Optics and Waves II PDF

Optics and Waves II Part of module: PYU22P20 David McCloskey, Ussher Assistant Professor for the Science of Energy and Energy Systems SNIAM 3.03C [email protected] 1...

Optics and Waves II Part of module: PYU22P20 David McCloskey, Ussher Assistant Professor for the Science of Energy and Energy Systems SNIAM 3.03C [email protected] 1  Course Objectives  To make the connection between Electricity, Magnetism and Light  To provide mathematical tools required to describe light as a wave in a vector field.  To explore vectorial properties of light such as polarisation, interference and diffraction  Apply this knowledge to understand applications of electromagnetic waves in modern technology and every day life. 2 In this course we are particularly interested in electromagnetic waves and their properties: 3  Course Structure Contact Hours (16 total) Course reading: 14 lectures TEXTBOOKS: Optics by Hecht, Addison Wesley, 1 revision lecture, 1 test University Physics Ch. 32,33,35,36 2 small group tutorials. OPTIONAL: Div Grad Curl and all that, H.M.Schey Thomas' Calculus: Early Transcendental, 12th Edition, George B. Thomas. Related Labs: JF Lab Exp 14: Interference and diffraction. Online content: SF Lab: The Michelson Interferometer, Polarized Light, Fourier Analysis Mastering Physics, Blackboard. Tutorials Assessment: Two 1 hour tutorials. Final lecture is CA: Mastering Physics problems tutorial on mastering End of year exam 2 questions, paper II Physics problems 4  Course in 4 parts I. Electromagnetism and Scalar Waves Maxwell’s Equations Sources of light Div History Grad Wave equation Phasors Coherence Curl Vector Fields II. Properties of Light 1:Polarisation Linear Degree of polarisation Optical Activity Jones Calculus Circular Phase retarders Polarisers Stokes Parameters Elliptical III. Properties of Light 2: Interference Superposition Coherence Length Interferometers Fabry Perot Fringes Visibility Thin films Localised Multiple beam Non-localised IV. Properties of Light 3: Diffraction Fresnel near-field, resolution Apertures Fraunhoffer Far-Field Diffraction Limit Image formation Spatial Filtering 5 Lect. Content Topic 1 Introduction : History, Maxwell’s equations, differential form I 2 Differential form: Maxell’s equations to wave equation Solutions: plane waves, Energy and I Power 3 Scalar waves : Equation, Phasor representation I 4 Sources of light: Natural, man made, coherent, incoherent. I Current theories that describe light and interaction with matter  Size scale much larger than wavelength. Geometrical optics  Polarisation not important Physical optics,  Size scale comparable to wavelength (branch of Classical  Polarisation important Electromagnetism)  Interference and diffraction effects Quantum optics  Certain problems require Semi-classical approach, e.g. lasing, blackbody radiation Quantum  Both Light and matter need to be Electrodynamics quantised (QED) Physical optics deals with the wave nature of light. 7  An Extremely Brief History of Theories Describing Light 300BC: Euclid recorded the law of reflection. Described light as stream of particles emitted from the eyes. ~1000 AD: Empirical study of refection 1621 Willebrord Snell : Law of refraction 1660 Francesco Maria Grimaldi: Coined term diffraction, showed then light spreads out when passing through an aperture. Danish Astronomer, Ole Römer shows light has a finite velocity from astronomical observations 1672-1704: Isaac Newton argues that light consisted of a stream of particles traveling in straight lines from a source. 1801 Thomas Young showed that light could undergo interference. This definitively showed that light was somehow a wave. 1860: James Clerk Maxwell saw a link between Electricity and Magnetism and thought it might have something to do with light. >1905: Einstein+Plank noticed that light interacted with matter in discrete quanta of packets of energy. 8  Todays Objectives  Understand course structure and content  Review of Maxwell’s equations in integral form  Maxwell-Ampere correction  Derive the speed of an electromagnetic disturbance. 