1 Electromagnetic waves Hecht, Chapter 2 Wednesday October 23, 2002

2 Electromagnetic waves: Phase relations Thus E and B are in phase since, requires that E B k

3 Irradiance (energy per unit volume) Energy density stored in an electric field Energy density stored in a magnetic field

4 Energy density Now if E = E o sin(ωt+φ) and ω is very large We will see only a time average of E

5 Intensity or Irradiance In free space, wave propagates with speed c c Δt A In time Δt, all energy in this volume passes through A. Thus, the total energy passing through A is,

6 Intensity or Irradiance Power passing through A is, Define: Intensity or Irradiance as the power per unit area

7 Intensity in a dielectric medium In a dielectric medium, Consequently, the irradiance or intensity is,

8 Poynting vector Define

9 For an isotropic media energy flows in the direction of propagation, so both the magnitude and direction of this flow is given by, The corresponding intensity or irradiance is then,

10 Example: Lasers Spot diameterIntensity (W/m 2 )Electric field magnitude (V/m) 2 mm1.6 X X µm (e.g. focus by lens of eye) 1.6 X X µm1.6 X X 10 6 Laser Power = 5mW nb. Colossal dielectric constant material CaCu 3 Ti 4 O 12,  = 10,000 at 300K Subramanian et al. J. Solid State Chem. 151, 323 (2000) Near breakdown voltage in water Same as sunlight at earth  o = X CV -1 m -1 (SI units)

11 Reflection, Transmission and Interference of EM waves

12 Reflection and Transmission at an interface 12 Normal Incidence – Two media characterized by v 1, v 2 incidenttransmitted reflected

13 Reflection and Transmission at an interface Require continuity of amplitude at interface: f 1 + g 1 = f 2 Require continuity of slope at interface: f 1 ’ + g 1 ’ = f 2 ’ Recall u = x – vt

14 Reflection and Transmission at an interface Continuity of slope requires, or,

15 Reflection and Transmission at an interface Integrating from t = -  to t = t Assuming f 1 (t = -  ) = 0 Then,

16 Amplitude transmission co-efficient (  ) Medium 1 to 2 Medium 2 to 1

17 Amplitude reflection co-efficient (  ) At a dielectric interface

18 Phase changes on reflection from a dielectric interface n 2 > n 1 n 2 <n 1 Less dense to more dense e.g. air to glass More dense to less dense e.g. glass to air  phase change on reflection No phase change on reflection

19 Phase changes on transmission through a dielectric interface Thus there is no phase change on transmission

20 Amplitude Transmission & Reflection For normal incidence Amplitude reflection Amplitude transmission Suppose these are plane waves

21 Intensity reflection Amplitude reflection co-efficient and intensity reflection

22 Intensity transmission and in general R + T = 1 (conservation of energy)

23 Two-source interference What is the nature of the superposition of radiation from two coherent sources. The classic example of this phenomenon is Young’s Double Slit Experiment a S1S1S1S1 S2S2S2S2 x L Plane wave ( ) P y 

24 Young’s Double slit experiment Monochromatic, plane wave Monochromatic, plane wave Incident on slits (or pin hole), S 1, S 2 Incident on slits (or pin hole), S 1, S 2  separated by distance a (centre to centre) Observed on screen L >> a (L- meters, a – mm) Observed on screen L >> a (L- meters, a – mm) Two sources (S 1 and S 2 ) are coherent and in phase (since same wave front produces both as all times) Two sources (S 1 and S 2 ) are coherent and in phase (since same wave front produces both as all times) Assume slits are very narrow (width b ~ ) Assume slits are very narrow (width b ~ )  so radiation from each slit alone produces uniform illumination across the screen Assumptions

25 Young’s double slit experiment slits at x = 0 The fields at S 1 and S 2 are Assume that the slits might have different width and therefore E o1  E o2

26 Young’s double slit experiment What are the corresponding E-fields at P? Since L >> a (  small) we can put r = |r 1 | = |r 2 | We can also put |k 1 | = |k 2 | = 2  / (monochromatic source)

27 Young’s Double slit experiment The total amplitude at P Intensity at P

28 Interference Effects Are represented by the last two terms If the fields are perpendicular then, and, In the absence of interference, the total intensity is a simple sum

29 Interference effects Interference requires at least parallel components of E 1P and E 2P We will assume the two sources are polarized parallel to one another (i.e.

30 Interference terms where,

31 Intensity – Young’s double slit diffraction Phase difference of beams occurs because of a path difference !

32 Young’s Double slit diffraction I 1P = intensity of source 1 (S 1 ) alone I 2P = intensity of source 2 (S 2 ) alone Thus I P can be greater or less than I 1 +I 2 depending on the values of  2 -  1 In Young’s experiment r 1 ~|| r 2 ~|| k Hence Thus r 2 – r 1 = a sin  r 2 -r 1 a r1r1r1r1 r2r2r2r2