OPTICAL Measurementswednesdayjuly 22Nd2015

OPTICAL MEASUREMENTSWednesdayJuly 22nd2015

Prof. Cesare Svelto4thExamYR 2014/2015

Time available1hr40minroom T.2.2at9.15

SURNAME: ______Name: ______(block capital)

MS Degreeand year: ______
POLIMI-ID number and readable Signature ______

POINTS (per exercise)(7+6+8+6+6=33p)

YOU MUST cross all subpointsyou gave, even partly, an answer[e.g. 1a), 1c), 1d) 2a), 2c), 3b) etc.].

YOU MUST be able solving all problems in order to conclude and deliver the test (no blank exercises).

SOLUTIONS

(25 min)Exercise 1

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1a)Provide a drawing of the 3level and 4levellasing schemes and explain in some detail how the LASER action can happen in these two different configurations.

Give specific numbers/information on the Nd:YAG levels and pump/lasing wavelengths and comment.

1b)List, also with brief numerical examples, the main properties of laser sources making them uncoparable to traditional light sources.

1c)Describe the causes of laser frequency instability in the case of a diodepumped solidstate laser.

How can we limit (passively) or reduce (actively) the laser frequency noise?

An Nd:YAG laser with linear cavity has an optical length variable between 20cmand50cm, with a temperature coefficient =(L/L)/T=4×106/°C. Calculate the laser frequency shift for a temperature variation of +5°C?

1d)Finesse(F) and Free Spectral Range(FSR) of a Fabry-Perot interferometer: provide definitions and formulas of these parameters in the case of an optical resonator made of two identical planeparallel mirrors.

Two mirrors, plane and parallel and each with a power reflectivity R=80%, are facing each other at the extreme points of the diameter of one circle havingarea 78.54m2. What is the FreeSpectralRange of the interferometer? What is the Full Width at Half Maximum of this Fabry-Perot transmission peaks?

21a)See the notes, slides, Book of the Course.

21b)See the notes, slides, Book of the Course.

31c)See the notes, slides, Book of the Course.

For the Nd:YAG laser, it is well known that the emission wavelength is 1064nm and hence its optical frequency is =c/282THz300THz. For a temperature variation of +5°C of the laser cavity, we shall have a relative length variation (L/L)=T=20×106. Since the relative frequency variation for a laser is equal, apart from the sign, to its relative optical cavity length variation, (/)=(L/L), we immediately obtain the corresponding absolute frequency shiftas =(L/L)5.64GHz6GHz.

31d)See the notes, slides, Book of the Course.

NamingRthe power reflectivity of the mirrors, set at a distance L, the Finesse isF=while the FreeSpectralRange is FSR=c/2L.

In the specific case discussed, we can dind the diameter of the circular suface of area A=78.54m2simply as D=(4A/)1/210m. Thus the Fabry-Perot has a Free Spectral Range FSR=c/2D=15MHz.

The Finesse is F=14and hencethe transmission linewidth (FWHM) is FWHM=FSR/F=1.07MHz1MHz.

(15 min)Exercise2

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2)An optical instrument for laser alignment uses a Nd:YLF laser source, frequency doubled in the green, with TEM00output beam and beam waist w0L=100µm. The collimating optics is a telescope with lenses of focal lengths f=20mm andF=200mm. We ant to achieve an optical beam with maximum collimation over a range of 500m.

2a)Evaluate the laser spot size in the center point and at the extreme limit of the measurement range.

2b)What is the plane divergence angle (div) after the telescope? And the corresponding solid angle (div)?

2c)State the magnification M of the telescope. At which distance from the laser shall we place the telescope in order to obtain the requested collimation?

42a)To achieve optimal collimation over a measurement range z*, we must have at the range center a beam waist with dimension w0===9.2mm (spot diameter 18.4mm).

At the extreme limits of the range the spot size is w=w0=13mm (spot diameter26mm).

32b)The laser beam divergence ngle in the measurement range is div=/(w0)=18rad.