9  Scalar and vector fields A scalar field has only a magnitude at each point and space, and can vary with time. We represent it as : ( x, t ) A vector field has a magnitude and direction at each point in space, which can vary with time. We represent it as : ( x, t ) x  ( x, y, z )  xiˆ  yjˆ  zkˆ Temperature is an example of a scalar field Wind velocity is an example of a vector field. T ( x, t ) v ( x, t ) 10 Electric and magnetic fields From electrostatics and magnetostatics, we know that charge particles and currents (moving charges) interact with each other by exerting a force. We found it convent to introduce the concept of electric and magnetic fields which tell us the magnitude and direction of this force at each point in space through the Lorentz force law F  q( E  v  B) The electric and magnetic fields are related to each other and must obey strict rules. We often consider them as a combined electromagnetic vector field. In this combined field each point in space has a vector for the electric field and a vector for the magnetic field. ( E , B)  ( E ( x , t ), B( x , t )) 11 Light is wave….. A wave in what? (Ch.33 UP, Ch. 3 Hecht) It was known that light had wave properties (interference and diffraction) from Young’s experiment in 1801 The connection between optics and electricity and magnetism had not yet been made. Enter James Clerk Maxwell @ 1865: Electricity and magnetism were popular areas of study in Physics in the 1800’s. It was known that an electric field could produce a spark of light. What if light was a both a wave and something to do electricity and magnetism? Maxwell noted that if he added a correction term to Amperes Law, all the experimental observations of the time could be explained using just four laws 1. Gauss’s Law 2. The fact that magnetic charges (monopoles) don’t exist 3. Faradays Law 4. Maxwell-Amperes Law Written in integral form these laws are: 12  The four laws 𝑄𝑒𝑛𝑐𝑙 Eq.1 ඾ 𝐸 ⋅ 𝑑 𝐴Ԧ = Gauss’s Law 𝜀0 𝛿𝑉 Eq.2 ඾ 𝐵 ⋅ 𝑑 𝐴Ԧ = 0 No magnetic monopoles 𝛿𝑉 𝑑Φ𝐵 Eq.3 ර 𝐸 ⋅ 𝑑𝑙Ԧ = − Faraday’s Law 𝑑𝑡 𝛿𝐴 Eq.4 ර 𝐵 ⋅ 𝑑 𝑙Ԧ = 𝜇0 𝐼𝑒𝑛𝑐𝑙 Ampere’s Law Not complete 𝛿𝐴 Is the magnetic flux 𝐼𝑒𝑛𝑐𝑙 = ඵ 𝐽.Ԧ 𝑑𝐴Ԧ Φ𝐵 = ඵ 𝐵. 𝑑𝐴Ԧ 𝐴 𝐴 Φ𝐸 = ඵ 𝐸. 𝑑𝐴Ԧ Is the electric flux 𝐴 These laws place restrictions on the allowed values of electric and magnetic fields and are known as governing equations. 13  Maxwell’s displacement current Consider application of Amperes Law to a charging capacitor using two equivalent cases: ර 𝐵 ⋅ 𝑑 𝑙Ԧ = 𝜇0 𝐼𝑒𝑛𝑐𝑙 𝛿𝐴 Case B Case A dl I encl In Case A In Case B ර 𝐵 ⋅ 𝑑 𝑙Ԧ = 𝜇0 𝐼𝑒𝑛𝑐𝑙 ර 𝐵 ⋅ 𝑑 𝑙Ԧ = 0 𝛿𝐴 𝛿𝐴 This is a contradiction! 14  This is a worrying inconsistency, how can we fix this? The key is to note that this is not steady state, so we will have time varying electric and magnetic fields while charging the capacitor. Consider the instantaneous charge on the capacitor: q (t )  Cv(t ) For a parallel plate capacitor 0 A C v(t )  E (t )d d Where A is the area of the plates and d the distance between them The instantaneous charge on the plate is therefore: q (t )   0 EA   0  E (t ) 15  The rate of change of charge building up on the plate is equal to the current: dE i (t )   0 dt So we see that a charging capacitor has a time varying electric flux which can act the same as a current To generalise Amperes Law to time varying fields we must therefore take account of this current ර 𝐵 ⋅ 𝑑 𝑙Ԧ = 𝜇0 𝐼𝑒𝑛𝑐𝑙 + 𝐼𝐷 𝛿𝐴 𝑑Φ𝐸 ර 𝐵 ⋅ 𝑑 𝑙Ԧ = 𝜇0 𝐼𝑒𝑛𝑐𝑙 + 𝜀0 𝑑𝑡 𝛿𝐴 16 This also explains the experimentally measured magnetic field between two plates of a capacitor upon charging During charging a magnetic field is generated exactly as if there was a current there! 17  The four laws: Maxwell’s equations in vacuum. 𝑄𝑒𝑛𝑐𝑙 Eq.1 ඾ 𝐸 ⋅ 𝑑 𝐴Ԧ = Gauss’s Law 𝜀0 𝛿𝑉 Eq.2 ඾ 𝐵 ⋅ 𝑑𝐴Ԧ = 0 No magnetic monopoles 𝛿𝑉 𝑑Φ𝐵 Eq.3 ර 𝐸 ⋅ 𝑑𝑙Ԧ = − Faraday’s Law 𝑑𝑡 𝛿𝐴 𝑑Φ𝐸 Eq.4 ර 𝐵 ⋅ 𝑑 𝑙Ԧ = 𝜇0 𝐼𝑒𝑛𝑐𝑙 + 𝜀0 Maxwell-Ampere’s Law 𝑑𝑡 𝛿𝐴 Φ𝐵 = ඵ 𝐵. 𝑑𝐴Ԧ Is the magnetic flux 𝐼𝑒𝑛𝑐𝑙 = ඵ 𝐽.Ԧ 𝑑𝐴Ԧ 𝐴 𝐴 Φ𝐸 = ඵ 𝐸. 𝑑𝐴Ԧ Is the electric flux 𝐴 These laws place restrictions on the allowed values of electric and magnetic fields and are known as governing equations. 18 Graphical Summary Gauss’ Law No magnetic monopoles 𝑄𝑒𝑛𝑐𝑙 ඾ 𝐸 ⋅ 𝑑 𝐴Ԧ = ඾ 𝐵 ⋅ 𝑑𝐴Ԧ = 0 𝜀0 𝛿𝑉 𝛿𝑉 dA dA E B + dA  dAnˆ 19  Faraday’s Law Maxwell-Ampere’s Law 𝑑Φ𝐵 𝑑Φ𝐸 ර 𝐸 ⋅ 𝑑𝑙Ԧ = − ර 𝐵 ⋅ 𝑑 𝑙Ԧ = 𝜇0 𝐼𝑒𝑛𝑐𝑙 + 𝜀0 𝑑𝑡 𝑑𝑡 𝛿𝐴 𝛿𝐴 Φ𝐸 = ඵ 𝐸. 𝑑𝐴Ԧ Φ𝐵 = ඵ 𝐵. 𝑑𝐴Ԧ 𝐴 𝐴 𝐼𝑒𝑛𝑐𝑙 = ඵ 𝐽.Ԧ 𝑑𝐴Ԧ 𝐴 dl dl B I encl 20  The speed of electromagnetic disturbances (For more detail see chapter 32 University Physics, p1218 ) Concept of a wavefront: Consider a plane parallel to yz moving in the +x direction (red) with velocity c. At any time, all points left of the plane experience E and B fields as shown. Any point to the right of the plane has zero electric and magnetic field. In time dt, the wavefront moves a distance cdt and sweeps out part of the blue rectangle. (as dt is small, E, B are constant over dt) In this time the Magnetic flux  B  B  A through the blue rectangle rises by d  B  BdA  B.a.c.dt 21 Dividing across by dt: dB  B.a.c dt 𝑑Φ𝐵 But Faradays law states: ර 𝐸 ⋅ 𝑑𝑙Ԧ = − 𝑑𝑡 𝐿 Taking ර 𝐸 ⋅ 𝑑𝑙Ԧ anti-clockwise (Right hand rule) around the perimeter of the blue 𝐿 square gives ර 𝐸 ⋅ 𝑑𝑙Ԧ = −𝐸𝑎 𝐿 (Remember E is zero to right of wavefront) E  cB ① So Ea  B.a.c Therefore: c is still an unknown velocity 22 Now to find c , we consider the same system but the blue rectangle is now in the xz plane. 