Assignment for Lectures

21
Assignment-1 Gaussian beams 1.Calculate the collimation range (confocal beam parameter) for three Gaussian beams with w o1 =3mm, w o2 =4cm and w o3 =30cm for the case of i) a CO 2 laser =10.6 m and b) a frequency doubled Nd: YAG laser =532 nm. Present the results in a table. (1.5 marks) Ans. Collimation Range (2z o ) = Collimation Range (in m) wo (in mm) CO2, λ=10.6 μm Nd:YAG, λ=.532 μm 3 5.33 106.29 4 9.48 188.97 300 53347.80 1062944. 88 2.Derive analytically the expression for the optimum w o of a monochromatic Gaussian beam in order to achieve the minimum spot size w(z) at a distance z away from the source. Apply the result to calculate w o for =488 nm and z=1000 km. What is the value for the spot size w(z) at that distance? (2 marks) Ans. For optimum w o differential of w(z) with respect to w o should be 0.

Transcript of Assignment for Lectures

Page 1: Assignment for Lectures

Assignment-1 Gaussian beams

1. Calculate the collimation range (confocal beam parameter) for three Gaussian beams with wo1=3mm, wo2=4cm and wo3=30cm for the case of i) a CO2 laser =10.6 m and b) a frequency doubled Nd: YAG laser =532 nm. Present the results in a table. (1.5 marks)

Ans. Collimation Range (2zo) =

Collimation Range (in m)

wo (in mm)

CO2, λ=10.6 μm

Nd:YAG, λ=.532 μm

3 5.33 106.294 9.48 188.97

300 53347.80 1062944.88

2. Derive analytically the expression for the optimum wo of a monochromatic Gaussian beam in order to achieve the minimum spot size w(z) at a distance z away from the source. Apply the result to calculate wo for =488 nm and z=1000 km. What is the value for the spot size w(z) at that distance? (2 marks)

Ans.

For optimum wo differential of w(z) with respect to wo should be 0.

For =488 nm and z=1000 km, minimum wo = 0.394m.

3. Calculate the size of the mode on the mirrors of a symmetric (R1=R2=1.5d) Nd:YAG laser (=1064 nm ) cavity with length d=30 cm. What should be the minimum size of the mirrors in order to have a loss of less than 0.001% per round trip? (2.5 marks)

Ans. For stable symmetric resonator minimum spot size should be at centre. Distance of mirror from centre is 0.5d.

Page 2: Assignment for Lectures

Using, for R(z)=1.5d at z=0.5d, d=30cm and =1064 nm we

get =.268mm

Radius of the mode, W(z)=.328mm

For .001% round trip loss the loss at each mirror should be 0.0005%. Radius of the aperture of mirror = .810mm

4. A common method for the evaluation of the spot size of a Gaussian beam consists of measuring the power transmission of the beam that is being partially obstructed by scanning a knife edge as shown in the following image. Show analytically how this measurement is possible. Assume that:

. Consider movement of the knife edge along the x-axis (4

marks)

Ans. The Intensity function of the Gaussian beam is same as the Gaussian distribution function in two variables which are independent of each other and hence can be separated into two independent Gaussian functions.

In knife-edge experiment intensity of the beam passing to the detector can be obtained by integrating the above equation for -∞<y<∞ and -∞<x<x1, (erf(x)). This will give the error function in the x-direction as a function of knife edge length. If we plot this function on graph, it will be a monotonically decreasing graph and the point at which the intensity falls to half of the maximum intensity will be the centre point. In Gaussian beams, the distance of wo/ relative to the centre corresponds to the radius of the circle containing 68.3% of the total intensity (equivalent to standard deviation). The two points

x

zwo

Page 3: Assignment for Lectures

where the intensity falls to 66.85% and 34.15% of the maximum are the points corresponding to wo/ distance from centre of the beam.

Page 4: Assignment for Lectures

Assignment 2- Pulse Propagation

Ans. Time-bandwidth product shows the uncertainty principle that exists in all the Fourier-transform pairs that is they cannot simultaneously be determined accurately. The very small temporal width (ultra-short pulse) will have very high bandwidth and vice versa. It is the product of the error/broadening in time and bandwidth which can be minimised to low values.

Ans.

On simplification of given expression, we get

In this the lifetime of the pulse , dependent on the distance given by the expression,

Page 5: Assignment for Lectures

Now, using , we get the variation of time width of the pulse with distance as

Ans.

(i)

The green line shows the straight line and the blue curve shows the variation of with respect to . The figure shows that as increases the pulse

broadening decreases.

Page 6: Assignment for Lectures

(ii)

For optimum differential of with respect to should be 0.

(iii)Using the above expression of for minimum pulse length at specific length.

