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 a S1S1S1S1 S2S2S2S2 x L Plane wave ( ) P y 

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 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

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

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 Interference

9 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,

10 Phasors Imagine disturbance given in the form =φ2-φ1=φ2-φ1=φ2-φ1=φ2-φ1 φ1φ1φ1φ1 φ2φ2φ2φ2 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 s2s2s2s2 S1S1S1S1 S ds

13 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

14 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 )

15 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’ ’’’’ ’’’’  

16 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

17 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.

18 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”

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 A’  z = 2t tan  ’

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

21 Reflection from a dielectric surface where, Since k 2 = n 2 k o k 1 =n 1 k o and n 1 sin  = n 2 sin  ’(Snells Law) Thus,

22 Reflection from a dielectric surface Since I 1 ~ I 2 ~ I o Then, I = 2I o (1+cos  ) Constructive interference Destructive interference  =  2m  = 2ktcos  ’ -  (here k=n 2 k o ) 2ktcos  ’ =  (2m+1)  ktcos  ’ =  (m+1/2)  2n 2 cos  ’ =  (m+1/2) o 2n 2 cos  ’ =  m o

23 Haidinger’s Bands: Fringes of equal inclination d n2n2n2n2 n1n1n1n1 Beam splitter Extendedsource PIPIPIPI P2P2P2P2  P x f Focalplane 1111 1111 Dielectricslab

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 Extended source Beam splitter x n n2n2n2n2 n

27 Wedge between two plates 1 2 glass glass air D y 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 Look at p and p + 1 maxima y p+1 – y p = 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 = o L/2n Δx air D y 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.