PHYS20312 Wave Optics - Section 3: Interference
3.1 Conditions for Interference
Consider the superposition of two plane waves:
� = �!exp � �!. � − �!� + �! +��exp � �!. � − �!� + �! but we observe
�. �∗ � = 2� where
∗ � � �. � = �� + �� + 2�!. �� cos �! − �! . � − �! − �! � + �! − �!
constants interference term
��. �� = 0 if wave are orthogonally polarised
Hence, we have our first condition for interference i.e. �!. �� ≠ 0
Now let’s concentrate on the interference term (and call it �!"#$% �, � for brevity) and re-write it as using the trigonometrical identity for the cosine of two angles:
�!"#$% �, � = cos ∆�. � + ∆�] cos ∆�� + sin [∆�. � + ∆�] sin[∆�� where
∆� = �! − �!; ∆� = �! − �!; ∆� = �! − �!.
Let’s now see what happens when we take the time average over a period �
∗ � � �. � = �� + �� + 2�!. �� �!"#$% �, �
!! ∗ but sin �� = cos �� = 0 for � ≫ � then �. � = 0 unless ∆� = �! − �! ≈ 0
Therefore, our second condition for interference is �! ≈ �!
If �! ≈ �! then cos �� ≈ 1 and sin �� ≈ 0
∴ �!"#$% �, � = cos ∆�. � + ∆� = cos ∆�. � cos ∆� − sin ∆�. � sin ∆�
If ∆� = � � then �. �∗ = 0 as above and so our third condition for interference is ∆� ≈ constant
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PHYS20312 Wave Optics - Section 3: Interference over �. Such beams are ‘coherent’ (more on this later) and can be achieved experimentally if the two beam originate from the same source. In contrast, it is very hard to synchronise to independent sources so that ∆� ≈ constant.
Summary of the conditions of interference:
1. �!. �� ≠ 0 i.e. non-orthogonal polarization 2. ∆� < ~�!! i.e. narrow bandwidth 3. ∆� ≈ constant over � i.e. coherent sources
3.2 Coherence
In general , waves that maintain a fixed phase relationship are coherent but we can divide this into two types of coherence: temporal and spatial
3.2.1 Temporal coherence
If ∆� ≈ 0 the the phase relationship will change over a period ∆�!! , as shown below for two similar frequencies:
Figure 1 Two similar get progressively out of phase as time advances (Image from Wikipedia).
For polychromatic light with a bandwidth, ∆�, we can define a coherence time:
1 � = ! ∆�
(3.1) and also a coherence length: �! = ��! (3.2) which is the characteristic path length difference between two beams that still allows interference to be observed.
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PHYS20312 Wave Optics - Section 3: Interference
3.2.2 Spatial coherence.
Spatially coherence light maintains a fixed phase relationship along a wavefront, as illustrated in Fig 2 below:
Figure 2 Spatial coherence (Image adapted from Wikipedia).
Spatial coherence is important if different parts of a wavefront are to interfere such as for ‘wavefront division interference’ but is less important for ‘amplitude division interference’ – see Fig 3 below. We will now look at each of these cases in turn.
Figure 3 Comparison of wavefront and amplitude division inteference.
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PHYS20312 Wave Optics - Section 3: Interference
3.2.3 Fringes and Fringe Visibility
For parallel polarized, monochromatic, coherent beams:
∗ � � �. � = �� + �� + 2�!�! cos � or
� = �! + �! + 2 �!�! cos � where
� = �! − �! . � + Δ� if cos � > 0 then we have constructive interference if cos � < 0 then we have destructive interference
As cos � varies between -1 and +1, we get a series of bright and dark band, which are known as “interference fringes”, as shown in the figure below.
Figure 4 Interference fringes.
