Electromagnetic waves: Interference

# Electromagnetic waves: Interference

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## Electromagnetic waves: Interference

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1. Electromagnetic waves: Interference Wednesday October 30, 2002

2. 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 Plane wave () P S1 y  x a S2 L

3. Interference terms where,

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

5. Young’s Double slit diffraction • I1P = intensity of source 1 (S1) alone • I2P = intensity of source 2 (S2) alone • Thus IP can be greater or less than I1+I2 depending on the values of 2 - 1 • In Young’s experiment r1 ~|| r2 ~|| k • Hence • Thus r2 – r1 = a sin  r1 r2 a r2-r1

6. Intensity maxima and minima Maxima for, If I1P=I2P=Io Minima for, If I1P=I2P=Io

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

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

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

10. Phasors • Imagine disturbance given in the form =φ2-φ1 φ2 φ1 Carry out addition at t=0

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

12. Other forms of two source interference Fresnel Biprism S1 S s2 d s

13. Other sources of two source interference Altering path length for r2 r1 r2 n With dielectric – thickness d kr2 = kDd + ko(r2-d) = nkod+ ko(r2-d) = kor2 + ko(n-1)d Thus change in path length = k(n-1)d Equivalent to writing, 2 = 1 + ko(n-1)d Then  = kr2 – kor1 = ko(r2-r1) + ko(n-1)d

14. Incidence at an angle Before slits Difference in path length a sin i i = a sin I in r1  After slits Difference in path length = a sin  in r2 a sin  Now k(r2-r1) = - k a sin  + k a sin i Thus  = ka (sin  - sini)

15. Reflection from dielectric layer n1 n2 n1 • Assume phase of wave at O (x=0, t=0) is 0 • Amplitude reflection co-efficient • (n1n2)  = 12 • (n2 n1) ’=21 • Amplitude transmission co-efficient • (n1n2)  =  12 • (n2 n1) ’=  21 • Path O to O’ introduces a phase change A ’ A’ O’  ’ O  t x = t x = 0

16. Reflection from a dielectric layer • At O: • Incident amplitude E = Eoe-iωt • Reflected amplitude ER = Eoe-iωt • At O’: • Reflected amplitude • Transmitted amplitude • At A: • Transmitted amplitude • Reflected amplitude

17. Reflection from a dielectric layer • At A’ A and ΔS1= z sin  = 2t tan ’ sin   z = 2t tan ’ Since, A’ The reflected intensities ~ 0.04Io and both beams (A,A’) will have almost the same intensity. Next beam, however, will have ~ ||3Eo which is very small Thus assume interference at , and need only consider the two beam problem.

18. Transmission through a dielectric layer • At O’: Amplitude ~ ’Eo ~ 0.96 Eo • At O”: Amplitude ~ ’(’)2Eo ~ 0.04 Eo • Thus amplitude at O” is very small O” O’

19. Reflection from a dielectric layer • Interference pattern should be observed at infinity • By using a lens the pattern can be formed in the focal plane (for fringes localized at ) • Path length from A, A’ to screen is the same for both rays • Thus need to find phase difference between two rays at A, A’. A  z = 2t tan ’ A’

20. Reflection from a dielectric surface A  z = 2t tan ’ A’ If we assume ’ ~ 1 and since ’ = || This is just interference between two sources with equal amplitudes

21. Reflection from a dielectric surface where, Since k2 = n2ko k1=n1ko and n1sin = n2sin’ (Snells Law) Thus,

22. Reflection from a dielectric surface Since I1 ~ I2 ~ Io Then, I = 2Io(1+cos) Constructive interference • =  2m = 2ktcos’ -  (here k=n2ko) 2ktcos’ = (2m+1) ktcos’ = (m+1/2) 2n2cos’ =  (m+1/2)o Destructive interference 2n2cos’ =  mo

23. Haidinger’s Bands: Fringes of equal inclination d n1 n2 Beam splitter P 1 x  1 f Extended source Focal plane Dielectric slab PI P2

24. Fizeau Fringes: fringes of equal thickness • Now imagine we arrange to keep cos ’ constant • We can do this if we keep ’ small • That is, view near normal incidence • Focus eye near plane of film • Fringes are localized near film since rays diverge from this region • Now this is still two beam interference, but whether we have a maximum or minimum will depend on the value of t

25. Fizeau Fringes: fringes of equal thickness where, Then if film varies in thickness we will see fringes as we move our eye. These are termed Fizeau fringes.

26. Fizeau Fringes Beam splitter n Extended source n2 n x

27. Wedge between two plates 1 2 glass D y glass air L Path difference = 2y Phase difference  = 2ky -  (phase change for 2, but not for 1) Maxima 2y = (m + ½) o/n Minima 2y = mo/n

28. Wedge between two plates Maxima 2y = (m + ½) o/n Minima 2y = mo/n D y air Look at p and p + 1 maxima yp+1 – yp = o/2n  Δx where Δx = distance between adjacent maxima Now if diameter of object = D Then L = D And (D/L) Δx= o/2n or D = oL/2n Δx L

29. Wedge between two plates Can be used to test the quality of surfaces Fringes follow contour of constant y Thus a flat bottom plate will give straight fringes, otherwise ripples in the fringes will be seen.