact, it is obvious that Eq. 1 doesnõt know anything about the boundary: according to it, the pulses would keep traveling, undisturbed, to the right and to the left. If we want to use the traveling wave solutions for times after the pulse bounces off the boundary, it is clear that we must modify Eq. 1. In the particular case of Fig. 1, it is easy to see what happens: If a section of the wave a(x-ct) hits the right boundary at time t 1, it is transformed into the wave R a(x+c(t-t 1 )), where R -1. Similarly, if a section of the wave a(x+ct) hits the left boundary at time t 1, it is transformed into the wave R a(x-c(t-t 1 )), where R -1. The constant R is called the coefficient of reflection. In the next few sections, we will develop a theory of the coefficient of reflection, allowing for cases in which R is not necessarily of magnitude 1. This occurs when the boundary is not a rigid wall. In fact, we will derive the most general expression for the coefficient of reflection by considering the case of two cords joint at one interface. In this case, a wave vibrating in a cord will propagate into the other cord, so that we will have to define a coefficient of transmission T. Reflection and transmission The wave equation at a discontinuity Let us consider two different cords (for example, cords with mass per unit length ρ 1 and ρ 2 ) with a common end at x 0. Each segment satisfies a wave equation of the form M 2 ξ M t v 2 1 M 2 ξ M x 2 M 2 ξ M 2 ξ v M t 2 2 M x 2 (2) phy 241 MenŽndez 169
(1) (2) x 0 Figure 2 Cord with a discontinuity at x 0. where v 1 is the wave speed in segment (1) and v 2 the wave speed in segment (2). Exactly at the boundary x 0 the wave equation is not satisfied because there is a discontinuity in the density. All quantities entering a differential equation must be continuous and differentiable. On the other hand, the displacement ξ, the cord velocity ξ/ t, and the transverse force T ξ/ x (see Chapter 6) are obviously continuous across the interface. Hence no matter what the solutions to the two wave equations are, we must have lim ε 6 0 lim ε 6 0 lim ε 6 0 ξ ( ε, t ) lim ξ ( + ε, t ) ε 60 M ξ M t ( ε, t ) lim M ξ ε 60 M t ( + ε, t ) T M ξ M x ( ε, t ) lim ε 6 0 T M ξ M x ( + ε, t ) (3) Let us consider the case of a wave traveling from the left to the right. Suppose that we write the solutions as ξ ( 1 ) A i sin ω ( 1 ) ( t (2) ξ A t sin ω ( 2 ) ( t where the subscripts ÒiÓ and ÒtÓ mean ÒincidentÓ and x v 1 x v 1 ), ) (4) phy 241 MenŽndez 170
Òtransmitted,Ó respectively. This equations contain four free parameters: A i, A t, ω (1), and ω (2). However, we should have five free parameters, since we must satisfy three boundary conditions plus we are free to select any frequency and any amplitude for the incident wave. Let us see where the problem arises. When we apply the first two boundary conditions at x 0, we obtain A i sin ω ( 1 ) t A t sin ω ( 2 ) t ω ( 1 ) A i cos ω ( 1 ) t ω ( 2 ) A t cos ω ( 2 ) t (5) which can be satisfied for A i A t, and ω (1) ω (2). It is not surprising that the transmitted frequency turns out to be equal to the incident frequency. We can consider the incident wave as an external force acting on medium (2). As we learned in Chapter 2, if a system is acted upon by a force of angular frequency ω, the system responds by oscillating at the same frequency. This is a very general result for systems that satisfy linear equations of motion. When we subject our proposed solutions to the third boundary condition, we find at x 0 T A i v 1 cos ω ( 1 ) t T A t v 2 cos ω ( 2 ) t. (6) This cannot be satisfied for A i A t unless v 1 v 2, in which case there is no discontinuity. Clearly, if there is a discontinuity our solutions in Eq. (4) cannot be valid. Something is missing. The missing term is precisely the reflected wave. In order to satisfy all boundary conditions we must therefore propose solutions of the form ξ ( 1 ) A i sin ω ( 1 ) ( t ξ ( 2 ) A t sin ω ( 2 ) ( t x v 1 x v 1 ) + A r sin ω ( 1 ) ( t + where the subscript ÒrÓ in the second term of ξ(1) means Òreflected.Ó Using these solutions we find that all boundary conditions can be satisfied provided ) x v 1 ) (7) phy 241 MenŽndez 171
