Lecture 15: Refraction and Reflection, Schemes and Mind Maps of Law

Download Lecture 15: Refraction and Reflection and more Schemes and Mind Maps Law in PDF only on Docsity! Matthew Schwartz Lecture 15: Refraction and Reflection 1 Refraction When we discussed polarization, we saw that when light enters a medium with a different index of refraction, the frequency stays the same but the wavelength changes. Using that the speed of light is v = c n we deduced that λ1n1 = λ2n2, so that as the index of refraction goes up, the wave- length goes down. The picture looks like this Figure 1. Plane wave entering and emerging from a medium with different index of refraction. Now let us ask what happens when light enters a medium with a different index of refraction at an angle. Since we know the wavelength of light in the two media, we can deduce the effect with pictures. The key is to draw the plane waves as the location of the maximum field values. These crests will be straight lines, but spaced more closely together in the medium with higher index of refraction. For example, if sunlight hits ice (or water), the picture looks like this Figure 2. Matching wavefronts demonstrates refraction. Orange lines represent the crests of waves (or the maximum amplitude 1 The bending of light when the index of refraction changes is called refraction. To relate the angles θ1 (the angle of incidence) to θ2 (the angle of refraction) we draw a triangle Figure 3. Light comes in from the air on the left with the left dashed blue line indicating one wavecrest with the previous wavecrest having just finished passing into the water. Thus the wavelength λ1 in medium 1 is the solid thick orange line on the bottom left, and the wavelength in the second medium λ2 is the thick solid orange line on the top right. Call the distance between the places where the wave crest hits the water along the water R (the thick green vertical line in the picture). The distance between crests is λ1 = R sinθ1 in the air and λ2=R sinθ2 in the water. Since R is the same, trigonometry and n1λ1=n2λ2 imply n1 n2 = λ2 λ1 = sinθ2 sinθ1 (1) This is known as Snell’s law. The same logic holds for reflected waves: R is the same and λ is the same (since n1=n2 for a reflection) therefore θ1 = θ2. This is usual the law of reflection: the angle of reflection is equal to the angle of incidence. For a fast-to-slow interface (like air to water), the angle gets smaller (the refracted angle is less than the incident angle). For a slow-to-fast interface (like water to air), the angle gets larger. Since the angle cannot be larger than 90◦ while remaining in the second medium, there is a largest incident angle for refraction when n2<n1. In equations, since the refracted angle satis- fies sinθ2 = n1 n2 sinθ1, we see that if n1 n2 sinθ1 > 1 there is no solution. The critical angle beyond which no refraction occurs is therefore θc= sin−1n2 n1 (2) For air n ≈ 1 and for water n = 1.33 so θc = 49◦. For incident angles larger than the critical angle, there is no refraction: all the light is reflected. We call this situation total internal reflection. Total internal reflection can only happen if n2<n1. Thus, light can be confined to a material with higher index of refraction but not a lower one. Total internal reflection is the principle behind fiber optics. A fiber optical cable has a solid silica core surrounding by a cladding with an index of refraction about 1% smaller. For example, the core might have n1≈ 1.4475 and the cladding n2= 1.444, so the critical angle is θc= 86◦ from normal incidence, or 4◦ from the direction of propagation. As long as the cable is not bent too much (typically the cable is thick enough so that this is very difficult), the light will just bounce around in the cable, never exiting, with little loss. Why might you want to send signals with vis- ible light rather than radio waves? 