𝑑Φ𝐸 Ampere’s Law says: ර 𝐵 ⋅ 𝑑 𝑙Ԧ = 𝜇0 𝐼𝑒𝑛𝑐𝑙 + 𝜀0 𝑑𝑡 𝐿 But if this is in free space, there are no currents and so I encl  0 23  𝑑Φ𝐸 Therefore ර 𝐵 ⋅ 𝑑 𝑙Ԧ = 𝜇0 𝜀0 𝑑𝑡 𝐿 In a time dt the wavefront propagates a distance c.dt. The electric flux through the blue rectangle during this time has increased by d  E  E.dA  E.b.c.dt dE Therefore  E.dA  E.b.c dt Taking ර 𝐵 ⋅ 𝑑 𝑙Ԧ anticlockwise around the perimeter of the blue square gives: 𝐿 ර 𝐵 ⋅ 𝑑 𝑙Ԧ = 𝐵. 𝑏 (B is zero to the right of the wavefront.) 𝐿 𝑑Φ𝐸 But since ර 𝐵 ⋅ 𝑑 𝑙Ԧ = 𝜇0 𝜀0  Bb  0 0 Ebc 𝑑𝑡 𝐿 B  0 0 Ec ② 24  We now have two equations relating E and B: 1 ① E  cB and ② E B 0 0c These can only both be true if : 1 1 c 1 c  2 c 0 0c 0 0 0 0 But 0 and 0 are constants which we can determine from simple experiments. So now we can find the speed of these electromagnetic waves as 0  4  107 Tm / A  0  8.85 1012 F / m c  299792458m / s c  3  108 m / s (0.07% off) 25 Example questions: An imaginary cubical surface of side L is in a region of uniform electric field E. Find the electric flux through each face of the cube and the total flux when (a) It is orientated with two faces perpendicular to R , and (b) when the cube is turned by an angle θ about a vertical axis. 26 Example derivation question: Q. Show using Maxwell Amperes Law and Faradays law that if a disturbance exists in the electromagnetic field it propagates at the speed of light A. Sketch the diagram and state your assumptions. Go through the derivation in the previous section. 1 Conclude that c 0 0 Note that this is equals to 3x108m/s , the speed of light 27  Today you have learned  The course structure and content  Revised Maxwell’s equations in integral form  Derived the Maxwell-Ampere correction  Derived the speed of an electromagnetic disturbance. 28 Section 2:The Differential Form of Maxwell’s Equations (The one with the maths in it) Derivation1 (If you wish to understand more, try reading :Div Grad Curl and all that, H.M.Schey, Basics also covered in Thomas’ Calculus. ) As you saw in PY2P10 Electronics, the integral form of Maxwell's equations are particularly useful for solving problems with certain symmetries, by performing surface or line integrals around imaginary surfaces or paths. There are however cases where you may not know the field over the entire surface or path, and a more local description would be useful. Consider Gauss’ law, one way to get a local description would be to take the limit as a spherical volume tends to zero around a point. 𝑄𝑒𝑛𝑐𝑙 lim ඾ 𝐸 ⋅ 𝑑𝐴Ԧ = lim =0 𝑉→0 𝑉→0 𝜀0 𝛿𝑉 Of course this goes to zero! 29 But if we re-write in terms of a charge density ρ, then 𝜌𝑉 ඾ 𝐸 ⋅ 𝑑 𝐴Ԧ = 𝜀0 𝛿𝑉 1 𝜌 Then this limit answer is finite, Taking the limit lim ඾ 𝐸 ⋅ 𝑑𝐴Ԧ = It is non-zero only in places where 𝑉→0 𝑉 𝜀0 𝛿𝑉 the charge density is non-zero. 1 We define divergence as 𝑑𝑖𝑣(𝐸) ≐ lim ඾ 𝐸 ⋅ 𝑑𝐴Ԧ 𝑉→0 𝑉 𝛿𝑉 div() is an operator which means apply the above limit to the vector field. The divergence of a vector field is a scaler field, which has a single positive or negative value at each point in space. 30  This doesn’t seem any more useful than the integral form! Lets look closer at this limit: S1 z Consider the specific case of a small cube in (x,y,z) z 3D Cartesian coordinates with volume x V= ∆𝑥∆𝑦∆𝑧 centred at (x,y,z). y S2 y x Consider the limit as Δz tends to zero. We are left with two surfaces S1 and S2 1 lim ඾ 𝑣Ԧ ⋅ 𝑑𝐴Ԧ = Δ𝑧→0 Δ𝑉 𝑆1 +𝑆2 Δ𝑧 Δ𝑧 𝑣𝑧 (𝑥, 𝑦, 𝑧 + ) − 𝑣𝑧 (𝑥, 𝑦, 𝑧 − ) = lim 2 2 Δ𝑥Δ𝑦 Δ𝑧→0 Δ𝑥Δ𝑦Δ𝑧 𝜕𝑣𝑧 = 𝜕𝑧 31 A similar procedure on each face gives: 1 𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑧 𝑑𝑖𝑣(𝑣) Ԧ ≐ lim Ԧ ඾ 𝑣Ԧ ⋅ 𝑑𝐴 = + + Δ𝑉→0 Δ𝑉 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝛿𝑉 It is useful to define the del operator as  ˆ  ˆ  ˆ       i j k  , ,  x y z  x y z  We can then abuse vector notation to help remember the form of div.  vx         vx v y vz div(v )   v   , ,  v y       x y z    x y z  vz  32 Now we have the differential form of Eq.1 and Eq.2 written in an easy to use format. 𝜕𝐸𝑥 𝜕𝐸𝑦 𝜕𝐸𝑧 𝜌 ∇∙𝐸 = + + = 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜀0 Similarly 𝜕𝐵𝑥 𝜕𝐵𝑦 𝜕𝐵𝑧 ∇∙𝐵 = + + =0 𝜕𝑥 𝜕𝑦 𝜕𝑧 Its physical interpretation of the divergence is a number which tells us if there is a source or sink at a particular point in space and how strong it is So Eq.1 tells us that the divergence of the electrical field is zero unless there is a charge there. Whereas Eq.2 tells us there are no sources or sinks of magnetic field. 33 Some examples of divergence. E  E0iˆ  E0 ˆj E  xiˆ  yjˆ E  0 E  2 34 E  x 2iˆ  y 2 ˆj   E  2x  2 y 35 What about equation 3 and 4? Consider the case of a time varying magnetic field perpendicular to a fixed area A: Lets start again by writing Faradays Law: dl Ԧ 𝜕(𝐵. 