ω ( 1 ) ω ( 2 ) A t 2 v 2 v 1 + v 2 A i A r v 2 v 1 v 1 + v 2 A i (8) We can thus define the reflection and transmission coefficients as A r A i R A t A i T v 2 v 1 v 1 + v 2 2 v 2 v 1 + v 2 (9) Reflection and transmission in three dimensions Let us now consider some limit cases. If the two velocities are equal, i.e., there is not discontinuity, we obtain T 1 and R 0. Hence the incident waves travels through x 0 unperturbed. Suppose now that the cord in medium (2) is infinitely heavy. Then we approach the boundary condition of problem 5.2, where the cord (1) is attached to a rigid wall. Since the wave speed in a cord is given by (T/ρ)1/2, we obtain v 2 0 for ρ 2. Hence we obtain R -1 and T 0, the result we anticipated from our standing-wave solution to problem 5.2. Hence we have rederived the expressions for the reflection and transmission coefficients for that particular problem, but we have achieved more: we now have general expressions for the coefficients that apply to a variety of boundary conditions and which are valid not only for transverse waves in a cord but for all vibrations that can be described by one-dimensional wave equations. Two-dimensional and three-dimensional waves at a boundary The above discussion applies to one-dimensional waves. On the other hand, many of the waves we have discussed, such as sound waves or electromagnetic waves, occur in a three dimensional space. There we are looking not for a function ξ(x,t) but for a function ξ(x,y,z,t). Moreover, the quantity ξ itself might be a component of a three-dimensional vector. phy 241 MenŽndez 172
For example, the electric field in an electromagnetic wave can be written as E (E x (x,y,z,t), E y (x,y,z,t), E z (x,y,z,t)), with each component satisfying a wave equation. To represent waves in these inherently three-dimensional systems, it is very convenient to use a vector notation. Hence we write the wave as ξ ( r, t ) ξ 0 sin ( k A r ω t ) () where ξ is a vector, r (x, y, z) is the position vector, and we have defined a wavevector k as a vector with the magnitude of the wavenumber k and the direction of propagation of the wave. For waves propagating in the x- direction, we have k (k,0,0), and k r kx, so that we obtain the usual expression. The advantage of Eq. () is that it remains valid no matter what the orientation of the axis is, since the dot product of two vectors is independent of the orientation of the axis system. It is customary to trace rays following the direction of propagation of a wave. These rays are parallel to the wavevector k. We will often use the expression Òa rayó instead of Òa plane wave.ó In the context of Eq. (), longitudinal waves are those for which the vector ξ (or ξ 0 ) is parallel to the wavevector k. Transverse waves are those for which the vector ξ (or ξ 0 ) is perpendicular to the wavevector k. In the case of electromagnetic waves, ξ represents either the electric or the magnetic fields, which are perpendicular to the direction of propagation represented by k. In the case of sound waves, ξ represents the displacement of the air. This displacement is longitudinal, so that ξ is parallel to k. Sometimes, the quantity ξ remains a scalar. For example we showed that the pressure fluctuations in a sound wave also satisfy a wave equation. Pressure is a scalar, so that in this case we have a wave of the form p (r,t) p 0 sin (k r-ωt). In the previous chapters we were able to use one-dimensional methods to treat three-dimensional waves because we considered the case of plane waves. For these waves, ξ is constant on a plane perpendicular to the direction of phy 241 MenŽndez 173
propagation. If the direction of propagation is x, this means that ξ is independent of y and z, so that our problem reduces to a one-dimensional one. For the case of reflections and transmissions, we need to pay more attention to the three-dimensional character of some waves. Even if we have a plane-wave incident on a surface, the direction of propagation of the reflected and transmitted wave do not coincide with the direction of propagation of the incident wave, so that the problem cannot be reduced to a one-dimensional one. Therefore, in addition to determining the reflection and transmission coefficients, we must determine the directions of the reflected and transmitted waves. These directions are given by the reflection and refraction laws. The reflection law Suppose that a ray hits a flat surface at an angle θ i with the normal as in Fig. 3 What is the reflected angle θ r Õ? θi θ ' r 11111 00000 Figure 3 The reflection of a three-dimensional wave, for example, light at a mirror. We note that the incident ray and the normal to the surface at the point of contact define a plane. This plane is called the plane of incidence. Given the symmetry of the problem, phy 241 MenŽndez 174