2 Section 1 For normal incidence the incident angle to the normal θ1 = 0 (see Fig. 8 below). Thus E⊥ = B⊥ = 0. That is, the electric and magnetic fields are both polarized in the plane of the interface to there simply is no ⊥ component. Without loss of generality, let’s take a plane wave of fre- quency ω moving in the ẑ direction with E~ in the ŷ direction. Then since B~ = 1 ω k~ ×E~ , the mag- netic field points in the x̂ direction. So the incident fields are E~ I =EIŷe iω ( t− 1 v1 z ) , B~ I =BIx̂e iω ( t− 1 v1 z ) (8) The transmitted and reflected waves are also moving normal to the surface (by Snell’s law), so we can write E~ T =ETŷe iω ( t− 1 v1 z ) B~ T =BTx̂e iω ( t− 1 v1 z ) (9) E~R=ERŷe iω ( t+ 1 v1 z ) B~R=−BRx̂e iω ( t+ 1 v1 z ) (10) with ET , ER, BT and BR the transmission and reflection coefficients to be determined. Note that we have flipped the sign on the z term in the phase since the reflected wave is moving in the −ẑ direction (that is, k~ flips). Since ωB~ = k~ × E~ , this means that B~ flips sign too, which explains the minus sign in Eq. (10). Since there is no ⊥ component the boundary condition E|| (1) =E|| (2) implies that EI +ER=ET (11) Similarly, 1 µ1 B|| (1) = 1 µ2 B|| (2) implies 1 µ1 BI − 1 µ1 BR= 1 µ2 BT (12) Since ∣ ∣B~ ∣ ∣= 1 v ∣ ∣E~ ∣ ∣ for any plane wave, BI = 1 v1 EI and BT = 1 v2 ET and Eq. (12) becomes 1 µ1v1 (EI −ER)= 1 µ2v2 ET (13) Eqs. (11) and (13) look just very much like the equations for transmission and reflection for a wave that we studied in Lecture 9. Solving them gives ER=EI Z2−Z1 Z2+Z1 , ET =EI 2Z2 Z2+Z1 (14) where Z = µv= µ c n = µ ε √ (15) For most materials, µ is pretty constant, so the form Z = µc 1 n , which says Z ∝ 1 n , is most useful What is the power reflected and transmitted? The incident power in a medium is given by the Poynting vector P~ = 1 µ E~ ×B~ times area A. So ∣ ∣P~I ∣ ∣= 1 µv ∣ ∣E~ I ∣ ∣ 2A= 1 Z ∣ ∣E~ I ∣ ∣ 2A (16) Thus the reflected power is PR= ER 2 Z1 A= ( Z2−Z1 Z2+Z1 ) 2EI 2 Z1 A= ( Z1−Z2 Z1+Z2 ) 2 PI (17) and the transmitted power is PT = ET 2 Z2 A= ( 2Z2 Z2+Z1 ) 2EI Z2 2 A= 4Z1Z2 (Z1+Z2)2 EI 2 Z1 A= 4Z1Z2 (Z1+Z2)2 PI (18) These satisfy PT + PR = PI as expected. Note that these equations hold for normal incidence only. Normal incidence 5 By the way, Eq. (15) implies that empty space has impedance of Z0 = µ0 ǫ0 √ = 376.7Ω. This number is very useful in broadcasting, since you want to impedance-match your antenna to air to get efficient signal transmission. Coaxial cables usually have Z = 50Ω so some circuitry is required to get the antenna impedance higher. Antennas and interference patters are the subject of Lecture 18. 4 Thin film interference Eq. (14) is ER=EI Z2−Z1 Z2+Z1 . Using Z ∝ 1 n , this becomes ER EI = n1−n2 n1+n2 (19) Thus for n2 > n1 the reflected electric field has a phase flip compared to the incoming field, and for n1<n2 there is no phase flip. So what happens when light hits a thin film? The film will generally have n2 = nfilm > nair = n1. Thus we get a phase flip at the top surface, but no phase flip at the bottom surface. The picture looks like this Figure 6. There is a π phase flip from reflections off the top surface, where n2 > n1, but not off the bottom surface, where n2<n1. So what happens if light of wavelength λ hits a film of thickness d? There will be two reflected waves, one off the top surface (A) and one off the bottom (B) which will interfere. Say the incoming electric field is EI cos(kz − ωt), pointing to the right. Then the wave A which reflects off the top surface (z=0) will be EA=REI cos(−ωt− π)=−REI cos(−ωt) (20) with R the reflection coefficient. This now points to the left. The wave which gets through will be ET cos ( 2π λ z − ωt ) , with T the transmission coefficient. This wave then goes to the bottom suface, reflects with no phase flip, and comes back and exits with no more phase flips. So when it exits, it is back to z=0 after having traversed a distance ∆z=2d. Thus it is EB=EIT 2R cos ( −ωt+2π 2d λ ) (21) The total wave at t=0 therefore Etot=EIR [ −1+T 2cos ( 4πd λ )] (22) where λ is the wavelength in the second medium. 6 Section 4 In the limit that d≪ λ the two reflections will be exactly out of phase. For T ≈ 1, there will be complete destructive interference and no reflection. But there will also be complete destruc- tive interference whenever cos ( 4πd λ ) =1, which is when d= λ 2 , 2λ 2 , 3λ 2 , 4λ 2 ··· ( complete destructive interference) (23) On the other hand if the two waves are completely in phsae there will be constructive interfer- ence. This happens when cos ( 4πd λ ) =−1 which is when d= λ 4 , 3λ 4 , 5λ 4 , ··· ( complete constructive interference) (24) If the material is much thicker than the wavelength of light, and not of completely uniform thickness, then there will be some constructive and some destructive interference and we won’t see much interesting. However, if the material has a well-defined thickness which is of the same order of magnitude as the wavelength of visible light, we will see different wavelengths with dif- ferent intensities. This happens in a soap film. If we put a soap film vertically, then gravity will