W9_parallel_resonance.eps

Week 13: Interference and Diffraction 421

• To understand this, note that when light reflects from an interface between a
  medium with a lower index of refraction (source) and a medium with a higher
  index of refraction (destination) the reflected waveinverts(shifts its phase byπor
  a half-wavelength). When light refects from an interface betweena medium with a
  lower index (source) moving towards a higher index (destination) the reflected wave
  doesnotinvert its phase. Note that we learned precisely these rules for wave pulses
  reflected from the interface between light string and heavier string or vice versa in
  the first part of this course.


• Second, thetransmittedlight that is partially reflected and partially transmitted
  at the first surface of the thin film has to travel to the second surface through the
  film (typically a distance given asd, not to be confused with the distance between
  two slits above) and then back to the first surface again, where the wave that is
  partially transmitted here recombines with the original reflected wave. The light
  that went into the film thus travels an (approximate) additional distance of 2d, and
  we can use the heuristic rule above to determine whether or not we get construc-
  tive interference (brightening of some given wavelength) or destructive interference
  (partial cancellation and dimming of some given wavelength),if we also account
  for the discrete phase shift(s) at the interfaces.


• Letn 1 < n 2 < n 3 orn 1 > n 2 > n 3 , where by convention we will use 123 to indicate
  the order of the media in the direction of the incoming light. Then there are either
  two phase shifts ofπ(first case) or no phase shifts ofπ(second case) at the two
  reflecting surfaces of the middle layer, and the phase difference is dueonlyto the
  path differencein the film medium with index of refractionn 2. The heuristic
  rule is then:

2 d=mλ′=mλ

n 2

Maxima

2 d= (m+

1

2

)λ′= (m+

1

2

)

λ

n 2

Minima

withm= 0,± 1 ,± 2 ,± 3 ...as usual. Note Well: the use ofλ′=λ/n 2 , the path

differencein the mediummust contain an integer number of wavelengths for the

reflected light that emerges back inton 1 to be in phase.

• A special result occurs whend≪λ. In this case there is “no” path difference, and
  the waves emerge in phase forallwavelengths. The surface becomes “shiny”. You
  can observe this when a drop of oil spreads out on water on dark pavement – at
  first there are many colors and then the surface takes on a silverygrey sheen.


• Letn 1 < n 2 > n 3 orn 1 > n 2 < n 3. Then there is only one phase shift ofπat the
  first surface (first case)orone phase shift ofπat the second surface (second case),
  and the total phase difference is that from the path difference plusan additional
  phase ofπ. This is equivalent to half a wavelength difference. The heuristic rule
  thenreverses:

2 d= (m+

1

2 )λ

′= (m+^1

2 )

λ

n 2 Maxima

2 d=mλ′=mλ

n 2

Minima

withm= 0,± 1 ,± 2 ,± 3 ....

• A second special result occurs whend≪λ. In this case there is “no” path difference,
  and the waves emerge exactlyout of phasebyπfor all wavelengths. The surface