DOCSLIB.ORG

Chapter 6 Light at Interfaces – Law of Reflection and Refraction

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Chapter 6 Light at Interfaces – Law of Reflection and Refraction Chapter 6 Light at Interfaces – Law of Reflection and Refraction Part I Introduction Law of Reflection and Refraction Law of Reflection and Refraction normal Law of reflection: i r i r n1 n2 Law of refraction “Snell’s Law”: sin n t i 2 sint n1 Incident, reflected, refracted, and surface normal in same plane Huygens’ Principle Every point on a wavefront may be regarded as a secondary source of wavelets Huygens’ Principle Law of refraction Fermat’s Principle The path a beam of light takes between two points is the one which is traversed in the least time Law of reflection Fermat’s Principle Law of refraction SO OP t vi vt h2 x2 b2 (a x)2 t vi vt To minimize t(x) with respect to x, dt x (a x) dx 2 2 2 2 vi h x vt b (a x) 0. sin sin Thus, i t , vi vt so, ni sini nt sint Fermat’s Principle For a ray passing through a multilayer (m layers) material s s s t 1 2 m v1 v2 vm Then 1 m t ni si , c i1 The summation is known as the optical path length (OPL) traversed by the ray. P (OPL) S n(s)ds. The route light travels having the smallest optical path length EM Wave Point of View Light is an electromagnetic wave, obeys the Maxwell’s equations Plane of incidence: formed by k and the normal of the interface plane normal EB k EM Wave Point of View i(tkr) E E0e i(tkr) B B0e E cB 1. What happens to and at the interface? 2. Can one derive law of reflection and refraction? 3. Or obtain more information? Part II Fresnel Equations Plane of Incidence Polarization of Light Always relative to plane of incidence TE: Transverse electric TM: Transverse magnetic s: senkrecht polarized p: plane polarized (E-field sticks in and out of the plane) (E-field in the plane) perpendicular, horizontal parallel, vertical Polarization of Light Linear polarized light decomposition y x Any polarization state can be described as y linear combination of these two: ikzt ikzty E E e x xˆ E e yˆ 0x 0 y x ix iy ikzt E E0xe xˆ E0 ye yˆ e “complex amplitude” contains all polarization info Waves at Interface boundary point Waves at Interface At the boundary point, phases of the three waves must be equal (k i r it) (k r r rt) (k t r tt) r is at the boundary This is true for any boundary point and time, so let’s take r 0 it rt tt i r t Thus, the frequencies are equal If we now consider t 0 k i r k r r k t r which means all three propagation vectors lie in the same plane Law of Reflection Since k i r k r r kir sini kr r sinr Both incident and reflected waves travel in same medium, ki kr Thus one obtains the law of reflection: i r Law of Refraction Since k r r k t r kr r sinr kt r sint Both reflected and transmitted waves travel in different media (same frequencies; different wavelengths!): kr / vr n1 / c kt / vt n2 / c which leads to the law of refraction: n1 sinr n2 sint Part III Fresnel Equations Maxwell’s Equations D – Displacement field D dS dV D H – Magnetic Intensity B B E dr dS E t t P – Polarization D D M – Magnetization H dr j dS H j t t - Magnetic permeability B dS 0 B 0 - Permittivity e - Dielectric Susceptibility m - Magnetic Susc. D E P P E o e – Conductivity 1 H B M M m H j – Current density o j E D E B H Boundary Conditions I B 0 A small cylindrical Gauss surface cross the interface gives The normal components of magnetic field are continuous across the interface. Boundary Conditions II D A small cylindrical Gauss surface cross the interface gives is the surface free charge density. If , then the normal component of D is continuous across the interface. But E may not unless will be the same. Boundary Conditions III B E t A small rectangular loop cross the interface gives The tangential component of the electric field is continuous across the boundary. Boundary Conditions IV D H j t A small rectangular loop cross the interface gives is the surface free current density. If , then the normal component of H is continuous