there is no reason why the reflected ray should deviate from the plane of incidence. Therefore, we can state The reflected ray is in the plane of incidence. Let us now compare the incident and reflection angles using Fig. 4 Figure 4 The reflection law. Since all the points of the surface behave the same way, ray R 1 ÕÕ and R 2 ÕÕ should be parallel (proof: slide the surface so that BÕ moves to A position). We can now write sin θ i BB' AB' sin θ ' r AA' ' AB' (11) But since the wave must take the same time t to go from B to BÕ and A to AÕÕ, because all points on a wave plane have the same time dependence, we can write BBÕ AAÕÕ v 1 t, where v 1 is the speed of the wave in the medium where the incident and reflected waves propagate. Hence sin θ i v 1 t/abõ sin θ r Õ. This implies θ i θ N r (12) This is obviously the result we expected from our physical intuition. The refraction law (SnellÕs law) We now must determine the direction of the transmitted wave, which is usually called the refracted wave. Using a similar argument as before, we can establish that phy 241 MenŽndez 175
the refracted ray is in the plane of incidence. Next we must find the refraction angle θ r. We refer to Fig. 5. We have Figure 5 The refraction law. sin θ i sin θ r BB' AB' AA' AB' (13) But now BBÕ v 1 t and AAÕ v 2 t, where v 1 is the speed of the wave in the medium where the incident wave propagates and v 2 the speed of the wave in the medium where the refracted wave propagates. Dividing the two expressions, we obtain sin θ i sin θ r v 1 v 2 (14) For electromagnetic waves, we can rewrite this expression if we recall that the speed of light in a medium equals the speed of light in vacuum c divided by the index of refraction n. Hence we obtain This is called SnellÕs law. Application I: apparent depth sin θ i n 2 sin θ r (15) phy 241 MenŽndez 176
A common observation is that objects immersed in liquids appear to be at a smaller depth than they really are. This can be understood with the diagram of Fig. 6 Air, n 1.00 11111111111 00000000000 10 Water, n 1.33 Figure 6 Real depth versus apparent depth. The distance to an object is determined by our brain by ÒcomputingÓ the intersection point of the rays that enter our two eyes. When a ray travels from a medium of high refraction index to a medium of lower refraction index, as is the case for an object immersed in water, the refraction angle is larger than the incident angle, as can be seen from SnellÕs law. Therefore, the intersection point of two rays that enter our eyes occurs closer to the surface than the real position of the object. This fools the brain into believing that the object is closer to the surface. Application II: total internal reflection From SnellÕs law, we obtain sin θ r n 2 sin θ i (16) Of course, sin θ r ² 1. However, if > n 2, Eq. (16) seems to allow for sin θ r > 1. This is impossible. If the incident angle θ i is such that according to Eq. (16) sin θ r > 1, then there is no refracted light: all the light is reflected. This phenomenon is called total internal reflection. The largest incident angle for which refraction can still occur is called the critical angle θ c. This angle can be calculated by setting sin θ r 1. phy 241 MenŽndez 177
We thus obtain sin θ c n 2 (17) For light traveling inside glass, which has a refraction index n 1.5, we obtain, assuming that the outside medium is air, sin θ c 1/1.5 0.67, which implies θ c 42¼. This is the principle of operation of optical fibers: provided light is incident on the surface at angles larger than 42¼, it is totally reflected inside the fiber and does not escape outside. Glass Air Figure 7 Optical fiber illustrating the principle of total internal reflection. Coefficients of reflection and refraction for electromagnetic waves We have already determined the directions of reflection and refraction (transmission) of electromagnetic waves, but we still have to determine the coefficients of reflection and refraction. We said above that Eqs. (9) are universally valid, so that we might expect them to be valid for electromagnetic waves. However, there is a fundamental problem related to the direction of propagation, which did not occur for the one-dimensional waves used to derive Eqs. (9). Consider the two situations depicted in Fig. 8 phy 241 MenŽndez 178
