Electromagnetic waves: Interference

1. Electromagnetic waves: Interference Wednesday November 6, 2002

2. 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

3. 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

4. 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.

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

6. 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

7. 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

8. 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.

9. Newton’s rings Used to test spherical surfaces Beam splitter

10. Newton’s rings Maxima when, y = (m+1/2) o/2n Gives rings, Rm2=(m+1/2)oR/n R- Rcos = y or, R2=(R-y)2+r2  R2(1-2y/R) + r2  i.e. r2 = 2yR R R y r

11. Reflection from dielectric layer: Antireflection coatings • Important in instruments such as cameras where reflections can give rise to spurious images • Usually designed for particular wavelength in this region – i.e. where film or eye are most sensitive

12. Anti-Reflection coatings 2 1 A. Determine thickness of film air n1 n2 film n1 < n2 < n3 n3 glass Thus both rays (1 and 2) are shifted in phase by  on reflection. For destructive interference (near normal incidence) 2n2t=(m+1/2) Determines the thickness of the film (usually use m=0 for minimum t)

13. Anti-Reflection coatings 2 1 air B. Determine refractive index of film n1 A’ A Near normal incidence Amplitude at A n2 film n3 glass Since  ’ ~ 1

14. Anti-reflection coating B. Determine refractive index of film Amplitude at A’ To get perfect cancellation, we would like EA = E A’ should be index of AR film

15. Multiple Beam interference • Thus far in looking at reflectivity from a dielectric layer we have assumed that the reflectivity is small • The problem then reduces to two beam interference • Now consider a dielectric layer of uniform thickness d and assume that the reflectivity is large e.g. || > 0.8 • This is usually obtained by coating the surface of the layer with a thin metallic coating – or several dielectric coatings to give high reflectivity • Or, one can put coatings on glass plates , then consider space between plates

16. Multiple beam interference Let 12 =  21= ’ 12=  21= ’ ’’ Eo (’)5’Eo Eo (’)3’Eo (’)7’Eo  n1 n2 ’ A B C D n1  (’)2’Eo (’)6’Eo ’ Eo (’)4’Eo

17. Multiple Beam Interference • Assume a (for the time being) a monochromatic source • , ’ small ( < 30o) usually • Now || = |’| >> , ’ • Thus reflected beams decrease rapidly in amplitude (from first to second) • But amplitude of adjacent transmitted beam is about the same amplitude • Amplitude of successfully reflected beams decreases slowly (from the second) • Thus treat in transmission where contrast should be somewhat higher • The latter is the configuration of most applications