Experiment 28. Fourier Transform Spectroscopysenior-lab/3YL/Expt_28.pdf · Experiment 28. Fourier...

Experiment 28. Fourier TransformSpectroscopy

c©School of Physics, The University of Sydney

Updated WJT August 11, 2014

General ReferencesHecht, E., Optics, 4th ed., Addison-Wesley, 2001

1 Safety

Follow the general laboratory safety rules when doing this experiment. The light sources used inthis experiment are safe, but some of them are very bright. Do not stare directly at any source whenit is illuminated.

28–2 SENIOR PHYSICS LABORATORY

2 Objectives

In this experiment you will investigate the temporal coherence properties of different light sourcesusing the Michelson interferometer. You will

• Learn how to align a Michelson interferometer

• Familiarise yourself with the concept of fringe visibility

• Use a Michelson interferometer as a real-time Fourier transform instrument to obtain theFourier transforms of the spectral intensity distribution for different light sources

• Use a Fast Fourier Transform (FFT) program to recover the spectral distributions of thesources

Prework for this experiment will be found in section (6.1).

3 Introduction

The interferometer developed by A. A. Michelson in the 19th century is still an important opticaltool that is extensively used in research and technology (see Experiment 24 for more information).In this experiment we focus on one significant aspect of the interferometer: its use as a spectrometer.

Fig. 28-1 : The layout of the Michelson interferometer used in this experiment. See the text for details.

Figure 28-1 shows the layout of the Michelson interferometer used in this experiment. The mirrorM1 has three adjustments:

• Two micrometers fixed to the mirror mount allow M1 to be tilted horizontally and vertically.

FOURIER TRANSFORM SPECTROSCOPY 28–3

• A third micrometer connected to a stepping motor that can be used to translate the mirroralong the optical axis.

The output of the interferometer can be viewed directly by eye or a photodetector can be used tomeasure the intensity. When the interferometer is properly adjusted the signal from the detector canbe used to record the “fringe pattern” as a function of the optical path difference (OPD) between thetwo arms of the interferometer.1 The fringe pattern has a sinusoidal component with a well definedperiod. The sinusoidal variation is modulated by a relatively slowly varying envelope function (seeFig. 28-2). In the next section we investigate why this behaviour occurs and how the interferometercan be used as a spectrometer.

4 Theory

If the interferometer in Figure 28-1 is carefully adjusted so the two mirrors are perpendicular to theoptical axis of the interferometer the light that reaches the detector will be uniformly illuminated.As the position of the mirror M2 is changed the field of view will vary from bright to dark due tointerference. Let Imax and Imin be the maximum and minimum intensities in this “fringe pattern.”The Michelson fringe visibility is defined to be

V (x) =Imax − Imin

Imax + Imin(1)

where x is the OPD between the two arms of the interferometer. It is easy to show that 0 ≤ V ≤ 1and in general the fringe visibility will depend on the optical path difference (OPD) x between thetwo arms of the interferometer. If the source is monochromatic V = 1 and will be independent ofx. For real light sources V ≤ 1 and becomes smaller as |x| increases. The visibility will equal 1when x = 0. The width of the visibility curve is called the coherence length of the light sources.Lasers, for example, can have a coherence length of many metres whilst an ordinary incandescentlight has a coherence length of ∼ 1µm.

Figure 28-2 is a plot of the output intensity of a Michelson interferometer as a function of the OPD.There are several features of this plot:

• The fringes are equally spaced. The period of the fringes (in terms of path difference) will bedenoted by L, which obviously has units of length.

• The fringe visibility V (x)is a maximum at x = 0.

• The fringe visibility is also the envelope function of the fringe pattern.

The mathematical form of the fringe pattern can therefore be expressed as

I(x) = I0[1 + V (x) cos{2πx/L}] (2)

It is usually more convenient to write this in complex form (we follow engineering practice and usej =√−1):

I(x) = I0[1 + V (x)e−2πjx/L]

It is always understood that the actual fringe pattern is the real part of I(x).1In a Michelson interferometer, if d1,2 is the distance between the beamsplitter and the mirror M1,2 the optical path

difference will be 2(d2 − d1). We could equally well define it to be 2(d1 − d2); it makes no difference as long as weadopt a consistent definition.


