1. 1 Electromagnetic waves: Two source Interference Monday November 4, 2002

2. 2 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’ ’’’’ ’’’’  

3. 3 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

4. 4 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.

5. 5 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”

6. 6 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  ’

7. 7 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

8. 8 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,

9. 9 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

10. 10 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

11. 11 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

12. 12 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.

13. 13 Fizeau Fringes Extended source Beam splitter x n n2n2n2n2 n

14. 14 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

15. 15 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

16. 16 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.