across the interface. E-fields at the Boundary For both electric and magnetic fields, the tangetial components must be continuous as boundary is passed All waves only have tangential electric fields: TE waves E E yˆ 0i i E E yˆ 0r r E0t Et yˆ complex field amplitudes Continuity requires: E i E r E t B-fields at the Boundary TE waves The magnetic fields: i(ki r t) Bi (Bi cosi xˆ Bi sini zˆ)e i(kr r t) Br (Br cosr xˆ Br sinr zˆ)e i(kt r t) Bt (Bt cost xˆ Bt sint zˆ)e Continuity requires: Bi cosi Br cosr Bt cost Same analysis can be performed for TM waves Boundary Conditions at the Interface TE waves TM waves E E i i Er Er n1 Bt B n2 t B B i Br i Br Et Et Ei Er Et Bi Br Bt Bi cosi Br cosr Bt cost Ei cosi Er cosr Et cost Fresnel Equations TE waves (s-polarization) TM waves (p-polarization) Get all in terms of E and apply law of reflection (i = r): Ei Er Et ni Ei ni Er nt Et ni Ei cosi ni Er cosi nt Et cost Ei cosi Er cosi Et cost For reflection: eliminate Et, separate Ei and Er, and take ratio: Er ni cosi nt cost Er nt cosi ni cost rTE r rTM r|| Ei ni cosi nt cost Ei nt cosi ni cost For transmission: eliminate Er, separate Ei and Et, take ratio… Et 2ni cosi Et 2ni cosi tTE t tTM t|| Ei ni cosi nt cost Ei nt cosi ni cost Fresnel Equations TE waves (s-polarization) TM waves (p-polarization) And together: nt t 1 r t|| 1 r|| ni Using Snell’s law, , one has tan( ) sin(i t ) i t r r|| tan( ) sin(i t ) i t 2sint cosi 2sint cosi t t|| sin(i t ) sin(i t )cos(i t ) Part IV Understand Fresnel Equations Perpendicular Incident When , then n n n n r i t r t i || n n ni nt t i 2n i 2ni t t|| ni nt ni nt Thus, At the air/glass interface, r r|| therefore t t|| The plane of incidence is no longer specified when i t 0 Grazing Incident When , but r 1 r|| 1 t t|| 0 Reflection and Transmission Coefficients nt Calculate r and t versus i, let: n ni cos n2 sin 2 2 2 2 i i n cosi n sin i r r cos n2 sin 2 || 2 2 2 i i n cosi n sin i 2cos t i 2ncosi 2 2 t || 2 2 2 cosi n sin i n cosi n sin i Mathematica plot of transmitteded/reflected light http://demonstrations.wolfram.com/FresnelEquations/ Reflection and Transmission Coefficients 1.0 t 0.5 || t r|| 0.0 Coefficients -0.5 r Brewster's angle 56.3o -1.0 0 20 40 60 80 Incident Angle i Reflection and Transmission Coefficients 1.0 0.5 r 0.0 41.8o r|| Coefficients -0.5 Critical angle Brewster's angle 33.7o -1.0 0 10 20 30 40 50 60 70 80 90 Incident Angle i Phase Shift k r t ˆ Ei Eie j k r t ˆ Er Ere j At the Air Glass interface One expect a reversal of sign in the electric fields, compared to ei 1 for . i.e., after reflection, the phase of the E-field of the reflected light changes with respect to the incident E-field Phase Shift We need to define phase shift for two cases: A. When two fields E or B are to the plane of incidence, they are said to be (i) in-phase ( = 0) if the two E or B fields are parallel and (ii) out-of-phase ( = ) if the fields are anti-parallel. B. When two fields E or B are parallel to the plane of incidence, the fields are (i) in-phase if the y-components of the field are parallel and (ii) out-of-phase if the y-components of the field are anti-parallel. Phase Shift (a) Bi , Bt (b) in phase, 0 Ei , Et E , E , E i r t all in phase 0 Ei , Er Bi , Br , Bt out of phase, Bi , Br Phase Shift Analogy between a wave on a string and an E-M wave traversing the air-glass interface. Phase Shift Analogy between a wave on a string and an E-M wave traversing the air-glass interface. Glass (n = 1.5) Air (n = 1) r 0 = 0 = 0 Air (n = 1) Glass (n = 1.5) = = 0 r 0 Phase Shift 1.0 0.5 t t|| r|| 0.0 Coefficients -0.5 r -1.0 0 20 40 60 80 Incident Angle i Phase Shift 1.0 0.5 r 0.0 Coefficients -0.5 r|| -1.0 0 10 20 30 40 50 60 70 80 90 Incident Angle i Reflectance and Transmittance From the geometry and total area A of the beam at the interface, the power (P) for the (i) incident, (ii) reflected and (ii) transmitted beams are: (i) Pi Ii Acosi (ii) Pr Ir Acosr (iii) Pt It Acost Cross-sectional Areas Reflectance and Transmittance P I Acos R r r r r 2 Pi Ii Acosi