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-5000 -4000 -3000 -2000 -1000 0 1000 2000 3000 4000 5000

path difference (nm)

rela

tive in

ten

sity

Fig. 28-2 : The fringe pattern observed with a non-monochromatic source. The fringe visibility given byMichelson’s formula V = (Imax − Imin)/(Imax + Imin) varies with optical path length. It is a maximumwhen the OPD is zero.

4.1 The visibility as a Fourier transform

If the light is strictly monochromatic V (x) = 1 for all values of x and the fringe signal will be

I(x) = I0[1 + e−2πjx/λ]

where λ is the wavelength of the light. The fact that λ appears as λ−1 complicates things and weuse instead the spectroscopic wavenumber2 σ = 1/λ:

I(x) = I(σ)[1 + e−2πjσx]

where we have explicitly indicated that the intensity can vary as a function of σ.

Real light sources are not monochromatic. If the light source is incoherent the phase between twodifferent wavelengths λ and λ + δλ fluctuates randomly and extremely rapidly. The result is thateach wavelength generates its own interference pattern independent of the other wavelengths and tofind the total fringe signal we simply integrate over all wavelengths:

I(x) =

∫ ∞−∞

I(σ)dσ +

∫ ∞−∞

I(σ)e−2πjσxdσ (3)

where we adopt the convention that the spectral intensity distribution I(σ) = 0 for σ < 0. Thefirst term is proportional to the integrated light intensity of the source while the second will berecognised as the Fourier transform of I(σ). It is convenient to write this result as

I(x) =

(∫ ∞−∞

I(σ)dσ

)[1 +

∫∞−∞ I(σ)e−2πjσxdσ∫∞

−∞ I(σ)dσ

]= I0[1 + S(x)] (4)

2The spectroscopic wavenumber should not be confused with the circular wavenumber k = 2π/λ which is frequentlyencountered in the theory of wave propagation. Various symbols are used for the spectroscopic wavenumber; the USNational Institute for Standards and Technology recommends σ and we follow that here.


where I0 is the integrated intensity and S(x) is the “fringe signal.” The fringe signal is the nor-malised Fourier transform of I(σ). The technique of determining the spectrum I(σ) by using aninterferometer to measure its Fourier transform is known as Fourier transform spectroscopy.

Let σ0 be the mean wavenumber of the spectral intensity distribution I(σ) and introduce the variables = σ − σ0. Let I(s) = I(σ0 + s); I(s) is a function that has the same shape as the spectraldistribution I(σ) but it is centred on s = 0. Change the variable of integration in Eq. (4) from σ tos = σ − σ0:

S(x) = e−2πjσ0x

∫∞−∞ I(s)e−2πjsxds∫∞

−∞ I(s)ds

= V (x)e−2πjσ0x (5)

This result shows that the the interference signal consists of an oscillatory term with frequency σ0.The fringe visibility is equal to the Fourier transform of I(s). The period of the fringe signal isL = σ−1

0 . In most cases3 we can assume that the mean wavelength λ0 ≈ σ−10 so the fringe period

is equal to the mean wavelength of the light.

It follows that the source intensity distribution, apart from an unimportant normalising factor, canbe found by taking the inverse transform of S(x):

I(σ) =

∫ ∞−∞

S(x)e+2πjσxdx (6)

This method of finding spectra is known as Fourier transform spectroscopy (FTS), and in section(4.2) a practical way to obtain the spectrum from the interferogram is discussed.

The Appendix gives some examples of the Fourier transforms for some commonly encounteredspectral distributions.

4.2 The Fast Fourier Transform

The fringe signal is recorded as a set of numbers (xi, Si), i = 1, 2, . . ., where Si is the signalmeasured at the OPD xi. The points xi are equally spaced. The data is thus a sampled version ofthe continuous function S(x). The Fourier transform (or its inverse) of a set of sampled data canbe found using a numerical procedure called the discrete Fourier transform (DFT). If the numberof data points N = 2m, where m is an integer, the discrete transform can be computed using analgorithm known as the fast Fourier transform (FFT). As the name suggests it is much faster tocompute than the DFT and it is widely used in many areas of numerical analysis (see Expt 3 formore information).

Whenever the DFT/FFT is used one must be aware that artefacts can be introduced which are theresult of the sampling process and are not physical. There are two different kinds of samplingartefacts:

• Aliasing can occur when the sampling is too coarse. Let δ be the sampling period. In thepresent experiment δ corresponds to the change in OPD for one step of the stepper motor.According to the Nyquist-Shannon sampling theorem the interferogram V (x) must not con-tain frequencies greater than (2δ)−1. This frequency is often called the Nyquist frequency.