𝐴) 𝜕𝐵 ර 𝐸 ⋅ 𝑑𝑙Ԧ = − =− 𝐴 𝜕𝑡 𝜕𝑡 𝐿 B (t ) 1 𝜕𝐵 lim ර 𝐸 ⋅ 𝑑 𝑙Ԧ = − 𝐴→0 𝐴 𝜕𝑡 𝐿 We define this limit as the curl of a vector field: 1 𝑐𝑢𝑟𝑙(𝑣) Ԧ ≐ lim ර 𝑣Ԧ ⋅ 𝑑𝑙Ԧ 𝐴→0 𝐴 𝐿 It is also called circulation, its physical interpretation is a vector which tells you the direction and magnitude of any rotation in the field 36 Using a similar argument of a small volume as used to derive divergence we can show that: 1 𝜕𝑣𝑧 𝜕𝑣𝑦 𝜕𝑣𝑥 𝜕𝑣𝑧 𝜕𝑣𝑦 𝜕𝑣𝑦 𝑐𝑢𝑟𝑙(𝑣) Ԧ ≐ lim ර 𝑣Ԧ ⋅ 𝑑𝑙Ԧ = − , − , − 𝐴→0 𝐴 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝐿 The form of curl is a bit trickier to remember than divergence, but this is where the del notation helps, as we just need to perform a cross product  vx          vz v y vx vz v y v y  curl (v )    v   , ,   v y    ,  ,                  x y z y z z x x x   z v iˆ ˆj kˆ    To remember cross product v  x y z vx vy vz 37 Example of a vortex v   yiˆ   xjˆ r  r cos(t )iˆ  r sin(t ) ˆj v  r sin(t )iˆ  r cos(t ) ˆj v   yiˆ   xjˆ   v  2 k   v  2 k 38 Another example E  cos( y )iˆ  sin( x) ˆj   E  kˆ(cos( x)  sin( y )) E ( x, y )  E 39 Example of vector field with both nonzero curl and divergence:  E ( x, y )  cos( y )  sin( x) E  x cos( y )iˆ  y sin( x) ˆj   E ( x, y )  y cos( x)  x sin( y ) E ( x, y ) 40 If you wish to understand more, try reading : Div Grad Curl and all that, Now we have both an integral and local form of the equations Integral form Differential form 𝑄𝑒𝑛𝑐𝑙 𝜌 ඾ 𝐸 ⋅ 𝑑 𝐴Ԧ = ∇∙𝐸 = Gauss’s Law 𝜀0 𝜀0 𝛿𝑉 ∇∙𝐵 =0 No magnetic ඾ 𝐵 ⋅ 𝑑𝐴Ԧ = 0 monopoles 𝛿𝑉 𝑑Φ𝐵 B Ԧ ර 𝐸 ⋅ 𝑑𝑙 = −  E   Faraday’s Law 𝑑𝑡 t 𝐿  E  Maxwell- ර 𝐵 ⋅ 𝑑 𝑙Ԧ = 𝜇0 𝐼𝑒𝑛𝑐𝑙 + 𝜀0 𝑑Φ𝐸   B  0  J encl   0  Ampere’s Law 𝑑𝑡  t  𝐿 41 Maxwell’s equations in differential form in absence of sources. (i.e. in vacuum) ∇∙𝐸 =0 Gauss’s Law No magnetic monopoles ∇∙𝐵 =0 B  E   Faraday’s Law t E Maxwell-Ampere’s Law   B  0 0 t 42 For the sake of completeness although it doesn’t appear explicitly in Maxwell’s equations we will discuss another operation known as the gradient The gradient applies to a scalar field and is given by:  ˆ  ˆ  ˆ grad ( )  ( )  i j k x y z (Again we can abuse our del notation to help remember its form ) So the gradient operates on a scalar field and returns a vector field. The vectors always point in the direction of steepest assent. A useful related concept is the directional derivative Dnˆ ( )  ( ) nˆ This tells us the rate of change of a scalar field in the direction n. Some examples of Gradient.   ( x2  y 2 )     2 2 Take a Gaussian scalar field in 2d:  ( x, y )  e   ( x, y )   ( x2  y 2 )      ( x2  y 2 )     2  2 2 2   2 xe ˆi  2 ye ˆj ( )( x, y )   2 2 2 2  Vector Identities We will invoke (without proof) an important vector identity that either you have seen already (TP), or will learn soon in your maths course (EP, Nano) Vector identity: Ԧ = 𝛻 𝛻 ∙ 𝑣Ԧ − 𝛻 2 𝑣Ԧ ∇ × (𝛻 × 𝑣) We have introduced a new symbol known as the Laplacian  2       2 f  2 f  2 f Where 2 f  2  2  2 x y z So  2 v    2vx ,  2v y ,  2vz    2 v  2 v  2 v  2 v  2 v  2 v  2 v  2 v  2 v   2v   2x  2x  2x , 2y  2y  2y , 2z  2z  2z   x y z x y z x y z   𝜕𝜓 𝜕𝜓 𝜕𝜓 Note ∇ ∙ 𝑣Ԧ = 𝜓 is a scalar field so ∇ (∇ ∙ 𝑣) = ∇ 𝜓 = + + 𝜕𝑥 𝜕𝑦 𝜕𝑧 Where ( ) is known as the gradient of  45  Derivation 2: Back to Maxwell’s Equations Examining Faraday’s law in differential form gives the first clues about the nature of light Faraday=> Taking curl of both sides B  (  B )  E     (  E )   t t Gauss’ Law   (  E )  (  E )   2 E ∇∙𝐸 =0 0 Ampere’s Law  (  B ) E  E   2   B  0 0 t t 2E 0 0 2   2 E  0 t But this is just a wave equation for each component of E with velocity: 1 c  2.9979...  108 m / s 0 0 46  1 c  2.9979...  108 m / s 0 0 In the 1800’s  0 and 0 are known constants from electrostatic and magnetostatic measurements. The finite speed of light c was also known from astronomical measurements. This velocity is so nearly that of light , that it seems we have strong reason to conclude that light itself (including radiant heat and other radiations if any) is an electromagnetic disturbance in the form of waves propagated through the electromagnetic field according to electromagnetic laws 47 A similar approach starting with Maxwell-Ampere Law will give: 1 2B   2 B0 Try to get this YOURSELF! c t 2 2 Remember with the notation we really mean 6 equations on scaler fields: 1 2E 1 2B   2 E0   2 B0 c t 2 2 c t 2 2 1  2 Ex  2 Ex 1  2 Bx  2 Bx  2 0  2 0 c t 2 2 x c t 2 2 x 1  Ey  Ey 2 2 1  By  By 2 2  0  0 c t 2 2 y 2 c t 2 2 y 2 1  2 Ez  2 Ez 1  2 Bz  2 Bz  2 0  2 0 c t 2 2 z c t 2 2 z 48 So we see that a disturbance in the electromagnetic field has to obey some specific rules. 