1 Electromagnetic waves: Interference Wednesday November 6, 2002

2 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

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 Extended source Beam splitter x n n2n2n2n2 n

6 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

7 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

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 R- Rcos  = y or, R 2 =(R-y) 2 +r 2  R 2 (1-2y/R) + r 2  R 2 (1-2y/R) + r 2  R R y r i.e. r 2 = 2yR Maxima when, y = (m+1/2) o /2n Gives rings, R m 2 =(m+1/2) o R/n

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 A. Determine thickness of film film glass air n1n1n1n1 n3n3n3n3 n2n2n2n2 n 1 < n 2 < n 3 Thus both rays (1 and 2) are shifted in phase by  on reflection. 1 2 For destructive interference (near normal incidence) 2n 2 t=(m+1/2) 2n 2 t=(m+1/2) Determines the thickness of the film (usually use m=0 for minimum t)

13 Anti-Reflection coatings B. Determine refractive index of film Near normal incidence Amplitude at A film glass air n1n1n1n1 n3n3n3n3 n2n2n2n2 1 2 A A’ Since   ’ ~ 1

14 Anti-reflection coating Amplitude at A’ B. Determine refractive index of film To get perfect cancellation, we would like E A = 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 =  ’ n1n1n1n1 n2n2n2n2 n1n1n1n1  ’’’’ ABCD  EoEoEoEo  ’  ’ E o  ’ E o (  ’) 3  ’E o (  ’) 2  ’E o (  ’) 6  ’E o (  ’) 4  ’E o (  ’) 5  ’E o (  ’) 7  ’E o

17 Multiple Beam Interference Assume a (for the time being) a monochromatic source ,  ’ small ( < 30 o ) 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