3Strictly speaking, this requires ∆λ� λ0.


In this experiment it is perhaps more intuitive to think in terms of the period of the highestfrequency. If xmax is the period of the highest frequency in units of length then the Nyquistcriterion takes the form xmax > 2δ.

• Leakage or spectral broadening occurs when not enough data points are taken. The Fouriertransform is an integral over the range (−∞,∞) but real data is always measured over somefinite range. Suppose that the interferogram is measured over the range (−X/2, X/2). Thespectral resolution will be 1/X wavenumbers. Spectral features will be broadened by approx-imately this amount, and it is not possible to resolve to spectral lines that have a separationless than this.

4.3 “Fellgett’s advantage”

Conventionally spectra are found by dispersing the light from the source using prisms or diffrac-tion gratings. Unlike FTS these techniques do not require moving optical elements with extremeprecision and from a practical point of view they are usually cheaper and easier to use.

However, when light is dispersed using, for example, a grating, one must effectively use a largenumber of detectors to cover the full spectral range compared to FTS, which only requires a singledetector.4

Because of the quantum nature of light photodetectors are necessarily noisy. If the mean countingrate is N photons per second the detected photocounts follow a Poisson distribution with mean N .At high counting rates this can be approximated by a Gaussian distribution having a mean of N anda standard deviation of N1/2. However, many detectors add additional noise, called detector noise,to the signal. This extra noise is largely due to thermal or Johnson-Nyquist noise in the detector.

In 1949 Fellgett showed (in his Ph.D. thesis) that if detector noise was the dominant source of noise,and a conventional spectrometer required m detectors, then the signal-to-noise ratio for FTS wasbetter by a factor of m1/2 compared to conventional spectroscopy. This result is called Fellgett’sadvantage; it is also called the multiplex advantage, since in FTS all the spectral channels areessentially multiplexed onto the single output channel of the interferometer.

Until quite recently the noise characteristics of infrared detectors were generally quite poor, andFTS was widely used for infrared spectroscopy. However, there are now photon-noise limited in-frared detectors available and conventional methods have become competitive with FTS. At longerwavelengths (submillimetre and millimetre parts of the spectrum) FTS is still used.

5 Apparatus

5.1 The Michelson interferometer

Figure 28-1 shows the layout of the Michelson interferometer used in this experiment. The mirrors,beamsplitter and compensator plate are mounted on a marble slab for stability. The apparatus ishoused in a box to exclude drafts and to exclude room light. The top cover can be lifted to inspect

4It is true that many grating spectrometers only have a single detector, but in this case the grating must be rotated inorder to scan the spectrum and one can show that this equivalent to using a large number of separate detectors.


the interferometer and to read the dial gauge (see below). Light from the source is split at thedielectric beamsplitter B. Approximately half the light is transmitted to the mirror M1 while theother half is reflected to the mirror M . A second glass plate called the compensator (C) is placedbetween B and M to ensure that the amount of glass in each path is the same.5

The two mirrors reflect the light back to B where the beams are recombined. The combined beamsthat emerge from the interferometer can be viewed directly by eye. Alternatively, a photodetectorcan be inserted into the output port of the interferometer.

The mirror M1 has three adjustments:

• Two micrometers fixed to the mirror mount allow M1 to be tilted horizontally and vertically.Always use “baby steps” when adjusting these micrometers!

• A third micrometer moves the mirror along the optical axis; i. e., either towards or away fromthe beamsplitter. It is connected to a stepper motor which is used to change the OPD (seebelow).

A dial gauge indicates the position of M1. With most of the light sources fringes will only be seennear the x = 0 position. This position is written on the cover of the interferometer (under the blacklight shield).

5.2 The stepper motor

The mirror M1 is moved by a stepper motor. This motor is connected to the micrometer via agearbox and each step of the motor is equivalent to a displacement of 24 nm or 48 nm of opticalpath. See the “Precautions” section below (5.6) for more information.

Note that “rotate left” increases the reading on the dial gauge (this actually decreases the OPDbetween the two arms of the interferometer). Because of backlash in the motor drive, if you reversethe motor direction it will take a few steps before the mirror starts to move in the correct direction.