Pt It Acost nt cost 2 T t Pi Ii Acosi ni cos i 2 2 E n cos E Therefore, or t t ot 1 R T Eoi ni cosi Eoi We can write this expression in the form of components and ||: n cos n cos 2 2 t t 2 t t 2 R r and R|| r|| , T t and T|| t|| ni cosi ni cosi Therefore, R T 1 and R|| T|| 1 Reflectance and Transmittance TM TE (p) (s) Reflectance and Transmittance for an Air-to-Glass Interface Reflectance and Transmittance Perpendicular polarization Parallel polarization 1.0 1.0 R R .5 .5 T T 0 0 0° 30° 60° 90° 0° 30° 60° 90° Incidence angle, i Incidence angle, i Reflectance and Transmittance for an Glass-to-Air Interface Reflectance and Transmittance 2 n n When 0, R R R t i i || nt ni 4nt ni and T T|| T 2 nt ni For an air-glass interface (ni = 1 and nt = 1.5), R = 4% and T = 96% The values are the same, whichever direction the light travels, from air to glass or from glass to air.
Load more

Recommended publications
  • Lecture 26 – Propagation of Light Spring 2013 Semester Matthew Jones Midterm Exam
    Physics 42200 Waves & Oscillations Lecture 26 – Propagation of Light Spring 2013 Semester Matthew Jones Midterm Exam Almost all grades have been uploaded to http://chip.physics.purdue.edu/public/422/spring2013/ These grades have not been adjusted Exam questions and solutions are available on the Physics 42200 web page . Outline for the rest of the course • Polarization • Geometric Optics • Interference • Diffraction • Review Polarization by Partial Reflection • Continuity conditions for Maxwell’s Equations at the boundary between two materials • Different conditions for the components of or parallel or perpendicular to the surface. Polarization by Partial Reflection • Continuity of electric and magnetic fields were different depending on their orientation: – Perpendicular to surface = = – Parallel to surface = = perpendicular to − cos + cos − cos = cos + cos cos = • Solve for /: − = !" + !" • Solve for /: !" = !" + !" perpendicular to cos − cos cos = cos + cos cos = • Solve for /: − = !" + !" • Solve for /: !" = !" + !" Fresnel’s Equations • In most dielectric media, = and therefore # sin = = = = # sin • After some trigonometry… sin − tan − = − = sin + tan + ) , /, /01 2 ) 45/ 2 /01 2 * = - . + * = + * )+ /01 2+32* )+ /01 2+32* 45/ 2+62* For perpendicular and parallel to plane of incidence. Application of Fresnel’s Equations • Unpolarized light in air ( # = 1) is incident
    [Show full text]
  • The Fresnel Equations and Brewster's Law
    The Fresnel Equations and Brewster's Law Equipment Optical bench pivot, two 1 meter optical benches, green laser at 543.5 nm, 2 10cm diameter polarizers, rectangular polarizer, LX-02 photo-detector in optical mount, thick acrylic block, thick glass block, Phillips multimeter, laser mount, sunglasses. Purpose To investigate polarization by reflection. To understand and verify the Fresnel equations. To explore Brewster’s Law and find Brewster’s angle experimentally. To use Brewster’s law to find Brewster’s angle. To gain experience working with optical equipment. Theory Light is an electromagnetic wave, of which fundamental characteristics can be described in terms of the electric field intensity. For light traveling along the z-axis, this can be written as r r i(kz−ωt) E = E0e (1) r where E0 is a constant complex vector, and k and ω are the wave number and frequency respectively, with k = 2π / λ , (2) λ being the wavelength. The purpose of this lab is to explore the properties the electric field in (1) at the interface between two media with indices of refraction ni and nt . In general, there will be an incident, reflected and transmitted wave (figure 1), which in certain cases reduce to incident and reflected or incident and transmitted only. Recall that the angles of the transmitted and reflected beams are described by the law of reflection and Snell’s law. This however tells us nothing about the amplitudes of the reflected and transmitted Figure 1 electric fields. These latter properties are defined by the Fresnel equations, which we review below.