E 111111111 000000000 B Chapter 9 B 111111111 000000000 E a) b) Figure 8 The two non-equivalent configurations for the same incident angle. In case a), the electric field lies in the plane of incidence, while the magnetic field is perpendicular to it, that is, out of the plane of the figure. We have represented this by a dot. In case b), it is the magnetic field which lies in the incident plane, whereas the electric field is perpendicular to it. This two situations are not geometrically equivalent except for the special case of normal incidence. Hence we might expect different reflection and transmission (refraction) coefficients for the two cases. If we define E π component of E in the plane of incidence, and E σ component of E perpendicular to the plane of incidence, we can define two reflection coefficients and two transmission coefficients: R π E N r, π E i, π, R σ E N r, σ E i, σ T π E r, π E i, π, T σ E r, σ E i, σ (18) Note that we donõt need to define reflection and transmission coefficients for the magnetic field because B is always perpendicular to E and its magnitude is given by B E/c. The four reflection coefficients can be obtained by straightforward but lengthy application of MaxwellÕs equations to phy 241 MenŽndez 179
the boundaries of the two materials. We will simply quote the results: R π cos θ r n 2 cos θ i cos θ r + n 2 cos θ i R σ cos θ i n 2 cos θ r cos θ i + n 2 cos θ r T π 2 cos θ i cos θ r + n 2 cos θ i T σ 2 cos θ i cos θ i + n 2 cos θ r (19) This expressions are quite complicated. They are best understood by studying some special cases. Let us consider, for example the case of normal incidence. In this case, all the angles that appear in Eqs. (17) are zero, so that we obtain R π R σ n 2 + n 2 T π T σ 2 + n 2 (20) It is understandable that R π R σ and T π T σ, since for normal incidence the π and σ directions are equivalent, as in the string case. The problems becomes effectively onedimensional, since the incident, reflected, and transmitted ray are all in the same direction. In fact, using v 1 c/ and v 2 c/n 2, we obtain R π R σ v 2 v 1 v 1 + v 2 T π T σ 2 v 2 v 1 + v 2 (21) which are identical to Eqs. (9)! Polarized light Most light sources, such as the sun and incandescent lamps, (but not lasers!) emit light with random directions of phy 241 MenŽndez 180
polarization. However, Eqs. (19) suggest that after passing through a medium (or being reflected by a medium) the two perpendicular polarization directions will not have equal magnitude, since their reflections coefficients are not equivalent, except in the case of normal incidence. BrewsterÕs angle We can now ask an interesting question. Is it possible to select an incident angle such that one of the two reflection or transmission coefficients becomes zero, so that we get linearly polarized light out of randomly polarized light? It is quite obvious from inspection of Eqs. (19) that we cannot have simultaneously T σ 0, T π 0, or T π 0, T σ 0. It is also possible to show (see homework problem) that we cannot have R σ 0, R π 0, because this violates SnellÕs law. The only possibility left is therefore R σ 0, R π 0.From R π 0 we obtain cos θ r - n 2 cos θ i 0. On the other hand, form SnellÕs law we know that sin θ i n 2 sin θ r. Solving for n 2 in this expression and substituting in the previous expression, we get sin θ r cosθ r sin θ i cosθ i, or sin 2 θ r sin 2 θ i (22) A possible, obvious solution to this equation is θ r θ i, but this violates SnellÕs law. However, the condition θ r π 2 θ i (23) is also a solution to Eq. (22). Let us see if this condition is compatible with SnellÕs law: sin θ i n 2 sin θ r n 2 sin Š Œ π 2 θ i sin θ i n 2 cos θ i tan θ i n 2 (24) phy 241 MenŽndez 181