1. It propagates at the speed of light 2. The electric field is much larger than the magnetic. It can also be shown using Maxwells equations that: 3. 𝑬 and B vectors are always in phase, 4. They must be mutually perpendicular 5. Both are perpendicular to the direction of motion. LIGHT AS A WAVE A particularly useful solution is the harmonic wave which has electric and magnetic fields E ( x, t )  E0 zˆ Sin( kx  t   ) B ( x, t )  B0 yˆ Sin(kx  t   ) Magnetic component Direction of motion Electric component We investigate this form in more detail in the next lecture 49 Example Questions: Q2. Show the relation between the integral and differential form of Guass’ Law. A. Derivation 1 in notes Q2. Use the differential form of Maxwell’s equations in free space to show that the electromagnetic field can support disturbances that satisfy the wave equation. You can use the vector identity :   (  v )  (  v )   2v A. Derivation 2 in class. Q3. Find the divergence and the curl of the vector field E  E0 x 2 cos( y )iˆ  E0 x 2 y cos( y ) ˆj Q4. Explain Maxwell’s contribution to the development of the understanding of the theory of light. Section 3: Description of Scalar waves Recap: What is a Wave? Surface of water Strings Sound Earth quakes Matter waves Gravitational Waves What is the physical definition of a wave? A Wave is a disturbance moving through space and time carrying energy and momentum Some revision: Wave’s can be transverse or longitudinal. A wave on a string is a displacement of the string perpendicular to the direction of motion Here the “disturbance” is just a local vertical displacement of the string Lets call this displacement, .  is a function of position along the string but is also a function of time. 52  Consider a person sitting on a bus which is traveling at a velocity v. We can describe his position using two different frames of reference. Frame S has a fixed origin outside the bus, and frame S’ has a fixed origin inside the bus. A position x in the frame S is related to the same position at x’ in frame S’ by the Lorentz transformation: x  x  vt x  x  vt We can imagine a waveform as a moving bus. 53 If we take a general wavefunction  we can derive a PDE to describe wave motion. We know we are looking for a function  such that    ( x, t )   ( x)   ( x  vt ) x  x  vt By the chain rule for differentiation,   x x  but  v t x t t       x   Therefore  v  v     t x x  x x x x  Differentiating again with respect to t gives  2        v  v  v   t 2 tx xt x  t  54 Which gives: This PDE is known as the differential  2 2   2 v wave equation and is completely t 2 x 2 general.  This equation applies to any wave. Equivalently any function that is a solution to this equation represents a wave.  . is known as the wavefunction. People like to write this like: 1  2  2  2 0 v t 2 2 x 55 Lets have a look at a specific one dimensional wave. At t=0,  can be thought of as the shape of the wave. (x,0) = (x) = (x’). The wave on the left has the general shape given by  e ax '2 If the wave moves to the right with velocity v we can find its equation just by replacing x’ with x-vt such that  e a ( x vt ) 2 Lets check if this satisfies the wave equation! 56 Harmonic Waves Harmonic waves result from a periodic disturbance. The most important form is a sine or cosine as using Fourier transform any periodic signal can be described as a sum of sin and cosine functions Here kx’ is in radians, x’ is in meters, so k has units of rad/m To turn this into a traveling wave just replace x’ with x-vt 57 So a simple traveling wave can be described by  ( x, t )  A Sin[k ( x  vt )]  2 1  2 Lets check that this is a solution to the wave equation:  2 2 x 2 v t    kA cos(k ( x  vt ))   kvA cos(k ( x  vt )) x t  2  2   k 2 A sin(k ( x  vt ))   ( kv ) 2 A sin( k ( x  vt )) x 2 t 2  2 1  2   2 2 x 2 v t .  ( x, t )  A Sin(k ( x  vt ))  A Sin( )   k ( x  vt ) where  is known as the phase (angle) 58 What is the constant k? Periodic waves repeat themselves in space after one wavelength.This means that the “disturbance”,  is the same at any position x and the position x+.  ( x, t )   ( x   , t ) But sin waves repeat after phase advances by 2π so if we want the sin wave to repeat every λ, then we should chose k such that 2 k   2 so k  k is called the wavenumber 59  In addition periodic waves repeat themselves in time after one period. This means that the “disturbance”,  is the same at any time t and another time t+.  ( x, t )   ( x, t   ) With the same logic we need to set: 2 kvT  2 but k  1 v So T /v f   T  Remember  ( x, t )  A Sin[k ( x  vt )] 2 x v 2 x  ( x, t )  A Sin[kx  kvt )]  A Sin[  2 t )]  A Sin[  2 ft )]    Usually we write:  ( x, t )  A Sin[kx  t )] ω is known as the angular frequency   2 f 60 Phase We have been talking about waves of the form:  ( x, t )  A Sin[kx  t )] However at t=0, x=0 we get =0. This is a special case, the “disturbance” is not necessarily zero at the origin. In some cases there may be a phase shift. This means the whole wave is shifted in the x direction at t=0, x=0. Then (0,0)0    Sin(  ) 4  So a sine is a shifted cosine, hence Cos( )  Sin(  ) 2 So in general :  ( x, t )  A Cos(kx  t   ) 2    2 Doesn’t matter if you use sin or cos. So to describe a wave uniquely we need an amplitude A and a total phase 61 Energy, power and momentum in light Both Electric and Magnetic fields carry energy. This energy is distributed over some region of space so we can think in terms of an energy density, U (energy per unit volume, J/m3). The energy density associated with an electric field (ie as in a vacuum capacitor) is given by: 0 uE  E2 [J/m3] University Physics p.918 2 The energy density associated with a magnetic field (ie as in a solenoid in vacuum) is given by 1 2 uB  B [J/m3] University Physics p.1157 2 0 Thus the total energy density associated 0 1 2 u  uE  uB  E2  B with a light wave is 2 2 0 1 1 E  cB c u  0E2  B2 0 0 0 62 Electromagnetic waves are traveling waves which transport energy through space. This energy flow can be quantified by considering the amount of energy flowing per unit time across a unit area, S (W/m2). Imagine a wave moving at speed, c, through an area A. 63 In time dt, the wave travels cdt, therefore only the energy in the cylindrical volume will flow through A. Energy u.c.dt. A S   u.c Time. Area dt. A 1 but u B2 and E  cB 0 1 Therefore S EB Is the magnitude of power flow per unit area 0 1 We can define a useful vector S E  B   0c 2 E  B 0 The magnitude of this vector is the power flow per unit area and it points in the direction of power flow It is known as the Poynting Vector, after British Physicist John Poynting. 64 For a harmonic monochromatic wave travelling in the x direction: E ( x, t )  E0 zˆ Sin(kx  t   ) B ( x, t )  B0 yˆ Sin(kx  t   ) Apply S   0 c 2 E  B (E and B mutually perpendicular and perpendicular to the direction of motion). If we assume that the energy flows in the direction of propagation of the wave, then S   0c 2 E0 B0 Sin 2 (kx  t   )iˆ This flow of energy fluctuates very rapidly in time. For visible light the frequency is in the region of 1014 Hz. This is too rapid to observe, what we see is an average over a long time. The average value of S  S is given by: S   0c 2 E0 B0 Sin 2 (kx  t   ) The average of Sin2 over many cycles is 0.5. 65  1 Since E  cB  S   0cE02 xˆ 2 The magnitude of the time averaged Poynting vector S is called the Irradiance I 1 I   0cE02 where I is a scalar 2 Also sometimes called Intensity, its units are W/m2 This is a very important property of light and it is what we perceive as brightness 66  Momentum of Light Since light transports energy it can also carry momentum. It can be shown that the magnitude of the momentum flow per unit volume is: dp S EB  2  dV c 0 c 2 Since the light covers a distance c.dt in time dt we can show the flow rate of electromagnetic momentum is given by : 1 dp S EB S Pr ad     A dt c 0 c c This is a force per unit area known as radiation pressure. The force is in the direction of light propagation. S A IA P0 The average force is given by: Frad  Pr ad A    c c c 2S For a totally reflective surface the momentum change is double so: Pr ad  c Remember for light reflecting off a surface this force will be doubled. 67 Devices utilizing radiation pressure Crookes Radiometer Solar sails Optical tweezers https://youtu.be/ju6wENPtXu8 68 Example: Researchers in the UK have proposed of putting extremely high power fiber lasers on satellites in order to deal with the problem of space debris. If the average output power of the laser is 10 kW predict the average recoil force on the satellite Frad. What velocity will a medium sized satellite (500kg ) be travelling after a 30 minute laser burst?. If the satellite needs to stay within 10m of its current position for stable orbit, how long from the start of the burst until the satellite position needs to be corrected? 