There are five motor speeds. Use 1 or 2 for visual observation, 3 for data acquisition and 5 to movethe mirror “rapidly,” e. g., when moving the mirror back to the x = 0 position.

5.3 Light sources

The light sources in this experiment are all solid-state devices. There is a solid state laser diode, adual-wavelength diode combination, several light-emitting diodes (LEDs) of different colours anda “white light” diode. The light sources are designed to slide into a tube located at the input ofthe interferometer. The light sources use a low voltage power supply that can be connected to thesource using the BNC connector on the rear of the housing.

5Unequal glass paths will introduce dispersion in the interferometer. This is discussed in more detail in Experiment24. In particular dispersion greatly reduces the fringe visibility and it is therefore important to ensure that the glass pathsin the two arms of the interferometer are equal.


5.4 Detector

The detector is a photodiode. The housing is designed to slide into a tube located at the output ofthe interferometer. The photodiode output is connected to a variable-gain analog amplifier which inturn is connected to an analog-to-digital converter in the computer.

5.5 Computer

A program called Interference Spectroscopy can be used to view and record the inter-ference signal from the photodetector. The fast Fourier transforms are done using QtiPlot.

5.6 Precautions

1. The optical surfaces of the mirrors, beamsplitter and compensating plate should never betouched.

2. Do not remove the main instrument cover. There is a hinged black cover on the top of theinterferometer. This can be opened to inspect the interferometer and to read the dial gauge.

3. The LED light sources must only be connected to the DC power supply furnished with theapparatus (-5VDC).

4. The photoelectric detector should be connected to the input of the amplifier, not to the ampli-fier’s output. Both connectors are BNCs and it is possible to make a mistake. If you are notsure how to connect things ask your tutor for help.

5. Do not look directly at any of the light sources.

6. Do not drive the mirror more than 2 mm from the x = 0 position. If this occurs power willbe cut off from the stepping motor and a buzzer will sound. Seek assistance from your tutoror the lab technician.

7. Please turn off the amplifier when not in use to conserve battery power.

6 Procedure

6.1 Prework

Note: These questions should be answered before coming to the lab. Answer the questions on asheet of paper and glue or staple the sheet in your logbook when you commence the experimentalwork.

1. From the step size of the stepper motor calculate the Nyquist frequency for the data recordedin this experiment. What optical wavelength corresponds to this frequency? Do you thinkaliasing might be a problem in this experiment?


2. A certain light source produces two spectral lines at 600 and 601 nm. If the equipment in thisexperiment is used to resolve these two lines what is the minimum number of data points thatyou need to take (remember it must be a power of 2)?

6.2 Alignment of the interferometer

To obtain good data it is essential that the interferometer be properly aligned. Any angular mis-alignment of M1 must be� α, where

α =√

2λ/D (7)

Here D = 40 mm is the diameter of M1 and λ is the wavelength. Even a small change in tempera-ture can cause a misalignment, so it is important to check this adjustment regularly.

The following procedure should be used to align the interferometer.

1. Using the stepper motor drive the mirror to the x = 0 position as read from the dial gauge.

2. Insert the WHITE LED source into the input port of the interferometer and visually inspectthe output port. You should see two crosses. Adjust the tilt of M1 until the two crosses aresuperimposed. If you can only see one cross the alignment is probably close to correct sodon’t touch anything!

3. Remove the WHITE LED and insert the RED LASER. Again observe the output beam. Youshould see some dark bands crossing the field of view. These are the interference fringes. Ifyou don’t see any fringes there are two possibilities:

• The interferometer is properly aligned. You can check this by touching the M1 mirrorholder or touching (not turning!) one of its alignment micrometers. As you apply gentlepressure you should see very broad fringes moving across the field of view.

• The interferometer is badly aligned. Go back to step (1)!

4. Adjust the micrometers on M1 (using baby steps!!) until the fringes are as wide as possible.Ideally the field of view will be uniformly illuminated and it will slowly blink on and off asyou apply gentle pressure with a finger to the mirror mount.