The corresponding solid angle is div=(div)2=1.1109sr.

32c)The telescope magnification is M=F/f=10.

Naming dthe distance between the laser (with output spot w0L=100µm) and the input of the telescope, in order to achieve an image with spot size w0=9.2mm in the center of the range (distance Z=500m from the telescope output), the following relation must hold:

w0

Hence, the requested distance dis obtainable as

d==0.54m=54cm

(25 min)Exercise 3

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3)We are working with the Michelson interferometer shown in the Figure, where M1 is the reference mirror and M2 the measurement mirror. The mirrors are ideal, with 100% reflectivity, and we have: L1=10cm, L2=2m. The interferometer is read using an HeNe laser (=632.8nm) with single longitudinal and transverse mode, having an output power of 5mW and linewidth =300kHz.

3a)Mirror M2 is moved by s=25m, getting closer to the laser. How many interferometric fringes do we observe in the waveform at the photodiode (FD) output?

3b)Which is the frequency of the interferometric signal if the mirror M2 moves with constant velocity v=20cm/s?

3c)We want to measure small vibrations of the mirror M2 (with amplitude much small than the laser wavelength), operating the interferometer in quadrature. What is the minimum vibration amplitude correctly detectable?

3d)Evaluate the fringes visibility if the beamsplitter (BS) has a power transmission of 66% and no losses.

3e)What will happen if the mirrors are still and we change in a continuous way the laser wavelength for a global deviation =0.05nm? Can we measure the length L2? If yes, write the measurement equation.

23a)The number of interferometric fringes is: N=s/(/2)=25m/0.3164m=79fringes80fringes.

23b)The frequency of the interferometric signal is: f=v/(/2), where vis the target velocity. We hence obtain: f=(0.2m/s)/(316.4109m)632kHz.

Alternatively, from the Doppler shiftDoppler/0=2v/cwe get f=Doppler=2(v/c)0= v/632kHz.

The factor 2 arises from the fact that the Doppler frequency shifti is seen twice: once from the target, when the e.m. wave impinges on it, and a secon time when the e.m. wave is emitted (diffused) from the moving target. In both cases the target is moving with velocity vand the Doppler shift is Doppler/0=v/c.

33c)The measurementof small vibrations is limited by the laser linewidth, since the interferometer is unbalanced (L1L2). The phase noise can be calculated as:

The interferometrica phase is =2k|L1-L2|=2|L1-L2|/(/2)=22|L1-L2|/cand if the frequency  undergoes fluctuations , the corrisponding phase noise is n=4|L1-L2|/c24mrad.

The minimum detectable amplitude of vibration can be obtained by the proportion:

smin:(/2)=n:2, hence getting: smin=(/2)n/2=1.2nm.

13d)Poiché il beamsplitter non ha perdite, e gli specchi sono ideali, il campo proveniente dal cammino di misura e quello proveniente dal cammino di riferimento hanno la stessa ampiezza. Quindi, la visibilità del segnale interferometrico è pari a 1.

23e)Since the phase shift between the measurement and reference arms is:

=2k|L1-L2|=2(2/)|L1-L2|

introducing a small wavelength variation we get a phase variation:

=-(4/2)|L1-L2|=-2981rad

This phase variation produces a number of interferometric fringes on the photodetector:

N=||/2474fringes.

Since this nmberof fringesNis proportional to the quantity |L1-L2|, knowing the length of the reference arm L1 and counting N, it is possibile to get the measurement arm length L2, as:

L2=L1N(/2)(/)

(20 min)Exercise 4

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4)Using a Laser Doppler Velocimeter e want to measure the speed of a fluid, seeded with small scattering particles, while flowing within a transparent pipe. The fluid velocity range is between 1cm/s and 10m/s and we want to achieve a measurement resolution v=5mm/s.

4a)If the projecting lens has a focal length f=400mm, do evaluate the offaxis separation (2h) of the two incident beams in order to have a 10m fringe spacing in the focusing/interference region?

