Interference and Colour Part III - Mathematics of Thin-Film Interference

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How can alternating layers in a beetle's shell give the beetle brilliant colour?

Big Ideas: 
  • The phase difference, $ \Delta \phi $ between two waves determines how the waves interfere and depends on path length difference, difference in 'inherent phase' and the wavelength of the waves
  • If the phase difference is even/odd multiples of $  \pi $ then constructive/destructive interference results.
  • The condition then on the path length difference that gives constructive/destructive interference will change depending on the presence of 'inherent' phase shifts
  • By having multiple reflections interfere constructively at a specific wavelength due to alternating thin-film layers brilliant colours can be created in nature.

We have talked in the article on diffraction gratings about the role of path length difference between two waves in determining how the waves will interfere when overlapped.  In particular, we have seen that the relation between path length difference and wavelength influences whether we get constructive or destructive interference, or something in between.  However in some situations such as thin film interference there are additional complications that may affect the conditions for constructive or destructive interference.  One complication has to do with the fact that the wavelength of light changes depending on what medium it is travelling in.  The wavelength of light in a medium, $ \lambda_{med} $, with index of refraction $ n $ is related to the wavelength of light in vacuum, $ \lambda_{med} $ by $ \lambda_{med} = \frac{\lambda_{vac}}{n} $.

 

Figure 1:  The wavelength of the light will be different travelling in a medium (shown shaded in blue) with index of refraciton different than for the medium shown in white.  The wavelength of light in a medium with index of rfraction $ n $ can be given as $ \lambda_{med} = \frac{\lambda_{vac}}{n} $ where $ \lambda_{vac} $ is the wavelength of the light in a vacuum.  In Figure 1 the extra path length still has to be odd multiples of half wavelengths to have constructive interference as seen in Figure 5 of the article on diffraction

Figure 1 shows if the extra path length difference takes place in a medium (shown in blue) with index of refraction $ n $ the wavelength of wave B will change in that medium.  The equations for constructive and destructive interference remain the same as we saw in our first article on diffraction except the wavelength used must be the wavelength of the light in the medium where the path length difference occurs.

$ \Delta x = m \lambda_{med} $$ m = 1, 2 ,3 ... $  (constructive)   Equation 1

$ \Delta x = (m+ 1/2)\lambda_{med} $, $  m = 0, 1, 2 ... $ (destructive) Equation 2

Figure2:  a)  Alternating layers of refractive index n1 and n2 can be used in nature to provide bright reflections at a specific wavelength.  b) TEM cross section of a Cincindela Scutellaris shell showing these multilayer reflectors are used in nature for coloration 1, 2. c) Picture of Cincindela scutellaris scutellaris1.  Photos courtesy of:  J. R. Soc. Interface 2009 6, S165-S184.

To calculate the thickness of each layer type as shown in Figure 2 that would give constructive interference for 635 nm (wavelength in air for the colour red) we need to account for the wavelength change in medium 1 with $ n_{1} = 1.5 $ and medium 2 with $ n_{2} =2 $.  Equation 1 would correctly predict that thickness $ t_{1} $ and thickness $ t_{2} $ should not be equal for $ n_{1} \neq n_{2} $.  We have made a step in the right direction but unfortunately, using equation 1 to calculate the thickness $ t_{1} $ and $ t_{2} $ for which wave B in Figure 2 would interfere constructively with wave A and C, respectively, would in fact be finding the thicknesses that would give destructive inteference! Why is that?

Consider, for example, Figure 3 which shows wave B starting with a rising slope while wave A starts with a falling slope.  As shown, even though the path length difference is half a wavelength wave A and B will interfere constructively!  Similarly a path length difference of one wavelength would, as shown, give destructive interference.  We see then there is more to account for in the condition for constructive or destructive interference than path length difference.  For this, we need to describe waves mathematically.

 

Figure 3:  In the top case wave A and B have a half a wavelngth difference and now constructive interference results.  In the bottom case a wavelength difference results in destructive interference.  This is opposite to what we had said earlier in the article on diffraction gratings where wave A and B were shown to be starting at the same stage in their oscillation cycle. 