69  FYI The greatest risk is to satellites in Low Earth Orbit, where there are around 20,000 tracked objects (including 600 operating satellites) &— around almost 2 million kg (4 million lbs) of space debris — everything from small screws and bolts to big chunks of spacecraft. All of this has the potential to hit and severely damage, or even destroy, a satellite. From 2010–2014 EUMETSAT had around 100 warnings of possible conjunctions (collisions) with debris objects and had to undertake five Collision Avoidance Manoeuvres to move Metop-A (three times) and Metop-B (twice) away from the trajectory of debris. Analysis of the trajectory of both the satellite and the debris showed that the risk in all cases was very large (more than 1:3000, with a maximum risk of 1:300) — anything higher than 1:10,000 is considered as too high a risk — so the satellites had to be moved. The actual manoeuvres were in every case quite small — each using only around 70 g of fuel — and for each Metop-A manoeuvre products for the Space Environment Monitor (SEM) and Global Ozone Monitoring Experiment (GOME) instruments were unavailable for a short period. Example: Researchers in the US have demonstrated that beams of light can be used to collect small particles in space. This can be use for example to sample material from a comets tail without damaging the satellite. If collimated light with power 200W and beam diameter 1cm illuminates a collection of reflective spherical micro particles with average diameter 100μm for 1 min, what is their final velocity ?The particles have an average density of 7g/cm3 (you can assume particles do not shadow each other and have cross sections equal to their geometrical cross section) 71 Electromagnetic waves-> transport energy between charged particles. E ( x, t )  E0 zˆ Sin(kx  t   ) B ( x, t )  B0 yˆ Sin(kx  t   ) F  q( E  v  B) Simple types of 3D Waves 1. Plane waves We can think of a plane wave as made up of an infinity set of planes moving in a direction in space. Each plane represents a surface where the E field is constant. This means the phase of the wave is constant over the plane. The set of planes above represent where the E fields are a maximum. 73  We have up to now considered waves travelling in only one dimension (the x direction) We won’t bother with the B part anymore, just assume where there is an E, there is a B field. E ( x, t )  E0 Sin( kx  t   ) In the real world, rays travel in 3 dimensions. We can generalise the above equation in 3 dimensions by E ( x, y, z , t )  E0 Sin(k x x  k y y  k z z  t ) This describes the electric field at the point (x,y,z) and time t, of a plane wave moving in one specific direction in space The direction is given by the vector k where k  k xiˆ  k y ˆj  k z kˆ 2 The magnitude of this vector is the wavenumber k  k  k x2  k y2  k z2   Any position on the wave is described by the position vector, r. r  xiˆ  yjˆ  zkˆ E (r , t )  E0 Sin( k  r  t   ) 74 Each plane can be thought of as a wavefront. Lines perpendicular to the wave front give the direction of propagation. These are called rays. This is the connection between geometrical optics and physical optics. Geometrical optics is the limit of Physical optics when the wavelength tends to zero. 2. Spherical Waves Waves don’t have to be plane waves. A localized light source or point source emits rays in all directions. This is known as a spherical wave. 75 Spherical waves must get weaker as they get further from the source, the Electric field must diminish to satisfy conservation of energy. This means the amplitude of the wave, Es must be a (diminishing) function of r E (r )  Es (r )Sin( k  r  t ) The total energy per second crossing any sphere a distance r from the source must be constant. The intensity [W/m2] is proportional to E2 1 I (r )   0cEs (r ) 2 2 The total energy per second crossing a sphere of radius r is 4 r 2 I  P0 where P0 is the power output of the source, which is constant. 76  1 Therefore 4 r  0cE0 (r ) 2  P0 2 2 2 P0 1 Rearranging: Es ( r )  2 4 0c r 2 2 P0 2 But is a constant and can be written as E0. Therefore: 4 0c E0 Es ( r )  r E0 So for a spherical wave: E ( x , t )  Sin( k  x  t ) r 77 3. Cylindrical waves With same logic we can analyze a cylindrical wave from a line source. E0 Tutorial Exercise: Show that: E (r )  Sin( k  r  t ) r 78 Example questions: A circular loop can of wire can be used as a radio antenna. If a 18cm diameter antenna is located 2.5km from a 95MHz source with a total power of 55kW, what is the maximum e.m.f induced in the loop? (You can assume that the plane of the antenna loop is perpendicular to the direction of the magnetic field and that the source radiates uniformly in all directions. ) 79 The Complex Representation We have shown that a plane wave can be described as either a sine or a cosine E  E0 Sin(k  x  t ) or E  E0 Cos(k  x  t ) There is however a third way using complex numbers Remember de Moivres theorem: ei  Cos   i Sin  then Cos   Re ei  So we can write a plane wave as:  E  E0 Cos(k  x  t )  E0 Re ei ( k x t )  Usually, for convenience we just omit the Re and write a wave as E  E0ei ( k x t ) 80  Why bother with