4b)What photoreceiver bandwidth is required?

4c)What kind of spectrum analyzer, and why, shall we use to detect the velocity (and frequency) profile?

4d)In a specific experiment, the fluid average velocity is 2m/s, with a Gaussian profile (v=0.1m/s).

Which is the corresponding average detected frequency?

What frequency resolution and span (fSTART and fSTOP) do we need for the spectral measurement?

24a)The two interfering optical beams should cross at an angle 2 such that

and hence==arcsin(32×103)32mrad1.8°

The two beams intersection angle is thus 2=3.6° and it must be 2=(2h)/f so that 2h=2f25mm.

34b)Given the dynamic range [vmin,vmax] needed for the measurement, the frequencies to be measured can be obtained from the equation:

so we getfD,min==1kHz andfD,max==1MHz

The photoreceiver shall hence have a bandwidth of 510fD,max=510MHz.

24c)For the frequency range up to 10MHz we need an heterodyne spectrum analyzer or a custom “fast” FFT spectral analysis after A/D conversion (fc>20MSa/s).

34d)The average detected frequency is fD,ave=vave/D=(2m/s)/(10m)=200kHz.

The required velocity resolution v=5mm/s requires a corresponding frequency resolutionfD=v/D=(0.005m/s)/(10µm)=500Hz. The distribution standard deviation corresponds to a frequency deviation fD=v/D=(0.1mm/s)/(10µm)=10kHz: therefore, the frequency span (taking the Gaussian profile at a width ±35) can be chosen as 200kHz ±3050kHz. We could choose a frequencyspan of 100kHz with fSTART=150kHz and fSTOP=250kHz.

(15 min)Exercise 5

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5)An ultracompact Optical Spectrum Analyzer(OSA) uses a diffraction grating to disperse different wavelengths of the incident spectrum toward a linear array CCD made of InGaAs (spectral sensitivity from 800nm to 1700nm). The CCD array, set at a distance L=30cm from the reticle, has 1024 pixel over a linear size DCCD=1cm. This OSA, for telecom applications, was designed to correctly measure optical signals spectra from 1200nm to 1610nm.

5a)Evaluate the dispersive power/coefficient (K=?mrad/nm) of the diffraction grating.

5b)Calculate the spectral resolutions, absolute and relative, at an operating wavelength=1550nm.

How do these figures compare to the spectral resolution of a good laboratory OSA?

5c)If the ultracompact OSA needs to be placed inside a PC board (having maximum length slightly greater than 10cm), how can we obtain the distance L=30cm between the grating and the CCD? Strengthen your answer with a schematic drawing of the laser beam path inside this OSA.

45a)The measurement range is =(1200nm-1610nm)=410nm. The diffracted beam from the reticle will diverge, after propagation by a length L, to a transverse dimension DCCD=L=(K)L. We hence obtain K=(DCCD/L)/=(1/30)/(410nm)0.08mrad/nm.

35b)The dimension of a single pixel on the CCD (spatial resolution) isd=DCCD/10249.8m10m. The corresponding (absolute) spectral resolution is =(d/L)/K=0.4nm, also achievable from =/1024=0.4nm, which is independent from the working wavelength. The relative spectral resolution, working at a wavelength =1550nm, is/=2.6104.

Agood laboratory OSA can achieve a spectral resolution OSA,Lab=2pmthus being 200 times better than the ultracompact OSA with =0.4nm.

35c)In order to achieve an overall lengthL=30cm having a linear space of approximately 10cm, we can use two “intermediate” reflections on two separate mirrors set at 10cm distance apart.The first mirror (set at 10cm from the grating) will reflect the beam exiting the grating toward the second mirror, which (set at10cmfrom the CCD) will reflect the beam toward the CCD. The same result can be achieved with more multiple reflections using more mirrors spaced less than 10cm.

Insert the DRAWING with the two reflections

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