Waves can be described by a sine or a cosine function, $ \sin (\phi) $ or $ \cos ( \phi) $.  Sine and cosine funcitons have angles as arguments.   Here these are not geometric angles but so called 'phase angles' that describe at what point on the sine or cosine function we are.  One wavelength $ \lambda $ corresponds to one full sine or cosine function period, or to an angle or 'phase' of $  2 \pi $ ( or 360 deg).  Figure 4 shows a graph of $ \sin (\phi) $, as well as $ \sin ( \phi - \alpha) $ where $ \alpha = \pi / 2, \pi, 2 \pi $.  Note that $ \sin ( \phi - \alpha) $ is just shifted over by $ \alpha $ compared to $ \sin ( \phi ) $.  For clarity, for $ \sin ( \phi - \alpha)  $ we would say the phase is $ \phi - \alpha $, i.e. whatever is inside the brackets.  For $ \alpha = \pi /2 $, $  \pi $, and $ 2 \pi $, this corresponds to a quarter, half and whole wavelgnth shift respectively.  You can easily generate these curves with your computer using a spreadsheet.  Try it!  Notice a shfit by $  2\pi $ gives the identical wave.  It follows that if a sine function is shifted by any multiple of $ 2 \pi  $ we would still get the same sine function.

Similarly to Figure 1 we can see that two waves will interfere constructively if their phases differ by $  2 \pi  $ or destructively if their phases differ by $ \pi $.

 

Figure 4:  The top graph shows $ \sin (\phi) $ vs $ \phi $ where $ \phi $ is an angle called 'the phase'.  $ \sin ( \phi) $ repeats itself when $ \phi $ increases by $  2 \pi $ radians ( or 360 deg).  Graphs of $ \sin ( \phi - \alpha) $ are shifted by $ \alpha $ to the right horizontally compared to $ \sin ( \phi) $.  A shift of $  \alpha = \pi/2 $, $ \pi $, $ \2 \pi $ corresponds to a quarter, half or whole wavelength shift, respectively.

Note that when we drew waves earlier we were drawing them as a function of position because we talked about differences in path length but the argument of a sine or cosine function is an angle or a 'phase' .  The connection is made by 'scaling' all distances to the wavelength by

$ \sin (\phi) = \sin ( \frac{2 \pi x}{ \lambda}) $

The right hand side says in mathematical form that if you change the position x by one wavelength $ \lambda $, the phase angle changes by $ 2 \pi $, exactly what we need.  So the wave pattern repeats after a distance $  x = \lambda $.  The term in brackets is now a phase in radians with no physical dimension.  Remember to put your calculator to radians when doing calculations.  So this gives us the correct wavelength  in our mathematical description.  It does not yet describe that waves travel.

For travelling waves described by $ \sin ( \phi) $, the 'phase' of the wave is

$ \phi = k x - \omega t + \phi_{0} $

 x is the distance travelled or position in space, $ k = \frac {2 \pi}{\lambda} $ is the 'wave number', and $ \omega = \frac {2 \pi}{T} $  where T is the period, t is the time, $ \phi_{0} $ is the 'inherent phase' of the wave which will determine the intial starting point in a wavelength of the wave at $  x = 0, t = 0 $

For simplicity of notation we will not write the subscript 'med' on $ \lambda $.  The term $  \omega t = 2 \pi t/ T $ is the term that makes the wave travel because time changes.  It does not influence the wave pattern described by the wavelength: the whole wave pattern just moves to the left with increasing time.  If you followed the discussion above on scaling the distances to the wavelength, you might be wondering why the time is scaled to a certain period T.  It turns out that it is related to the wave speed ( or 'propagation' speed) $  v_{p} $ due to the relation

$ v_{p} = \frac{\lambda}{T} $

which is true for all waves. 