The angle that satisfies the last condition in Eq. (24) is called Brewster angle. For an air-glass interface, with n glass 1.5, the Brewster angle corresponds to tan θ i 1.5, or θ i 56¼. In other words, if light with arbitrary polarization is incident upon a piece of glass at an angle of 56¼ to the normal, the reflected light is completely polarized perpendicular to the plane of incidence. Thus the light that reaches our eyes after reflection on distant horizontal surfaces tends to be polarized horizontally. This explains the use of polarizing sunglasses, which block preferentially the light that is polarized horizontally. These glasses are important to sportspeople, such as those who practice sky, since most of the light they receive is reflected from the snow. E i,π E E r,σ i,σ Er,π E r,σ Figure 9 Light incident on a piece of glass under BrewsterÕs angle condition. Notice that we have not indicated the components of the magnetic field, which are perpendicular to the respective components of the electric field. phy 241 MenŽndez 182
Figure 9 represents the conditions under which reflection under BrewsterÕs angle occurs. Notice that the reflected and refracted rays are perpendicular to each other. This is a direct consequence of Eq. (23).We have indicated above that the transmitted ray cannot be completely polarized. However, since the σ component of the light is preferentially reflected, the ratio σ/π becomes smaller in the transmitted beam. If we stack several layers of glass with small gaps in between, after several reflection/refraction events the transmitted light will be almost completely polarized in the π direction. phy 241 MenŽndez 183
Problems 1. (Alonso 32.6) Copper and steel wires of the same radii are joined, making a long string. Find T and R at the junction for waves propagated along the string. Let the common radius be 1 mm. Assuming that the incident wave has a frequency of Hz, that the amplitude is 2 cm, and that the tension is 50 N, write the equations for the incident, reflected, and transmitted waves. (The density of copper is 8.89 3 kg m -3, of steel 7.80 3 kg m -3 ). 1.(Alonso 32.4) A ray of light falls on a wedge of glass with n 1.6. Calculate the minimum angle α so there is total reflection at the second surface. 37 11111111111111 00000000000000 α 111111111111 000000000000 1111111111 0000000000 11111111 00000000 111111 000000 1100 3. Consider a piece of glass with parallel faces and thickness d. Light is incident on one of the faces at an arbitrary angle. Show that the ray emerging at the other surface is parallel to the original incident ray, but displaced a certain distance. Calculate this distance. 4. (Alonso 32.9) a) Calculate the angle the Sun must be above the horizontal so that sunlight reflected from the surface of a calm lake is completely polarized. (n water 1.33) b) What is the plane of the electric field vector in the reflected light? 4. (Alonso 32.19) When a plane electromagnetic wave falls perpendicularly on a plane surface separating a medium of index from a medium of index n 2, we have shown that the coefficients of reflection and refraction are R ( - n 2 )/( + n 2 ) and T 2 /( + n 2 ). Discuss what may be said about the intensity and phase (how the direction of the electric field changes) of the reflected and transmitted (refracted) waves for the cases > n 2, < n 2, and n 2. 5. (Alonso 32.20) a) Light falls perpendicularly on a glass plate (n 1.5). Find the coefficients of reflection and transmission. b) Repeat the calculation if the light is passing from the glass into the air. c) Discuss in each case the changes in phase. (Hint: see the previous problem). 6. Show that if R σ 0 and R π 0, one obtains tan θ i tan θ r, which violates SnellÕs law. 7. Consider light traveling in air and incident on a piece of glass, whose index of refraction is n 1.50. Graph, as a function of the incident angle, the fraction of light intensity that is reflected and transmitted for two cases a) the light has E σ 0, Ε π 0, and b) the light has E π 0, Ε σ 0 (Hint: remember that the intensity of a wave is proportional to the square of the amplitudes. For the graphs, include the reflected intensity and the transmitted intensity in the same graph, so that you need a total of two graphs). Indicate BrewsterÕs angle on the graphs. Also discuss the conditions for which a piece of glass will act as a good mirror. phy 241 MenŽndez 184