another representation? We will see in the next sections on interference and diffraction that using the complex representation can considerably simplify calculations. In particular multiplication and division of two exponential terms simplifies to addition and subtraction of their phases eA ( A B ) e A.e B  e( A B )  e eB As we saw in the previous example differentiation and integration of exponentials is also particularly easy to remember  ax 1 ax e  ae ax  e dx  a e ax x 81 Example questions: i ( kx t ) Show that the electric field E  E0 e satisfies the wave equation. What is the speed of this wave in terms of k and ω? Phasor diagrams We can consider the exponential notation as a rotating vector on an Argand diagram. We call this a Phasor, and it leads to some simple graphical solutions of complex interference problems Im i z  eit Im  z  sin(t )   t[rad ]  Re -1 1 i Re  z  cos(t ) In particular when we are adding waves of different amplitude and phase, the exponential notation doesn’t provide simplification, but on the Argand diagram it is equivalent to vector addition. The two phasors (although not technically vectors), can be added tip to end as in the case of vectors  Phasor representations Harmonic wave Sum of two harmonic waves, same frequency, different phase y  y0ei ( t ) y  y0ei ( t )  y1ei (1 t )  y2ei (2 t ) Re[y] Re[y] Re[y] Re[y] Im[y] Im[y] In general its tricky to do maths on moving things so we note that we can write y  y0ei ( t )  y0ei eit  y0eit When adding phasors, we can set t=0, find the solution and then add back in the time dependence For example: i Using a phasor diagram find the resultant phasor E  E0 e  E1  E2 i i from the addition of E1  E01e 1 and E2  E02 e 2. φ2 φ1 If 1   2 rad and 2   4 rad, then what is the resultant amplitude in terms of A1 and A2? Section 4: Sources of Electromagnetic radiation We have shown that light is a self sustaining propagating disturbance in the electromagnetic field. From Maxwell’s equations we show that an electric and magnetic fields are generated from charge and moving charge (current) respectively. We find that propagating electromagnetic disturbances can be caused by accelerating charged particles. The term “accelerating” includes both charges which change velocity and charges which change direction. So where do electromagnetic waves come from? 86 Radiation from linear accelerating charges 87  Synchrotron radiation: Bright X-ray source f =3×10 16 Hz - 3×10 19 Hz λ=0.01-10 nm These devices use accelerating charges directly to generate bright sources of electromagnetic radiation High brilliance> 1018photons/s/mm2/mrad2/0.1%BW, Wide tunability in energy/wavelength. Pulsed light emission λ= 12.5cm, 6cm) Satellite Antennae: f= 30MHz-3GHz => λ= 10 cm- 1 m 95 Radio and TV antennae (AM : f= 540 to 1600 kHz => λ= 187m-555m FM : f= 88 to 108 MHz => λ= 1.6-3.4km) 96 Receivers Note antennae are also used to collect electromagnetic radiation and convert back to useable electrical signal It is the connection between Electromagnetism and wave optics that allows us to send and receive signals wirelessly, and enables global communication networks. 97  Coherent and Incoherent sources Coherent sources have a fixed phase relationship in space and time. Real sources do not emit light as infinitely long harmonic waves. A nice sine wave is emitted for approximately 1 ns before the phase of the wave shifts abruptly. Thus the phase is constant for only a short period of time. The light emitted during this time is called a wavetrain, wavepacket or wavegroup. This time can be thought of as a coherence time τc E  E0 cos(kx  t   (t ))  (t ) can make random discontinuous jumps The distance the wave travels in this time is the coherence length, c  c c Both these properties are measures of the temporal coherence of the source. For a source with  c  1ns , c  0.3m τc and l c for typical sources are given below. Source Coherence Coherence length lc time τc Natural Light ~ 3 fs ~ 900 nm Mercury Arc Lamp ~1 ns ~0.03 m Kr discharge lamp ~1 ns ~0.3 m Stabilised He-Ne Laser ~1 μs ~300 m (This is why Lasers are generally used for interference work.) Another type of coherence is spatial coherence. Consider light emitted from a extended source composed of uncorrelated dipoles. If the source has appreciable size we loose spatial coherence. This is because different parts of the source can emit with different initial phase. Spatially and temporally coherent Partially coherent The distance over which spatial coherence is maintained is known as the coherence area Ac We will see later that in order to observe interference we need a source with reasonable temporal and spatial coherence.