If we have two travelling waves A and B, wave A will have $ \sin ( \phi_{A}) $ with $ \phi_{A} = kx_{A} - \omega t + \phi_{0,A} $ and another wave B will have $  \sin (\phi_{B}) = k x_{B} - \omega t + \phi_{0,B} $.  The phase difference between the two waves is

$ \Delta \phi = (k x_{A} - \omega t + \phi_{0,A}) - (k x_{B} - \omega t + \phi_{0,B}) = k \Delta x + \Delta \phi_{0} $   (Equation 3)

where $ \Delta x $ is the path length differnece which we discussed earlier and $ \Delta \phi_{0} $ is the 'inherent phase difference' with $ \Delta \phi_{0} = \abs{\phi_{0,A} - \phi_{0,B}} $.  As you can see, the $ \omega t $ term cancels.  The phase difference relation does not change with time for two waves with the same wavelength.  There are however instances where the inherent phase difference $  \Delta \phi_{0} $ is not zero.  Different starting points is one example and picking up an extra $ \pi $ phase shift upon reflection is another example.  This brings us back to thin film interference.

Because extra shifts of $  2 \pi $ in any one of the interfering waves does not affect the interference between the waves we can say most generally that the condition for constructive interference is that the phase difference $  \Delta \phi $ equal even multiples of $  \pi $

$ \Delta \phi= \phi_{A} - \phi_{B} = 2 m \pi $, $  m = 0,1,2,... $ (constructive interference)   (equation 4)

For destructive interference the phase difference has to be odd multiples of $ \pi $,

$ \Delta \phi = \pi + 2 m \pi = (2m+1) \pi $$  m = 0,1,2... $ (destructive interference) (equation 5)

These conditions on phase difference are the most fundamental underlying conditions for constructive and destructive interference.  We can show that our previous conditions for constructive and destructive interference given in the article on diffraction gratings follow from equation 4 and 5.  For the case where $  \Delta \phi_{0} = 0 $, equating equation 3 with 4 gives

$ \Delta \phi = k \Delta x = \frac{2 \pi \Delta x}{\lambda} = 2 m \pi $, $  m = 0, 1, 2, ... $ (constructive interference)

or more simply

$  \frac {2 \pi \Delta x}{\lambda}= 2 m \pi $, $  m = 0, 1, 2, ... $

Dividing by $ 2 \pi $ and multiplying both sides by $ \lambda $ we have

$ \Delta x = m \lambda $, $  m = 0,1,2,... $ (constructive interference)  (equation 6)

For $ \Delta \phi_{0} = 0 $ equating equations 3 and 5 and following the same steps gives

$ \Delta x = (m + 1/2) \lambda $, $  m = 0,1,2... $ (destructive interference)  (equation 7)

We have seen these conditions on path length difference before in our discussion of diffraction gratings when discussing Figure 2.  The inherent phase difference $  \Delta \phi_{0}= 0  $ for reflection from different points on the same surface becase all the waves originating from the point sources start off 'in phase', meaning  at the same place in their wavelength cycle.  Similarly we saw that in figure 2 wave A and B started off in phase with $  \Delta \phi_{0} = 0 $

In figure 1 the waves started out shifted by $ \pi $ with respect to each other.  When $ \Delta \phi_{0} = \pi $ equating equation 3 with 4 or 5 now gives

$ \frac{2 \pi x}{\lambda} + \pi = 2 m \pi $, $  m = 0, 1,2, ... $  (constructive interference)

$ \frac{2 \pi x}{\lambda} + \pi = (2m +1) \pi $, $  m = 0, 1, 2, ... $ (destructive interference)

which upon division by $  2 \pi $ and multiplication by $ \lambda $ gives

$ \Delta x = ( m + 1/2) \lambda $, $  m = 0, 1,2,... $  (constructive interference)  (equation 8)

$ \Delta x = m \lambda $, $  m = 0, 1,2,... $ (destructive interference) (equation 9)

We can see that equation 8 and 9 which hold for $ \Delta \phi _0 = \pi $ are the same as for equation 6 and 7 for $ \Delta \phi_{0} = 0 $, except swapped. 

Taking into account the inherent phase difference $ \Delta \phi_{0} $ becomes very important when considering thin film interference.  For a wave traveling from medium 1 to medium 2 shown in Figure 5, if  n2 > n1, the reflected ray instantly gets shifted forward by half a wavelength in its cycle ( or we can say experiences a $ \pi $ phase shift) compared to the incident wave at the boundary where reflection occurs.  This does not occur if n1 < n2.  Transmitted waves do not experience this shift either regardless of which of n1 or n2 is bigger.

