Physics 1502: Lecture 27 Today’s Agenda Announcements: –Midterm 2: Monday Nov. 16 … –Homework 07: due Friday this week Electromagnetic Waves –Energy and Momentum in Waves Optics –Waves, Wavefronts, and Rays –Reflection –Index of Refraction

f()x x f(x x z y

E & B in Electromagnetic Wave Plane Harmonic Wave: where: y x z Nothing special about (E y,B z ); eg could have (E y,-B x ) Note: the direction of propagation is given by the cross product where are the unit vectors in the (E,B) directions. Note cyclical relation:

Velocity of Electromagnetic Waves The wave equation for E x : (derived from Maxwell’s Eqn) Therefore, we now know the velocity of electromagnetic waves in free space: Putting in the measured values for  0 &  0, we get: This value is identical to the measured speed of light! –We identify light as an electromagnetic wave.

The EM Spectrum These EM waves can take on any wavelength from angstroms to miles (and beyond). We give these waves different names depending on the wavelength. Wavelength [m] Gamma Rays Infrared Microwaves Short Wave Radio TV and FM Radio AM Radio Long Radio Waves Ultraviolet Visible Light X Rays

Energy in EM Waves / review Electromagnetic waves contain energy which is stored in E and B fields: The Intensity of a wave is defined as the average power transmitted per unit area = average energy density times wave velocity: Therefore, the total energy density in an e-m wave = u, where =

Momentum in EM Waves Electromagnetic waves contain momentum. The momentum transferred to a surface depends on the area of the surface. Thus Pressure is a more useful quantity. If a surface completely absorbs the incident light, the momentum gained by the surface is, We use the above expression plus Newton’s Second Law in the form F=dp/dt to derive the following expression for the Pressure, If the surface completely reflects the light, conservation of momentum indicates the light pressure will be double that for the surface that absorbs.

The Poynting Vector The direction of the propagation of the electromagnetic wave is given by: This wave carries energy. This energy transport is defined by the Poynting vector S as: –The direction of S is the direction of propagation of the wave –The magnitude of S is directly related to the energy being transported by the wave: The intensity for harmonic waves is then given by:

The Poynting Vector Thus we get some useful relations for the Poynting vector. 1.The direction of propagation of an EM wave is along the Poynting vector. 2.The Intensity of light at any position is given by the magnitude of the Poynting vector at that position, averaged over a cycle. I = S avg 3.The light pressure is also given by the average value of the Poynting vector as, P = S/cAbsorbing surface P = 2S/cReflecting surface

Generating E-M Waves Static charges produce a constant Electric Field but no Magnetic Field. Moving charges (currents) produce both a possibly changing electric field and a static magnetic field. Accelerated charges produce EM radiation (oscillating electric and magnetic fields). Antennas are often used to produce EM waves in a controlled manner.

A Dipole Antenna V(t)=V o cos(  t) x z y time t= E time t=  /2  E time t=  /  one half cycle later

dipole radiation pattern oscillating electric dipole generates e-m radiation that is polarized in the direction of the dipole radiation pattern is doughnut shaped & outward traveling –zero amplitude directly above and below dipole –maximum amplitude in-plane proportional to sin(  t)

Receiving E-M Radiation receiving antenna One way to receive an EM signal is to use the same sort of antenna. Receiving antenna has charges which are accelerated by the E field of the EM wave. The acceleration of charges is the same thing as an EMF. Thus a voltage signal is created. Speaker y x z

Lecture 27, ACT 1 Consider an EM wave with the E field POLARIZED to lie perpendicular to the ground. y x z In which orientation should you turn your receiving dipole antenna in order to best receive this signal? C) Along E a) Along S b) Along B

Loop Antennas Magnetic Dipole Antennas The electric dipole antenna makes use of the basic electric force on a charged particle Note that you can calculate the related magnetic field using Ampere’s Law. We can also make an antenna that produces magnetic fields that look like a magnetic dipole, i.e. a loop of wire. This loop can receive signals by exploiting Faraday’s Law. For a changing B field through a fixed loop

Lecture 27, ACT 2 Consider an EM wave with the E field POLARIZED to lie perpendicular to the ground. y x z In which orientation should you turn your receiving loop antenna in order to best receive this signal? a) â Along S b) â Along B C) â Along E

3-D Representation Waves, Wavefronts, and Rays Consider a light wave (not necessarily visible) whose E field is described by, This wave travels in the +x direction and has no dependence on y or z, i.e. it is a plane wave. Wave Fronts RAYS

EM wave at an interface What happens when light hits a surface of a material? Three Possibilities –Reflected –Refracted (transmitted) –Absorbed incident ray reflected ray refracted ray MATERIAL 1 MATERIAL 2

Geometric Optics What happens to EM waves (usually light) in different materials? –index of refraction, n. Restriction: waves whose wavelength is much shorter than the objects with which it interacts. Assume that light propagates in straight lines, called rays. Our primary focus will be on the REFLECTION and REFRACTION of these rays at the interface of two materials. incident ray reflected ray refracted ray MATERIAL 1 MATERIAL 2

Reflection The angle of incidence equals the angle of reflection  i =  r, where both angles are measured from the normal: ·Note also, that all rays lie in the “plane of incidence” ii rr ii rr x EiEi ii ErEr rr ·Why? »This law is quite general; we supply a limited justification when surface is a good conductor, Electric field lines are perpendicular to the conducting surface. The components of E parallel to the surface of the incident and reflected wave must cancel!!

