1 Electromagnetic waves: Reflection, Transmission and Interference Monday October 28, 2002

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

3 Intensity reflection Amplitude reflection co-efficient and intensity reflection

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

5 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 

6 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

7 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

8 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)

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

10 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

11 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.

12 Interference terms where,

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

14 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

15 Intensity maxima and minima Maxima for, Minima for, If I 1P =I 2P =I o

16 Fringe Visibility or Fringe Contrast To measure the contrast or visibility of these fringes, one may define a useful quantity, the fringe visibility:

17 Co-ordinates on screen Use sin  ≈ tan  = y/L Then These results are seen in the following Interference pattern Interference

18 Phasor Representation of wave addition Phasor representation of a wave E.g. E = E o sin  t is represented as a vector of magnitude E o, making an angle  =  t with respect to the y-axis Projection onto y-axis for sine and x-axis for cosine Now write,

19 Phasors Imagine disturbance given in the form =φ2-φ1=φ2-φ1=φ2-φ1=φ2-φ1 φ1φ1φ1φ1 φ2φ2φ2φ2 Carry out addition at t=0

20 Other forms of two-source interference Lloyd’s mirror screen S S’

21 Other forms of two source interference Fresnel Biprism s2s2s2s2 S1S1S1S1 S ds

22 Other sources of two source interference n Altering path length for r 2 r1r1r1r1 r2r2r2r2 With dielectric – thickness d kr 2 = k D d + k o (r 2 -d) = nk o d+ ko(r 2 -d) = nk o d+ ko(r 2 -d) = k o r 2 + k o (n-1)d = k o r 2 + k o (n-1)d Thus change in path length = k(n-1)d Equivalent to writing,  2 =  1 + k o (n-1)d Then  = kr 2 – k o r 1 = k o (r 2 -r 1 ) + k o (n-1)d

23 Incidence at an angle iiii  a sin  a sin  i Before slits Difference in path length After slits Difference in path length = a sin  I in r 1 = a sin  in r 2 Now k(r 2 -r 1 ) = - k a sin  + k a sin  i Thus  = ka (sin  - sin  i )

24 Reflection from dielectric layer Assume phase of wave at O (x=0, t=0) is 0 Amplitude reflection co- efficient  (n 1  n 2 )  =  12  (n 2  n 1 )  ’=  21 Amplitude transmission co-efficient  (n 1  n 2 )  =  12  (n 2  n 1 )  ’=  21 Path O to O’ introduces a phase change n2n2n2n2 n1n1n1n1 n1n1n1n1 A O’ O t x = 0 x = t A’ ’’’’ ’’’’  

25 Reflection from a dielectric layer At O:  Incident amplitude E = E o e -iωt  Reflected amplitude E R = E o e -iωt  At O’:  Reflected amplitude  Transmitted amplitude At A:  Transmitted amplitude  Reflected amplitude

26 Reflection from a dielectric layer A A’  z = 2t tan  ’ and ΔS 1 = z sin  = 2t tan  ’ sin  At A’ Since, The reflected intensities ~ 0.04I o and both beams (A,A’) will have almost the same intensity. Next beam, however, will have ~ |  | 3 E o which is very small Thus assume interference at , and need only consider the two beam problem.

27 Transmission through a dielectric layer At O’: Amplitude ~  ’E o ~ 0.96 E o At O”: Amplitude ~  ’(  ’) 2 E o ~ 0.04 E o Thus amplitude at O” is very small O’ O”