Ans. By chirping and compressing pulses we concentrate the energy which increases the peak pulse power with average power remaining the same. The figure shows the basic schematics of effect of pulse compression. On decreasing the pulse width the peak intensity increases but the average intensity remains the same. Due to average power/intensity being the same, which is a measure of heat produced in the system, there is no damage to the instrument.

time

Page 7: Assignment for Lectures

Assignment 3- Fibre Communications

1. What are the advantages of DPSK modulation over OOK? What are multilevel modulation formats and what is their attraction? Give some examples from the literature of optical communication system demonstrations that have used such formats and describe the benefits that they have allowed. (~300 Words) [6]

Ans. The advantages of differential phase-shift keying (DPSK) modulation over on-off keying (OOK) modulation are:1. They require much lower optical power than intensity modulated signals (OOK).2. They have higher receiver sensitivity than OOK signals.3. DPSK signals have higher tolerance to nonlinear effects.4. DPSK signals are more resilient to additive noise.

Achieve higher data rate using same bandwidth at the cost of immunity to noise

2. a. Draw the eye diagram of a NRZ-OOK signal. Show how the rise and fall time of the modulator used to generate this signal can be measured using this diagram.

Ans. The eye diagram for a NRZ-OOK signal with be

b. Do the same for an NRZ-DPSK signal, when detected directly using a photodiode. Similar to (a), how does the finite rise/fall time of the data modulator affect the appearance of the eye diagram?

Ans. The finite rise and fall of the time shows the exclusion of higher frequencies in the Fourier transform signal. Higher frequencies correspond to the sharpness of edges in time domain.

c. Draw the constellation diagram for the QPSK and 8-QAM format and show example bit combinations assigned to each symbol. Sketch how the intensity and phase of the signal would change with time for each of these symbols. [1+1+2]

Ans. There will be no change in the intensity for quadrature phase-shift keying (QPSK) signals.

Peak to average power ratio of the constellation diagram should be minimum in order to minimize the effect of amplifier non-linearity (saturation). Higher

PAPR demands that the amplifier be linear over larger range of input.

Page 8: Assignment for Lectures

Assignment 4- QPM Materials

i) Calculate the PPLN period required to quasi phase match 1047nm to 523.5nm at 152 centigrade.

Ans. For first order quasi phase matching,

Using Sellmeier’s equation for PPLN to obtain refractive index,

We get =2.246 and =2.162

From this, =0.502 or =12.5

ii) Calculate the expected acceptance bandwidth of a 2.5cm long PPLN crystal for this process - assuming a plane wave interaction.

Ans. The acceptance bandwidth is 2.8nm.

iii) For this interaction estimate the approximate fundamental power required to yield 40mW of second harmonic using the formula in the hand-out.

Ans. Taking the conversion efficiency of PPLN =2%/(W*cm) as given, the power required for 40mW yield in 2.5cm long crystal will be=(4/(2.5*2))Watts= 0.8W

Page 9: Assignment for Lectures

iv) Calculate the PPLN period required for an OPO pumped with 1064nm, and giving an idler wavelength of 3.79 microns.

Ans. The period of PPLN required for first order QPM in OPO is

Where, coefficients p, s and i represents pump, signal and idler respectively of the optical parametric oscillator. Using energy conservation equation, we can get the signal wavelength.

Using Sellmeier’s equations for T=152oC,

=2.1619, =2.1457 and =2.0713

=0.219 or =28.65

Page 10: Assignment for Lectures

Assignment -6 Laser Physics

1. For a system in thermal equilibrium calculate the temperature at which the rates of spontaneous and stimulated emission are equal for a wavelength of 9μm, and the wavelength at which these rates are equal at a temperature of 4500K.

Ans. For spontaneous emission rate to be equal stimulated emission rate, ratio of spontaneous to stimulated emission should be 1.

or

For =9 , T=2309K and for T=4500K, =4.6

2. If 0.25% of the light incident onto a medium is absorbed in 1mm, what fraction is transmitted by the medium if it is 0.1m long? Calculate the absorption coefficient of the medium.

Ans. Assuming exponential decay in the medium,

At l=1mm, I/Io=.9975. Using this,

Fraction transmitted at l=100mm = = (0.9975)100=.7785

Absorption coefficient, =2.5m-1

3. If the irradiance of a beam of light increases by 20% after a round trip through a gain medium, which is 0.3m long, calculate the small signal gain coefficient k, assuming no losses.

Ans. Assuming exponential function of the gain, ,

In round trip, l=0.6m, I/Io=1.20. Using this,

Page 11: Assignment for Lectures

Absorption coefficient, =0.3038m-1

4. Given that the lifetime of the upper level of the 632.8nm transition in the HeNe laser is 1.0 x 10-7 s, calculate the degree of population inversion required to give a gain coefficient of 0.06 m-1 ignoring line broadening effects.

Ans.

Putting the values in this to obtain , we get =

5. In a ruby laser (λ=694.3 nm) the Ruby crystal is 0.1m long and the mirrors’ reflectance are 0.99 and 0.9. Given that the losses are 9% per round trip, that the spontaneous lifetime of the upper laser level is 3 x10-3s, that the linewidth is 1.5 x 1011Hz and that the refractive index is 1.78, calculate the threshold gain coefficient and population inversion.

Ans. Threshold gain is the gain when total loss is equal to the gain from the medium.

Round trip loss=9%

=0.471 m-1

Where, is the cavity round trip loss per unit length.