If �! = �! = �! then � � = 2� + 2� cos � = 4� cos! ! ! ! 2
If �! ≠ �! we can define a measure of the depth of modulation of the light intensity which is known as the ‘fringe visibility’, �, (or alternatively as the ‘contrast’ or ‘modulation’):
�!"# − �!"# 2 �!�! � = = �!"# + �!"# �! + �!
(3.3)
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PHYS20312 Wave Optics - Section 3: Interference
3.3 Wavefront Division Interference.
3.3.1 Young slits.
This was a very important experiment historically because it was the first demonstration of the wave nature of light. Young used a candle flame as his light source which produced broadband, unpolarized and incoherent light i.e. met none of the conditions for interference identified above.
How did he get his experiment, shown below, to work?
• The first slit ensured good spatial coherence at the double slits so that they become coherent sources • The path length difference (pld) to the screen was very small and so: o instantaneous polarization was the same o the pld < coherence length
Figure 5 Young's double slit experiment.
The double slits produce cylindrical wavefronts and so the combined field is given by
1 1 � = �! exp � ��! − �!� + �! + �! exp � ��! − �!� + �! �! �! and so
� � ∗ �� �� 2�!. �� �. � = + + cos � �! − �! − ∆�� + ∆� �! �! �!�!
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PHYS20312 Wave Optics - Section 3: Interference
For coherent sources with �! > �! − �! and for �! ≈ �! ≈ �
! � �! + �� 2�!. �! �. �∗ = + cos � � − � � � ! !
Moreover, if the slits are equal width then �! = �! and hence �! = �! = �! then
�! � � = 4 cos! � 2 where � = � �! − �! .
We can find an expression for � by considering figure 6:
Figure 6 Calculating � for Young's slits.
Hence � 2� � � = ��� = �� sin � ≈ ��� ≈ �� = � � � �
Thus,
�! ��� � = 4 cos! � ��
(3.4)
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PHYS20312 Wave Optics - Section 3: Interference
3.3.2 Lloyd’s Mirror.
Consider the interference between a direct beam and its reflection, as shown in Figure 7.
Figure 7 Lloyd's mirror.
This is a variation on Young’s slits, with one source and its reflection replacing the two slits. However, a difference now arises because one of the two rays undergoes a reflection.
The Fresnel reflection coefficients for s- and p-polarized light are (see Hecht Chapter 4):
sin �! − �! �! = − sin �! + �!
tan �! − �! �! = tan �! + �! for �!~90° and �! < �! then �! < 0 and �! < 0 (See Hecht Ch 4) i.e. �� and �� are of opposite sign for both polarizations. In other words, a � phase change occurs on reflection under these conditions (N.B. exp �� =-1).
Thus, now
� = � �! − �! + �
Hence,
�! ��� �! ��� � = 4 cos! + � = 4 sin! � �� � ��
(3.5)
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PHYS20312 Wave Optics - Section 3: Interference
3.3.3 Multiple slits.
Let’s now extend our analysis to � slits, as in Figure 8:
Figure 8 Interference from N slits.
Consider the path lengths from each of the slits to �:
�!� = �!
�!� = �! + ��
�!� = �! + 2�� etc etc until:
�!!!� = �! + � − 1 ��
Thus for the �th path length, �!, we have
�! = �! + ��� for � = 0,1 ⋯ � − 1 but �� ≈ ��
∴ �! = �! + ��� for � ≫ �, �! ≈ �
∴ �! = � + ���
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PHYS20312 Wave Optics - Section 3: Interference
The field from the �th slit at P is
1 �! = �! exp ���! �! and hence the total field at P is
!!! 1 � = �! exp ���! � !!! !
For � ≫ � then �! ≈ � for all �. We can thus replace �! with � in the denominator but not in the argument of the exponential because that is sensitive to changes in �! of the order of �.
Hence, !!! �! � = exp ���! � !!!
!!! �! � = exp �� � + ��� � !!!
!!! �! � = exp ��� exp ����� � !!!