    [Show full text]
  • Electromagnetic Waves
    Electromagnetic Waves Lecture34: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Electromagnetic wave at the interface between two dielectric media We have so far discussed the propagation of electromagnetic wave in an isotropic, homogeneous, dielectric medium, such as in air or vacuum. In this lecture, we woulddiscuss what happens when a plane electromagnetic wave is incident at the interface between two dielectric media. For being specific, we will take one of the medium to be air or vacuum and the other to be a dielectric such as glass. We have come across this in school in connection with the reflection and transmission of light waves at such an interface. In this lecture, we would investigate this problem from the point of view of electromagnetic theory. Let us choose the interface to be the xy plane (z=0). The angles of incidence, reflection and refraction are the angles made by the respective propagation vectors with the common normal at the interface. 1 We have indicated the propation vectors in the apprpriate medium by capital letters I, R and T so as not to confuse with the notation for the position vector and time t. The principle that we use to establish the laws of reflection⃗ and refraction is the continuity of the tangential components of the electric field at the interface, as discussed extensively during the course of these lectures. Let us represent the component of the electric field parallel to the interface by the superscript . We then have, ∥ 0 exp + 0 exp = 0 exp ∥ ∥ This equation� ���⃗ ⋅ must⃗ − remain� valid at� all����⃗ points⋅ ⃗ − in the� interface � and ����⃗ at⋅ ⃗all− times.� That is obviously possible if the exponential factors is the same for all the three terms or if they differe at best by a constant phase factor.
    [Show full text]
  • Experiments for B. Tech. 1St Year Physics Laboratory Department Of
    Experiments for B. Tech. 1st Year Physics Laboratory Experiment No. 12 Brewster’s Angle Aim : To determine the of Brewster’s angle for glass using a polarized monochromatic light source. Apparatus Required: He laser, dial fitted polarizer, photo detector, micro ammeter, rotational mount, glass plate, constant power supply . Formula Used : Brewsters angle for a given pair of medium is 푛 θB = 휃 = 푡푎푛 ( ) 푛 Where n1 and n2 are refractive index of medium 1 and 2 respectively Principle: When light moves between two media of differing refractive index (n), some of the light is reflected from the surface of the denser material. This reflected ray’s intensity changes with change in the incident angle (휃) at the interface of two mediums. At one specific angle of incidence of light only perpendicular vibrations of electric field vectors are reflected whereas parallel vibrations are restricted or polarized. This loss in light intensity is due to polarization by reflection and the angle of incidence for which reflected ray is polarized is called the Brewster's angle θ (also known as the Polarization angle). B This phenomenon of polarization by reflection is illustrated in the figure below. Figure1. Polarization by reflection and Brewster's angle (θ ) B Department of Physics, IIT Roorkee © 1 Experiments for B. Tech. 1st Year Physics Laboratory The fraction of the incident light that is reflected depends on both the angle of incidence and the polarization direction of the incident light. The functions that describe the reflection of light polarized parallel and perpendicular to the plane of incidence are called the Fresnel Equations.