make it denser at the bottom. The result is the following: Note at the very top, the film is black because there is complete distructive interference for all wavelengths. Similar patterns can be seen in soap bubbles. Color due to thin-film interference is known as iridescence. The color of many butterflies and the gorgets of hummingbirds is due to iridescence. Thin film interference 7 6 Transmitted power To interpret the Fresnel coefficients, we would like to know not just the amplitude of the wave transmitted, but the intensity of the light that passes through, or equivalently the power. Recall that the power in a plane wave in the vacuum is P = cǫ0 ∣ ∣E~ ∣ ∣ 2. For a plane wave in a medium, ε changes and only the component of the velocity moving into the medium is relevant, so this becomes P = v cosθǫ ∣ ∣E~ ∣ ∣ 2 = cosθ ε µ √ ∣ ∣E~ ∣ ∣ 2 = cosθ Z ∣ ∣E~ ∣ ∣ 2. Thus the fraction of power reflected for ver- tical and horizontal polarizations is: PR vert PI vert = ( α− β α+ β ) 2 , PR horiz PI horiz = ( αβ − 1 αβ+1 ) 2 , (37) For the fraction of power transmited, PT vert PI vert = cosθ2 cosθ1 Z1 Z2 ( 2 α+ β ) 2 =αβ ( 2 α+ β ) 2 (38) and PT horiz PI horiz = cosθ2 cosθ1 Z1 Z2 ( 2 αβ+1 ) 2 =αβ ( 2 αβ+1 ) 2 (39) One can check that for either polarization, PT + PR = PI and that these equations agree with Eqs. (17) and (18) for normal incidence ( α=1, β= Z1 Z2 ) . For example, the air-glass interface has β = 1.5. Then the transmitted and reflected power in vertical and horizontal polarizations are Figure 9. Transmitted and reflected power as a function of incident angle for the two polarizations. We can also plot the ratio of power reflected in vertical to horizontal polarizations: Figure 10. Ratio PR vert /PR horiz of power reflected vertically polarized to horizontally polarized light, . From these plots we see there is an angle where the vertical polarization exactly vanishes. This angle is called Brewster’s angle, θB. We can see from the plots that for the glass-air interface θB≈ 56◦. At this angle, the reflected light is completely polarized. 10 Section 6 What is the general formula for θB? From Eq. (37) we see that PR vert = 0 when α = β. That is cosθ2 cosθB = ǫ2µ1 ǫ1µ2 √ (40) For most materials µ1 ∼ µ2 ≈ µ0. Then we can use n = µǫ √ to get cosθ2 cosθB = n2 n1 . Using also n1sinθB=n2sinθ2 we can solve for θB giving tanθB= n2 n1 (41) For the air-water interface, θB = tan−11.5= 56.3◦. One can see from Figure 10 that one does not have to be exactly at this angle to have little reflected vertical polarization. Angles close to θB work almost as well. What is going on physically at Brewster’s angle? We know that Snell’s law n1sinθ1= n2sinθ2 is always satisfied. At Brewster’s angle, when θ1= θB, we found also n1cosθ2= n2cosθ1. Dividing these two equations gives cosθ1sinθ1= cosθ2sinθ2, or equivalently sin(2θ1) = sin(2θ2) when θ1= θB (42) Now, by definition θ1 and θ2 are both between 0 and π 2 , so 2θ1 and 2θ2 are between 0 and π. Thus, sin(2θ1) = sin(2θ2) has two solutions. Either θ1 = θ2, which corresponds to n1 = n2 so the light just passes through, or π 2 − 2θ1=2θ2− π 2 (see Fig 11, left) which simplifies to θ1+ θ2= π 2 (43) Figure 11. The Brewster’s angle happens when sin(2θ1) = sin(2θ2) which corresponds to π 2 − 2θ1= 2θ2− π 2 as can be seen in the left figure. This means that the transmitted and reflected waves are perpendic- ular. Thus the reflected vertically polarized light (shown on the right) cannot be produced by the motion of particles in the surface. Hence the reflected vertically polarized light vanishes at Brewster’s angle. Working out the geometry, as in Fig. 11, we see that the transmitted and reflected waves are perpendicular at Brewster’s angle. Since the reflected wave has to be produced by the motion of particles in the surface we can understand why there is no reflection at Brewster’s angle: parti- cles moving in the surface can only produce light polarized transverse to their direction of motion. In Lecture 17, we will understand in more detail how accelerating changes produce elec- tromagnetic fields and waves. The famous Harvard dropout Edwin Land started a corporation called Polaroid that made its first fortune with polarizing sunglasses. These glasses remove glare from reflection by removing the horizontally polarized light. He made his second fortune was made with the Polaroid camera. In 1973 Land donated the money for the construction of the Science Center (which some people say looks like a camera...). Transmitted power 11