This is a geometric series so
�! 1 − exp ����� � = exp ��� � 1 − exp ����
���� ���� ���� exp −� exp � − exp −� �! 2 2 2 � = exp ��� ��� ��� ��� � exp −� exp � − exp −� 2 2 2 which yields
���� sin! �! 2 � = ��� � sin! 2
(3.6)
Figure 9 shows equation (3.6) plotted out for various values of �
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PHYS20312 Wave Optics - Section 3: Interference
Figure 9 Inteference patterns for N slits.
3.4 Amplitude Division Interference.
3.4.1 The Michelson Interferometer. In this device, light from a source is split into two parts, each of which travel down a separate arm before being recombined so that they interfere – see Figure 10.
Figure 10 The Michelson Interferometer.
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PHYS20312 Wave Optics - Section 3: Interference
Figure 11 The compensator plate.
Without the compensator plate, the two beams pass through unequal thicknesses of glass – see Figure 11. This means the that the optical path length difference is a function of wavelength because the refractive index of glass depends on wavelength. The compensator plate equalises the path through glass and this removes the wavelength dependence of the path length difference.
We can analyse the Michelson Interferometer by first drawing an equivalent diagram – see Figure 12.
Figure 12 Equivalent diagram for the Michelson Interferometer.
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PHYS20312 Wave Optics - Section 3: Interference
Path length difference �� = � cos �
Phase difference � = ��� + � where an additional phase lag, �, has been included because one beam undergoes more reflections than the other.
Bright fringes will occur when � = 2��, where � is an integer i.e. ��cos�! + � = �2� where �! is the angle at which the �th bright fringe occurs.
Hence, 2� �cos� = �2� − � � !
�cos�! = �� − �� where �� �� = 2�
• These equations correspond to a set of fringes (one fringe for each value of �). • These fringes will form rings if the mirrors are well-aligned due to symmetry.
• The greatest value of � corresponds to the central fringe (NB �cos�! is a maximum for �! = 0°). • If � is decreased then the rings will shrink towards the centre. ! • The central ring vanishes when ∆� = ! • At � = 0 the central fringe covers the entire screen
For the central fringe, �! = 0°: � = �!� − ��
For the �th fringe from the centre:
�cos�! = �! − � � − ��
Combining these together:
� � = � 1 − cos�!
!! for small �, cosθ ≈ 1 − !
2�� ∴ � = ! �
(3.7).
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PHYS20312 Wave Optics - Section 3: Interference
3.4.2 The Fourier Transform Spectrometer.
The Michelson interferometer can also be used as a spectrometer. In this case, a detector is used to monitor the intensity at the centre of the fringe pattern (i.e. for � = 0°) whilst one of the mirrors is moved along the beam direction – see Figure 13.
Figure 13 The Fourier Transform Spectrometer.
First consider what happens when a monochromatic light source is used:
Field at detector,
� = �!exp ��2�! + �!exp ��2�!
∴ � � = 2�! 1 + cos �� where � = 2 �! − �! as before. Hence, for monochromatic light there will be cosine variation of � with � – as shown in Figure 14.
Now, consider polychromatic light with a spectrum � � such that
! �! = � � d� !
(� is a scalar because we are only considering one direction i.e. � = 0°).
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PHYS20312 Wave Optics - Section 3: Interference
Thus the total intensity at the detector (see Figure 14) is
! � � = 2 � � 1 + cos �� d� !
! � � = 2�! + 2 � � cos �� d� !
!! but the second term can be recognised as an inverse cosine Fourier transform, �! , i.e.
! !! 2 �! � � = � � cos �� d� � !
!! ∴ � � = 2�! + ��! � �
So, if we measure � � and then find �! � � :
!! �! � � = �! 2�! + �! ��! � �
�! � � = 0 + � � �
�! � � = � � � i.e. calculating the Fourier transform of � � yields the spectrum � � .
Figure 14 Output from a Fourier Transform spectrometer of (left) monochromatic and (right) polychromatic light.