    [Show full text]
  • A Study of Reflection and Transmission of Birefringent Retarders
    JUST, Vol. IV, No. 1, 2016 Trent University A study of reflection and transmission of birefringent retarders James Godfrey Keywords Optics — Computational Physics — Theoretical Physics Champlain College 1. Interference in Rotating Waveplates The transmitted intensity of along each axis must be con- sidered separately, since equation (1-1) dictates that the re- 1.1 Research Goal flectance of the fast- and slow-axis components of the incident The objective of this project was to obtain a realistic theoreti- light will be different. The reflectance along the fast axis (Ro) cal prediction of the transmission curve for linearly-polarized and slow axis (Rs) are may be expressed: light incident normal on a [birefringent] retarder that behaves 2 as a Fabry-Perot etalon. The birefringent retarder considered n f − ni is a slab of birefringent crystal cut with the optic axis in the Ro = n f + ni face of the slab, and with parallel faces, and the waveplate has 2 no coating of any kind. ns − ni Rs = (1-2) Results of this project hoped to possibly explain the results ns + ni of work done by a previous student, Nolan Woodley, for his Physics project course in the 2013/14 academic year. In his Assuming the sides of the etalon are essentially paral- project, Woodley took many polarimeter scans which had a lel, the retarder may now be treated as a low-finesse (low- roughly sinusoidal shape [as they should], but adjacent peaks reflectance, R << 1) Fabry-Perot etalon. Their coefficients of of different heights. These differing heights were completely finesse (Fo and Fs, respectively) will also be different: unexplained, and not predicted by theory.
    [Show full text]
  • Polarization Lecture Outline
    1/31/20 Polarization Lecture outline • Why is polarization important? • Classification of polarization • Four ways to polarize EM waves • Polarization in active remote sensing systems 1 1/31/20 Definitions • Polarization is the phenomenon in which waves of light or other radiation are restricted in direction of vibration • Polarization is a property of waves that describes the orientation of their oscillations Coherence and incoherence • Coherent radiation originates from a single oscillator, or a group of oscillators in perfect synchronization (phase- locked, or with a constant phase offset) • e.g., microwave ovens, radars, lasers, radio towers (i.e., artificial sources) • Incoherent radiation originates from independent oscillators that are not phase-locked. Natural radiation is incoherent. 2 1/31/20 Why is polarization important? • Bohren (2006): ‘the only reason the polarization state of light is worth contemplating is that two beams, otherwise identical, may interact differently with matter if their polarization states are different.’ • Interference only occurs when EM waves have a similar frequency, constant phase offset and the same polarization (i.e., they are coherent) • Used by active remote sensing systems (radar, lidar) Classification of polarization • Linear • Circular • Elliptical • ‘Random’ or unpolarized 3 1/31/20 Linear polarization • Plane EM wave – linearly polarized • Trace of electric field vector is linear • Also called plane-polarized light • Convention is to refer to the electric field vector • Weather radars usually
    [Show full text]
  • Fresnel's Equations
    Prof. Raghuveer Parthasarathy University of Oregon Physics 352 – Winter 2008 5 R EFLECTION AND T RANSMISSION (FRESNEL’ S E QUATIONS) The law of reflection (θri= θ , where r and i refer to reflected and incident rays – see Problem Set 3) and Snell’s Law ( nniittsinθ = sinθ , where t refers to the transmitted ray) give the directions of reflected and transmitted rays at boundaries (Figure 5.1). What are the amplitudes of the electromagnetic waves? In other words, how much light is reflected and transmitted? G Figure 5.1. Reflection and refraction at an interface. The incident wave (wavevector ki ) is G G reflected (wavevector kr ) and transmitted (wavevector kt ). What are the amplitudes of the reflected and transmitted waves? Similar questions arise when considering other sorts of waves hitting boundaries – for example waves on the 1D strings we examined last quarter, incident at a boundary between two media with different speeds. In all these situations, transmission and reflection are analyzed by considering the boundary conditions imposed by the junction. As we’ll see, our analysis of electromagnetic waves will reduce in the appropriate limit to that of simple strings. To consider the general case of a plane electromagnetic wave hitting a surface at some angle θi (with respect to the normal), we’ll have to separately consider the components with electric field perpendicular and parallel to the plane of incidence. (The incident, reflected, and transmitted rays all lie in the plane of incidence, POI, which also includes the normal to the surface.) First, some statements that follow from a study of electromagnetism.
