1 Electromagnetic waves: Multiple beam Interference Wed. Nov. 13, 2002

2 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

3 Multiple beam interference Evaluate Thus,

4 Multiple beam interference Now,

5 Multiple beam interference Now recall the definition of the intensity of an electromagnetic wave Thus, is the intensity distribution in the focal plane of the lens.

6 Multiple beam interference Fringe pattern

7 Multiple beam interference Maximum intensity when,

8 Multiple beam interference Thus maxima are circles in focal plane of lens – or rings The maximum intensity And intensity distribution is,

9 Multiple beam interference Minima intensity when, Intensity distribution, If R > 0.9 I min <<I max

10 Multiple beam interference Let the contrast, or co-efficient of finesse be defined by, Then the transmitted light is described by and Airy function, The same analysis for the reflected light gives

11 Multiple beam interference: Transmission curves R=0.87 R=0.046 R=0.18 F=200 F=1 F=0.2

12 Multiple beam interference: Reflectivity curves R=0.87 R=0.046 R=0.18 F=200 F=1 F=0.2

13 Observation of fringe patterns source f1f1 f2f2 Bright circles Screen or eye

14 Typical parameters in an experiment Consider a quartz slab (n 2 =1.5) of thickness d ~ 0.5 cm The condition for constructive interference requires, For light of wavelength o = 500 nm, incident at a small angle , i.e.  ’ also small, and m is large:

15 Spectroscopy applications: Fabry- Perot Interferometer Assume we have a monochromatic light source and we obtain a fringe pattern in the focal plane of a lens Now plot I T along any radial direction Let I MAX =I M The fringes have a finite width as we scan mm-1m-2 order m-3

16 Fabry-Perot Interferometer Full width at half maximum = FWHM, is defined as the width of the fringe at I=(½)I M Now we need to specify units for our application Let us first find  such that I = ½ I M

17 Fabry-Perot Interferometer We want, This gives,

18 Fabry-Perot Interferometer m m-1  = 2m  = 2(m-Δm)  I= ½ I M  = 2(m-1) 

19 Fabry Perot Interferometer Thus at I = ½ I M sin(  /2) = sin (m – Δm)    sin Δm  Assume Δm is small and sin Δm   Δm  Thus FWHM ~ Fraction of an order occupied by fringe

20 Fabry-Perot Interferometer The inverse of the FWHM is a measure of the quality of the instrument. This quality index is called the finesse It is the ratio of the separation between the fringes to the fringe width

21 Fabry-Perot Interferometer Not that  is determined by the reflectivity If R ~ 0.90  = 30 R ~ 0.95  = 60 R ~ 0.97  = 100 In practive, can’t get much better than 100 since the reflectivity is limited by the flatness of the plates (and other factors of course)

22 Fabry-Perot Interferometer Now consider the case of two wavelengths ( 1, 2 ) present in the beam Assume 1  2 and 1 < 2 Increase 2, dashed lines shrink e.g. order m-1 of 2 moves toward mth order of 1 Eventually (m-1) 2 =m 1 This defines the free spectral range m m-1 m

23 Fabry-Perot Interferometer m( )= 2 or mΔ = Δ FSR = /m Now since We have, e.g. =500 nm, d = 5mm, n=1  Δ FSR = 25(10 -2 )mm = 0.25Å