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PHYS20312 Wave Optics - Section 3: Interference
3.4.3 Thin Films.
The reflectivity of dielectric interfaces is low (e.g. for glass-air it is ~4%) which means that rays reflected more than once can be neglected and we need only consider the interference of rays reflected from the front surface and once from the back surface – see Figure 15.
Figure 15 Interference from a thin dielectric film.
From Figure 15, we can see that the optical path length difference, �, is given by
� = �� + �� �! − ���! but from the geometry shown in Figure 15 and using Snell’s law: � �� = �� = cos�!
�! �� = �� sin �! = �� sin �! �! and 1 �� tan � = ! � 2
Combining the second two of these we get
! sin �! �! �� = 2� cos�! �!
! 2�!� sin �! 2�!� ! ∴ � = − 2��! = 1 − sin �! = 2�!� cos�! cos�! cos�! cos�!
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PHYS20312 Wave Optics - Section 3: Interference
!! The phase difference, � = � − � ! where the – � is due to the difference between low �-to-high � reflections and high �-to-low � reflections (see Hecht 4.6.3).
Hence,
4� � = � � cos� − � � ! !
(3.8).
• � = 2�� corresponds to constructive interference • � = 2� + 1 � corresponds to destructive interference • Thin films of constant thickness are used to make: o dichroic mirrors o antireflection coatings • In practise, multiple layers are often used (Hecht 9.7) • Thin films are also found in nature: o oil on water (� varies) o soap bubbles (� and � vary) o tapetum lucidum
3.4.4. Multi-beam interference. Let’s consider what happens when we coat each interface so that it is highly reflective ; now we have to include multiple reflections from the front and back surface – see Figure 16.
Figure 16 Multi-beam interference from two highly reflected surfaces.
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PHYS20312 Wave Optics - Section 3: Interference
To analysis this case, we need first to derive Stokes’ relations – consider Figure 17.
Figure 17 Stokes' relations.
Returning to the problem of multi-beam interference we write down an expression for each of the reflected rays:
�!! = �!�
! ! �!! = �!�� � exp ���
! !! �!! = �!�� � exp �2�� etc etc and hence for the �the ray
! ! !!!! �!" = �!�� � exp ���� where � = 2�� cos �! (no � phase change here because both reflections are high �-to-low �).
Hence, the total reflected field is
!!! ! ! ! ! ! �! = �!" = �!� + �!�� � exp ��� 1 + � exp ��� + � exp ��� + ⋯ !!! where from Stokes’ relations we have used �! = �! !. The final bracket on the right-hand side is a geometric sum to infinity and so we can write
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PHYS20312 Wave Optics - Section 3: Interference
��!�!exp ��� � = � � + ! ! 1 − �!exp ���
Using �! = −�
1 − exp ��� ��! + �! � = �� ! ! 1 − �!exp ��� but ��! + �! = 1
1 − exp ��� � = �� ! ! 1 − �!exp ���
Hence,
! �! ∗ ! 1 − exp ��� 1 − exp −��� 2� 1 − cos �� = �!�! = � ! ! = ! ! �! 1 − � exp ��� 1 − � exp −��� 1 − 2� cos �� + �
Now using reflectance, � = �! = 1 − �, where the transmittance, � = ��!, and also utilising the trig ! identity 1 − cos � = 2 sin! : !
�� �� 4� sin! 4� sin! �! 2 2 = = �� �� �! 1 + �! − 2� 1 − 2 sin! 1 − � ! + 4� sin! 2 2
Defining the Coefficient of Finesse, �:
4� � = 1 − � !
(3.9)
Hence
�� � sin! �! 2 = �� �! 1 + � sin! 2 and for the total transmitted intensity, �!
�! �! 1 = 1 − = �� �! �! 1 + � sin! 2
(3.10).
Equation (3.10) is known as the Airy function and is plotted in Figure 18.