    [Show full text]
  • 13. Fresnel's Equations for Reflection and Transmission
    13. Fresnel's Equations for Reflection and Transmission Incident, transmitted, and reflected beams Boundary conditions: tangential fields are continuous Reflection and transmission coefficients The "Fresnel Equations" Brewster's Angle Total internal reflection Power reflectance and transmittance Augustin Fresnel 1788-1827 Posing the problem What happens when light, propagating in a uniform medium, encounters a smooth interface which is the boundary of another medium (with a different refractive index)? k-vector of the incident light nincident boundary First we need to define some ntransmitted terminology. Definitions: Plane of Incidence and plane of the interface Plane of incidence (in this illustration, the yz plane) is the y plane that contains the incident x and reflected k-vectors. z Plane of the interface (y=0, the xz plane) is the plane that defines the interface between the two materials Definitions: “S” and “P” polarizations A key question: which way is the E-field pointing? There are two distinct possibilities. 1. “S” polarization is the perpendicular polarization, and it sticks up out of the plane of incidence I R y Here, the plane of incidence (z=0) is the x plane of the diagram. z The plane of the interface (y=0) T is perpendicular to this page. 2. “P” polarization is the parallel polarization, and it lies parallel to the plane of incidence. Definitions: “S” and “P” polarizations Note that this is a different use of the word “polarization” from the way we’ve used it earlier in this class. reflecting medium reflected light The amount of reflected (and transmitted) light is different for the two different incident polarizations.
    [Show full text]
  • Lecture Notes - Optics 3: Double Refraction, Polarized Light
    Lecture Notes - Optics 3: Double Refraction, Polarized Light • Experiment: observations with optical calcite. • Light passing through a calcite crystal is split into two rays. This process, first reported by Erasmus Bartholinus in 1669, is called double refraction. The two rays of light are each plane polarized by the calcite such that the planes of polarization are mutually perpendicular. For normal incidence (a Snell’s law angle of 0°), the two planes of polarization are also perpendicu- lar to the plane of incidence. • For normal incidence (a 0° angle of incidence), Snell’s law predicts that the angle of refraction will be 0°. In the O E case of double refraction of a normally incident ray of light, at least one of the two rays must violate Snell’s Law as we know it. For calcite, one of the two rays does indeed obey Snell’s Law; this ray is called the ordinary ray (or O-ray). The other ray (and any ray that does not obey Snell’s Law) is an extraordinary ray (or E-ray). • For ordinary rays the vibration direction, indicated by the electric vectors in our illustrations, is perpendicular to the ray path. For extraordinary rays, the vibration direction is not perpendicular to the ray path. The direction perpendicular to the vibration direction is called the wave normal. Although Snell’s Law is not satisfied by the ray path for extraordinary rays, it is satisfied by the wave normals of extraordinary rays. In other words, the wave normal direction for the refracted ray is related to the wave normal direction for the incident ray by Snell’s Law.