Index of Refraction The wave incident on an interface can not only reflect, but it can also propagate into the second material. Claim the speed of an electromagnetic wave is different in matter than it is in vacuum. –Recall, from Maxwell’s eqns in vacuum: –How are Maxwell’s eqns in matter different?    ,    ·Therefore, the speed of light in matter is related to the speed of light in vacuum by: ·The index of refraction is frequency dependent: For example n blue > n red where n = index of refraction of the material:

Refraction How is the angle of refraction related to the angle of incidence? –Unlike reflection,  1 cannot equal  2 !! »n 1  n 2  v 1  v 2 but, the frequencies (f 1,f 2 ) must be the same  the wavelengths must be different! Therefore,  2 must be different from  1 !! 11 22 n1n1 n2n2 1 2

Snell’s Law From the last slide: 11 11 1 2 L n1n1 n2n2 11 22 22 22 22 The two triangles above each have hypotenuse L  But,

Lecture 27, ACT 3 Which of the following ray diagrams could represent the passage of light from air through glass and back to air? (a) (b)(c)

Lecture 27, ACT 3 Which of the following ray diagrams could represent the passage of light from air through glass and back to air? (a) (b)(c)

Lecture 27, ACT 4 Which of the following ray diagrams could represent the passage of light from air through glass and back to air? (a) (b) (c) (d)

Prisms Index of refraction frequency ultraviolet absorption bands white light prism A prism does two things, 1.Bends light the same way at both entrance and exit interfaces. 2.Splits colours due to dispersion.

Prisms 11 22 Entering For air/glass interface, we use n(air)=1, n(glass)=n 33 44 Exiting

Prisms 11 22 33 44 Overall Deflection At both deflections the amount of downward deflection depends on n (and the prism apex angle,  ). The overall downward deflection goes like,  ~ A(  ) + B n Different colours will bend different amounts ! 

Lecture 27, ACT 5 White light is passed through a prism as shown. Since n(blue) > n(red), which colour will end up higher on the screen ? ? ? A) BLUE B) RED

LIKE SO!In second rainbow pattern is reversed

Total Internal Reflection –Consider light moving from glass (n 1 =1.5) to air (n 2 =1.0) ie light is bent away from the normal. as  1 gets bigger,  2 gets bigger, but  2 can never get bigger than 90  !! In general, if sin  1  sin  C  (n 2 / n 1 ), we have NO refracted ray; we have TOTAL INTERNAL REFLECTION. For example, light in water which is incident on an air surface with angle  1 >  c = sin -1 (1.0/1.5) = 41.8  will be totally reflected. This property is the basis for the optical fibers used in communication. incident ray reflected ray refracted ray 22 11 rr GLASS AIR n2n2 n1n1 2

ACT 6: Critical Angle... air n =1.00 glass n =1.5 air n =1.00 cc water n =1.33 glass n =1.5 water n =1.33 cc Case I Case II An optical fiber is cladded by another dielectric. In case I this is water, with an index of refraction of 1.33, while in case II this is air with an index of refraction of Compare the critical angles for total internal reflection in these two cases a)  cI >  cII b)  cI =  cII c)  cI <  cII

ACT 7: Fiber Optics air n =1.00 glass n =1.5 air n =1.00 cc water n =1.33 glass n =1.5 water n =1.33 cc Case I Case II The same two fibers are used to transmit light from a laser in one room to an experiment in another. Which makes a better fiber, the one in water ( I ) or the one in air ( II ) ? a)  Water b)  Air

Problem You have a prism that from the side forms a triangle of sides 2cm x 2cm x 2  2cm, and has an index of refraction of 1.5. It is arranged (in air) so that one 2cm side is parallel to the ground, and the other to the left. You direct a laser beam into the prism from the left. At the first interaction with the prism surface, all of the ray is transmitted into the prism. a)Draw a diagram indicating what happens to the ray at the second and third interaction with the prism surface. Include all reflected and transmitted rays. Indicate the relevant angles. b)Repeat the problem for a prism that is arranged identically but submerged in water.

Solution At the first interface  =0 o, no deflection of initial light direction. At 2nd interface  =45 o, from glas to air ? Critical angle: sin(  c )=1.0/1.5 =>  c = 41.8 o < 45 o Thus, at 2nd interface light undergoes total internal reflection At 3rd interface  =0 o, again no deflection of the light beam A) Prism in air B) Prism in water (n=1.33) At the first interface  =0 o, the same situation. At 2nd interface now the critical angle: sin(  c )=1.33/1.5 =>  c = 62 o > 45 o Now at 2nd interface some light is refracted out the prism n 1 sin(  1 ) = n 2 sin(  2 ) => at  2 = 52.9 o Some light is still reflected, as in A) ! At 3rd interface  =0 o, the same as A)