5. The final check is done using the data acquisition (DA) system, which is described next. If theDA system indicates that there is still a misalignment, you will need to repeat the procedurehere. It is not advisable to try improving the DA signal by adjusting the tilt mirrors. The mostaccurate way to do the alignment is visually

Question 1: The tilt micrometers are located 40 mm from the centre of the mirror. One divisionon the micrometer is equal to a displacement of 0.01 mm. Calculate the angular tilt produced byturning the micrometer by one division. Compare this to the tolerance α given by Eq. (7), assuminga wavelength of 625 nm. To what fraction of a division on the micrometer does this correspond?

Question 2: Draw a sketch of the output of the interferometer when the two beams are properlyaligned and the beams are interfering constructively. Next to it draw a sketch showing the samething but with the OPD changed by λ/2; i. e., with the beams interfering destructively.Draw another sketch showing the pattern you would see if there were three fringes across the field


of view. Now draw a sketch of the pattern you would see if the OPD was changed by λ/2.Finally, with the aid of these sketches explain why it is so important to have the interferometercorrectly aligned.

6.3 Data acquisition and data analysis

In this part of the experiment you will use the RED LED light source to familiarise yourself withthe data acquisition (DA) and data analysis packages on the computer.

1. Place the RED LED in the input port of the interferometer and the photodetector in the out-put port. Make sure they are properly connected (the LED to the -5VDC supply and thephotodetector to the amplifier). The output of the amplifier should be connected to the com-puter. Likewise there is a “SYNC” signal from the stepper motor controller that also shouldbe connected to the computer.

2. Turn on all the equipment.

3. Log on to the computer as Student.

4. Run the program Interference Spectroscopy (an icon is located on the desktop).From the menu chooseAction→Start Data Acquisitionor just hit the “A” key.

5. You should see a trace appear in the data acquisition window on the computer. The verticalaxis is the signal intensity (in arbitrary units). The horizontal axis is the OPD in nanometres(the zero of the scale is arbitrary). Set the vertical position of the trace (or the average valueof the trace, if there are fringes present) close to the middle of the range using the coarse andfine gain adjustments on the amplifier box. You should start with the coarse gain set to itslowest value. If you can’t get the trace in the middle of the range with the fine gain adjustmentincrease the coarse gain by one step and repeat the procedure. Keep doing this until the traceis satisfactory.

6. Using motor speed 3 record some fringes. They should have good contrast; that is, the fringevisibility should be ∼ 0.5 or greater. If not, the interferometer alignment will need to bechecked.

7. To stop data acquisition, first stop the stepper motor then from the program menu chooseAction→Stop Data Acquisitionor hit the “P” key.

8. There is a cursor (a vertical red line) that can be displayed by positioning the mouse over apoint on the display and clicking. The x position and intensity of the point under the cursorwill be displayed.

9. Save your data after every successful run by using the menu itemFile→ Save DataThe data will be saved in a binary format file with the extension .ifs. These files can beread by the Interference Spectroscopy program. It is good practice to make yourown subdirectory in My Documents and keep all your files there.


10. To use the data with other programs you will need to “export” the data. There are two options:

• The Export All Data option exports the data as a standard text file that can be read withprograms like Notepad, QtiPlot and Excel. For example, you can import the datainto QtiPlot or Excel, plot the data and print them.

• The Export for FFT option is used when you analyse the interferograms using the FFTapplication. To save data for FFT analysis do the following:Place the cursor over the x = 0 point on your data set (this is the point where thevisibility is a maximum). ChooseAction→Set Originfrom the menu or just hit the “O” key. Next chooseFile→Export for FFTThis will automatically save as much data as possible such that the number of datapoints is a power of 2 and with the origin at x = 0.

Save a copy of the fringe plot in your notebook and record the measured fringe visibility.

C1 .

6.4 Investigation of various light sources

Throughout this section you should make printed copies of all the fringe signals, Fourier transforms,etc., and put these in your logbook.

6.4.1 Monochromatic light

Using the RED LASER as the light source record a few fringes. Use Eq. (1) to calculate the fringevisibility and estimate the uncertainty in your result. The visibility should be > 0.6. If it is not theinterferometer is probably misaligned. Check the alignment and repeat your measurements.

Save two sets of fringe data. Save a “short” set consisting of a few fringes (no more than 10 fringes).Also save a “long” set with a large number of fringes. You will need these later.

Using the information in Appendix A.1 calculate the wavelength of the laser. Estimate the uncer-tainty in your result, recalling that one step of the motor corresponds to a 48 nm change in OPD.