Threshold population inversion

Page 12: Assignment for Lectures

Assignment-7 (Solid-State Lasers)

Figure 1 shows a simple fibre-coupled Nd:YAG laser source operating at 1064nm. The laser system comprises a fibre-coupled diode pump source, a Nd:YAG laser and an optical arrangement for coupling the light from the Nd:YAG laser into a single-mode optical fibre.

The fibre-coupled pump diode has a maximum output power of 4W at 808nm and produces an output beam with a nearly Gaussian transverse intensity profile of beam radius, 25µm, and far-field beam divergence (half-angle) of 0.4 rad. The output from the delivery fibre is collected by a lens of focal length, f1, and focussed into the centre of the Nd:YAG crystal by a lens of focal length, f2.

The Nd:YAG laser employs a simple resonator comprising three mirrors (M1, M2 and M3) and a 2mm long Nd:YAG rod mounted in a heat-sink next to mirror M3. M1 has a radius of curvature of 150mm and high reflectivity (100 %) at the lasing wavelength (1064nm). M2 also has high reflectivity at 1064nm, and, in addition, high transmission (~100%) at the pump wavelength (808nm). M3 has a reflectivity of 90% at 1064nm and 0% at 808nm, and serves as the output coupler. The total resonator length (i.e. from M1 to M3) is 80mm, and the round-trip loss (excluding the output coupling loss) is 5%

The output from the Nd:YAG laser is collected by a lens of focal length, f3, and focussed into a single-mode fibre for beam delivery by a lens of focal length, f4. The fibre has a step-index core of radius 4µm and numerical aperture of 0.1. The positions all the lenses are as shown in figure 1.

1. Estimate the beam quality factor (M2) for the fibre-coupled pump source. If f1=20mm and f2=120mm, then calculate the waist radius and the Rayleigh range of the pump beam in the Nd:YAG rod.

Ans. Far-field beam divergence=

Initial waist raidus =25μm

38.88

Beam radius after f2 = =150 μm

Far-field beam divergence, = 0.067 rad

Page 13: Assignment for Lectures

2. Derive an expression for the TEMoo beam waist radius in the Nd:YAG rod and calculate its value for the resonator described above.

Ans. TEMoo beam waist radius =24.06 μm

3. Write down a suitable expression for threshold pump power for the Nd:YAG laser (see lecture notes) and calculate its value.

Ans.

4. Estimate the maximum output power for the Nd:YAG laser. You may assume that the spatial overlap factor ηPL=1.

5. Derive an expression for the optimum output coupler transmission for the Nd:YAG laser and calculate its value. Estimate the maximum output power using the optimum output coupler transmission.

6. If f3=100mm, then what value for f4 should be used to ensure efficient launching of the TEMoo output from the Nd:YAG laser into the fibre. (Note: You may neglect the effect of spherical aberration).

You may assume the following for the Nd:YAG laser:

Absorption coefficient for pump light at 808nm (αp) = 300m-1

Upper laser level lifetime (τ) = 230µs

Stimuled emission cross-section (σe) = 3.4×10-23m2

Refractive index of Nd:YAG = 1.82

Page 14: Assignment for Lectures

Assignment-8 (Waveguides)

Using the effective index method, design a strip-loaded waveguide, as in figure 1, which is at the cut-off point for propagation of a second TE mode at 1.5µm by the following steps:

a) Draw how the structure can be split into 3 guides corresponding to x<-w/2, -w/2<x<w/2, and x>w/2, where x=0 corresponds to the centre of the strip. (1 mark)

Ans. The waveguide structure is seen in y-direction, with high refractive index layer surrounded by the low refractive index layer. The layer structure in y-direction is dependent on the x-direction. This can be effectively seen as three waveguides.

Refractive index variation in x and y directions

b) Show that the normalised frequency parameter for which the central guide no longer guides 2 modes is ν=π. (1 mark)

c) Calculate the depth of the core, d1, at which the second mode in the y-axis is just cut-off. (1 mark)

d) Using the normalised dispersion curve, estimate the effective index for the fundamental mode. (1 mark)

e) What is the spot size of the fundamental mode in the y-axis? Spot size is here defined as the 1/e2 half-width of intensity. (5 marks)

d1d2

w

x

y

z

n=1.515

n=1.52

n=1.515

n=1.0

X

Y

d1d2

w/2-w/2

Page 15: Assignment for Lectures

f) If we approximate that the physical extent of the mode is 3 times the spot size, how thick must we make the strip, d2, in order to avoid significant mode intensity at the air interface? (1 mark)

g) Using the normalised dispersion curves, estimate the effective index of the fundamental TE mode of the side guides of the split structure drawn in part a). (2 marks)

h) Draw the effective structure that enables the guidance in the x-axis to be solved. (1 mark)

i) Calculate the width, w, of the strip at which the 2nd mode is just cut-off. (2 marks)

j) Calculate the effective index of the fundamental mode (neff)0,0 . (1 mark)

k) Calculate the spot size of the fundamental mode in the x-axis. (4 marks)

Page 16: Assignment for Lectures

Figure 1: Fibre-coupled Nd:YAG laser