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PHYS20312 Wave Optics - Section 3: Interference
Figure 18 The Airy function for various values of coefficient of finesse, F.
We can characterise Airy functions by 3 parameters:
1. Free Spectral Range, ∆� (see Figure 18) 2. Resolution, �� (see Figure 18) ∆� 3. Finesse, � = ��
Free Spectral Range, ∆� ∆� is the free difference between adjacent maxima of the Airy function.
Maxima occur for �� �� = 2 where � is an integer.
!! Sub. in for � = 2�� cos � and for � = � : ! ! ! 2� �� = �� cos � � � ! ! where �! is the frequency corresponding to the �th maximum.
Thus �� �! = 2�� cos �!
For the next maximum (�~constant):
� + 1 � �!!! = 2�� cos �!
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Hence, � ∆� = �!!! − �! = 2�� cos �!
(3.11).
Resolution, �� �� is defined is the full width at half maximum of an individual peak.
Half the maximum amplitude corresponds to when
1 1 = �� 1 + � sin! 2 2 i.e. �� 1 sin = 2 �
For large values of � the transmission peaks are sharp (see Figure 18) and so the half maximum !" !" points will occur at a small angle,∆ , to the condition for a maximum, = �� ! ! i.e. at half maximum �� �� = �� + ∆ 2 2 and hence
�� 1 sin �� + ∆ = 2 �
�� �� 1 sin �� cos ∆ + cos �� sin ∆ = 2 2 � but sin �� = 0 and cos �� = 1
�� 1 sin ∆ = 2 �
!" !" !" and since ∆ is small sin ∆ ≈ ∆ , so ! ! !
�� 1 ∆ ≈ 2 �
!" !" The angle ∆ corresponds to the shift in from the maximum to the half maximum point of the ! ! peak i.e. the half width half maximum. The resolution is defined as the full width half maximum (FWHM) i.e. from the half max on one side of the peak to the half max point on the other, and so corresponds to twice this value.
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PHYS20312 Wave Optics - Section 3: Interference
Thus, FWHM corresponds to the angle:
�� 2 ∆ ≈ 2 !"#$ �
!" Sub. in for � = 2�� cos � and recognising that a change in angle : corresponds to a change in ! ! frequency, ��, i.e. the resolution, we get:
1 2� 2 2�� cos �! �� ≈ 2 � � or
� �� = ���� cos �!
(3.12).
Finesse
The finesse, , is the ratio of the free spectral range to the resolution:
!! ∆� � � � = = �� 2�� cos �! ���� cos �!
� � � � � = = 2 1 − �
(3.13)
Finesse is plotted as a function of � in Figure 19 below.
Figure 19 Finesse as function of reflectance, R. (Image from Wikipedia)
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PHYS20312 Wave Optics - Section 3: Interference
The Fabry-Perot etalon. The Fabry-Perot etalon is a device that uses multi-beam interference from two highly reflective surfaces to enable high resolution spectroscopy. Its design is shown in Figure 20.
Figure 20 The Fabry-Perot etalon.
Here the gap is air filled and � = 1 and so the condition for a bright fringe becomes:
2� �� = � cos � � ! where �! is the resonant ray angle for �. Note that for a divergent (or convergent) source of light a range of ray angles will be incident on to the etalon.
For small �!
! 2 �! � ≈ � 1 − � 2
At �! = 0°
2 � = � ! �
Letting � = �! − �
! �! � − � = � − � ! ! �
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PHYS20312 Wave Optics - Section 3: Interference
�� ∴ � = ! �
If these rays pass through the focal point of a lens then they will form an image of a series of concentric rings of radius, �! , as shown in Figure 21:
�! ≈ ��!
�� � ≈ � ! �
Hence, the concentric rings get closer together for increasing � because the resonant radii, �! ∝ �
r θ
f
Figure 21 Imaging etalon rings.
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