6.4.2 Two wavelength source

Change the light source to 2 RED LASERS. This source consists of two solid state lasers havingslightly different wavelengths. Because of the way the diodes are mounted in the housing thecontribution from each diode to the total amount of light reaching the photodetector will change ifthe light source is moved or rotated. Do not change the position of the source when taking yourdata!

Question 3: Based on the discussion in Appendix A.2 what do you expect the fringe pattern to looklike (make a sketch)? Explain how you can use this pattern to determine the two wavelengths of thediode lasers.


Using the DA system record a sufficient amount of data to allow you to calculate the wavelengthsof the diode lasers, and estimate the uncertainty in your results.

Question 4: What is the largest uncertainty in this measurement?

Question 5: Measure the peak-to-peak amplitude of the envelope where it is a maximum and whereit is a minimum. Use this information to estimate the contribution of each diode to the total signalfor your data set.

C2 .

6.4.3 Gaussian light sources

The sources labeled INFRARED, RED, YELLOW, GREEN, BLUE, and VIOLET LED are all lightemitting diodes. Their light output is, to a good approximation, Gaussian (see Appendix A.3). Oneproperty of a Gaussian function is that its Fourier transform is also a Gaussian, so the envelope ofthe fringe signal will be Gaussian. In optics it is often convenient to measure the width of suchGaussian sources in terms of the Full Width at Half Maximum (FWHM) and we will do so here.

Choose any two of the light sources and record their interference signals. Make sure you includeenough data points; i.e., extend your measurements on both sides of the x = 0 position until thevisibility drops to zero. As shown in Appendix A.3, if ω is the FWHM of the fringe envelope theFWHM of the spectral intensity distribution, ∆σ, can be determined. Likewise, the fringe periodcan be used to determine the central wavenumber of the distribution, σ0.

Use your interferograms to calculate σ0 and ∆σ for the two light sources you chose. Finally, convertyour results from wavenumbers to wavelengths.

Question 6: “If the FWHM in wavenumber is ∆σ and λ = 1/σ then the FWHM in wavelength willbe approximately 1/∆σ.” What is the fallacy in this argument? Derive a more correct relationshipbetween ∆λ and ∆σ.

C3 .

6.5 Fourier transform spectroscopy

You will now determine the spectral distribution of the light sources by taking the Fourier transformof your data numerically.

1. Export your data using the Export for FFT option (see section 6.3)

2. Run QtiPlot and chooseFile→Import ASCIIto import your data.

3. The first column will have the OPD in nanometres and the second column will be the mea-sured intensity in arbitrary units. The fringe pattern has the form

I(x) = I0[1 + V (x) cos(2πσ0x)]


If we were to take the Fourier transform of the raw data (Eq. 2) there would be a very largesignal at σ = 0. This is due to the constant term (the Fourier transform of a constant valueis a delta function at σ = 0). Because of the way the FFT works the whole spectrum willbe scaled to this peak and the interesting part of the spectrum will practically disappear. Tofix this, calculate the average value of the second column and subtract it from your data. Theprocedure for doing this is as follows:

(a) Click the right mouse button on the second column and choose the Statistics on Columnscommand. A new table should be displayed.

(b) Copy the mean value from the third column.

(c) Delete the table with statistics and now click the right mouse button on the empty spacein the table with your data. From the popup menu choose Add New Column. A newempty column should appear on your spreadsheet.

(d) Click the right mouse button over this new column and choose Set Column Values. . .

(e) Set column values as “col(“Intensity”)-(Ctrl+V)”, where Ctrl+V means to hold downthe Control key while pressing the V key. This will paste the value of the averageintensity that you copied from the statistics table.

The result of this procedure is that your new column contains the (real) fringe signal S(x).

4. To find the FFT of your data follow these instructions:

(a) Highlight the new column and choose Analysis→FFT. . . . A table with FFT data anda plot window should appear. If the table “Forward FFT Values” is not visible useView→Project Explorer and double click over the hidden table.Note that the x-axis on the plot is labeled Frequency (Hz) because QtiPlot assumesthat the data was taken in the time domain. Rename this to Wavenumber (nm−1).

(b) To view the data as a function of wavelength instead of wavenumber add a new columnin the FFT window and set its value to be the reciprocal of “col(“Frequency”)”.

(c) Copy the full range of values (300 to 1200 nm) from the new column.