 

Figure 5:  The top and bottom graphs show a wave traveling from medium 1 to medium 2.  Part of the wave gets transmitted to medium 2 and part gets reflected back into medium 1.  For n2 > n1  (the bottom case) the reflected wave (dashed) will be shifted forward by a half-wavelength while for n1 < n2 this does not occur.

Considering wave C and F in figure 6 we include these possible $ \pi $ phase shifts in the inherent phases of wave C and F.  If both or neither of wave C and F experience a $ \pi $ phase shift  (n1 > n2 > n3 or n1 < n2 < n3) then $ \Delta \phi_{0} = 0 $.  If only one of the waves experiences a $ \pi $ phase shift (n1 > n2 < n3 or n1 < n2 > n3) then $ \Delta \phi_{0} = \pi $.

 

Figure 6:  The inherent phase difference $ \Delta \phi_{0} = 0 $ if neither or both of the rays C and F pick up a $ \pi $ phase shift upon reflection and equation 6 and 7 describe the path length differences that give constructive and destructive interference.  If either C or F picks up a $ \pi $ phase shift upon reflection (but not both) then $ \Delta \phi_{0} = \pi $ and equation 8 and 9 describe constructive and destructive interference.

Returning to the multi-layers in Figure 2, because the indices of refraction n1 and n2 for the different  layers are alternating we wil have $ \Delta \phi_{0} = \pi $ for waves A and B, and for waves B and C and so on.  Equation 8 describes the condition of path length difference in this situation for constructive interference.  To find the minimum thickness $ t_{1} $ (setting m =1) for which wave A and B will intefere for 635 nm (wavelength given in air) we let $ \Delta x $ in equation 12 be $ 2t_{1} $ and solve for $ t_{1} $.  This gives

$ t_{1} = \frac{\lambda_{air}}{4n_{1}} =\frac{635 \times 10^{-9}}{(4)(1.5)} = 106 \mathrm{nm} $

Using equation 8 again and a path length difference of $ 2 t_{2} $ between wave B and C in figure 2 we find the minimum thickness $ t_{2} $ that gives constructive interference for 635 nm (in air) to be

$ t_{2} = \frac{\lambda_{air}}{4n_{2}} = \frac{635 \times 10^{-9}}{(4)(2)} = 79 \mathrm{nm} $.

By having alternating layers of melanin with thickness 79 nm and protein with thickness 106 nm, the red colour of 635 nm would have all the reflections A, B, C and D... interfere constructively and a bright reflection at this wavelength would be observed.  Other visible wavelengths may interfere partially constructively but not with the same intensity as for 635 nm.

In this article we have seen the concept of phase difference and its relation to path length difference and inherent phase difference.  We have also seen that phase difference between two waves determines how the waves will interfere.  From the fundamental conditions on phase difference that give constructive and destructive interference we were able to derive conditions for path length difference in various situtions such as diffraction gratings and thin film interference.  The discussion started  in our first article by developing a pictoral intuition of waves and interference and led to a full mathematical description.

The examples given in the articles concerning beetle coloration are a very small subset of the phenomena of structural coloration.  There are many examples of thin-film interference, diffraction gratings or photonic crystals, in nature such as in coloration of birds such as hummingbirds 3, peacocks 4, 5, and pigeons 6 and various butterflies 7 as well as other insects. The reflections seen from animals eyes at night and even the silvery appearance of fish are also due to interference of light caused by microscopic structures 8, 9.  Gems such as opal also display structural coloration 10. Of course the most common non-nature example of thin-film interference is the array of colours seen on a soap bubble or on a film of oil.  For simplicity we considered the case where the incident light was normal to the thin-fim surfaces. When the angle of incidence is different from normal incidence, equations 6-9 still apply however the path length difference will no longer simply be twice the thickness of the thin-film.  This means reflections occuring at different angles with respect to the bubble surface will interfere constructively at different wavelengths 11, 12.  The multiple colours seen on a soap bubble are due to the reflections reaching your eye coming from different angles and also explains why if you change your  position with respect to the bubble the colors change. 

In conclusion microscopic layers or structures can give rise to what is known as structural coloration where colour is caused by the interference of light rather than by apsorption by pigments.

 

 

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