    [Show full text]
  • CHAPTER 8 INTERACTION of LIGHT and MATTER Point in the Direction of Propagation
    Chapter 8 Interaction of Light and Matter 8.1 Electromagnetic Waves at an Interface A beam of light (implicitly a plane wave) in vacuum or in an isotropic medium propa- gates in the particular fixed direction specified by its Poynting vector until it encoun- ters the interface with a different medium. The light causes the charges (electrons, atoms, or molecules) in the medium to oscillate and thus emit additional light waves thatcantravelinanydirection(overthesphereof4π steradians of solid angle). The oscillating particles vibrate at the frequency of the incident light and re-emit energy as light of that frequency (this is the mechanism of light “scattering”). If the emit- ted light is “out of phase” with the incident light (phase difference ∼= π radians), then the two waves interfere destructively and the original beam is attenuated.± If the attenuation is nearly complete, the incident light is said to be “absorbed.” Scattered light may interfere constructively with the incident light in certain directions, forming beamsthathavebeenreflected and/or transmitted. The constructive interference of the transmitted beam occurs at the angle that satisfies Snell’s law; while that after reflection occurs for θreflected = θincident. The mathematics are based on Maxwell’s equations for the three waves and the continuity conditions that must be satisfied at the boundary. The equations for these three electromagnetic waves are not diffi- cult to derive, though the process is somewhat tedious. The equations determine the properties of light on either side of the interface and lead to the phenomena of: 1. Equal angles of incidence and reflection; 2. Snell’s Law that relates the incident and refracted wave; 3.
    [Show full text]
  • Geometrical Optics
    Geometrical Optics http://physics.tamuk.edu/~suson/html/1402/optics.html Geometrical Optics We have seen that light is an electromagnetic wave. We have also seen that waves can be reflected and refracted by various surfaces placed in the path of the wave. How can we apply this to light? To a high degree of accuracy, we can imagine that light waves move in a straight line, provided that we do not place any obstacle or aperture in the path of the wave that has the same magnitude of dimensions as the wavelength of the wave. If we also disregard the effect of edges on the light wave, we have the special case known as geometrical optics. In this case, we view the light wave as a ray which is always normal to the surface of the wave front. Consider a light ray that is incident on a plane glass surface. We see that the beam is partially reflected and partially transmitted. The transmitted beam is bent both upon entering the surface and upon exiting it. Let q1 be the angle the incident beam makes with the normal to the surface of the glass. Also, let q1' be the angle the reflected beam makes with the normal, q2 be the angle of the beam in the glass surface and q3 be the angle of the beam that exits the glass surface. By experiment, we find that the reflected and refracted rays lie in the plane formed by the incident ray and the normal to the surface at the point of incidence.
    [Show full text]
  • Sec. 16. Reflection and Refraction of Light at the Interface Between Two Dielectrics
    16. Reflection and Refraction of Light. Fresnel's Formulas 121 When light is reflected from a surface (see Sees. 16 and 20), the most strongly reflected regions of the spectrum are those which are absorbed most strongly upon passing through the bulk of the material. Hence the colour of a substance due to selective reflection is in addition to the colour of the same substance due to selective absorption. Example 15.1. Express the group velocity (15.40) as a function of the refractive index and wavelength. Find the group velocity in water for A. = 656.3 nm if the refractive index of water at 20 °C is 1.3311 for this wavelength, and 1.3314 for A.= 643.8 nm. Since OJ = 2rrc/(n'A.), k = 2rr/A., we obtain from (15.40) 2 2 v = d [2nc/(n'A.)] = c (nd'A. + A.dn)/(n A. ) = + dn). 2 1 g d (2n/A.) . d'A./A. n n d'A. (15.44) Substituting numerical values into this equation, we obtain 8 3 x 10 ( 656.3 0.0003) vg = 1 + ---- m/s = 2.28 x 108 m/s. 1.3311 1.3311 12.5 SEC. 16. REFLECTION AND REFRACTION OF LIGHT AT THE INTERFACE BETWEEN TWO DIELECTRICS. FRESNEL'S FORMULAS It is shown that the boundary BOUNDARY CONDITIONS. Dielectrics have different conditions for- the electric and values of permittivity. The behaviour of a wave at a boundary magnetic field vectors of waves is completely determined by the boundary conditions for the completely determine the laws field vectors of the waves.
    [Show full text]