(d) Open a new table and paste the copied data to the first column.

(e) Copy the corresponding data from the the “Amplitude” column in the FFT window andpaste it into the second column in the table.

(f) Select both columns and use Plot→Line+Symbol from the menu to plot the spectrum.

(g) Change the range by changing the scale of the x-axis so that it better fits a particularspectrum.

(h) In a similar way, print out the plots of the FFT as a function of wavenumber.

5. Print and paste all your spectra in your logbook.

Estimate the period of the highest frequency in each of your interferograms and use the Nyquistcriterion to decide whether or not your data may be affected by aliasing.

From the tabulated data you can also estimate the spectral resolution of the FFT algorithm. Comparethis resolution with the width of the spectral features that you measured. Are your data affected by“leakage?” Measure the FWHM for the RED laser for both the “short” and “long” data sets thatyou recorded. Are the two widths the same or different? Explain your observations.


Use the position and width of the main peak in the FFT to estimate the mean wavelength, meanwavenumber, FWHM in wavelength and FWHM in wavenumber. Compare these with the valuesyou obtained by assuming that the spectral distributions are Gaussian.

C4 .

6.6 Optional: The white LED

Change the light source to the WHITE LED and obtain its spectrum as you did for the other LEDsources and analyse your results as before. First, estimate the FWHM of the spectrum by assumingthat it is a Gaussian.

Question: Do you think that this is a good assumption? Have a close look at the recorded interfer-ogram before answering!

Use the FFT to find the spectral distribution.

Question: What is the explanation for the observed profile? Hint: From your previous work youknow that LEDs generally emit light in a comparatively narrow range. Can you identify a featurethat may correspond to the LED? What is its wavelength compared to any other features in thespectrum? You should now be able to give a plausible explanation of the operation of a “white”LED.

C5 .


Appendices

A Some examples of Fourier transforms

We use Eq. (5) to calculate the interference pattern expected for some simple situations.

A.1 Monochromatic light

In the case of monochromatic radiation of wavelength λ0 the mean wavenumber will be σ0 = λ0

andI(s) = δ(s)

where δ(s) is the Dirac delta-function. It follows that V (x) = 1 and the interference signal will be

S(x) = e−2πjx/λ0

The real signal will be proportional to cos(2πx/λ0).

A.2 Two wavelengths of equal intensity

We assume that the light consists of two “lines” of equal intensity. The real fringe signal will besimply

S(x) = [cos(2πσ1x) + cos(2πσ2x]/2 (8)

This is completely analogous to the formula for acoustic beats, with σ1,2 corresponding to thefrequencies of the two sound waves and x corresponding to t. We can use the standard trigonometricsum-to-product formula to write this as

S(x) = cos

(2πσ1 − σ2

2x

)cos

(2πσ1 + σ2

2x

)The fringe frequency is σ0 = (σ1 + σ2)/2 and is modulated by a slowly varying cosine functionthat has a frequency equal to (σ1 − σ2)/2.

Let X be the distance between two successive maxima (or minima) of the visibility. The full periodof the visibility function will be 2X and its frequency will be (2X)−1. It follows that

σ1 − σ2 = X−1

Suppose that n fringes span the distance xn. The fringe frequency will be n/xn and

σ1 + σ2

2=

n

xn

We can now solve for σ1 and σ2 (or λ1 and λ2):

σ1 =1

λ1=

n

xn+

1

2X(9)

σ2 =1

λ2=

n

xn− 1

2X(10)


A.3 A Gaussian spectrum

Suppose that the light source has a Gaussian intensity distribution:

I(σ) = exp

{−4 ln 2

(σ − σ0)2

∆σ2

}

where σ0 is the mean wavenumber and ∆σ is the full width at half maximum (FWHM). The Gaus-sian has the very useful property that its Fourier transform is also a Gaussian and the fringe signalcan be easily calculated:

S(x) = e−π2x2∆σ2/4 ln 2e−2πjσ0x

We can write this result asS(x) = e−4 ln 2(x2/ω2)e−2πjσ0x

where ω is the FWHM of the interference signal. Comparing these last two equations allows us todetermine the FWHM of the intensity distribution from ω:

∆σ =4 ln 2

πω−1 = 0.883ω−1 (11)

The units of ω and ∆σ are reciprocal: if ω is measured in nanometres ∆σ will have units of nm−1.

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