Chapter One General Introduction 1.1 Introduction

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Laser technology is one of the most rapidly developing areas in modern technology, when the laser was invented in 1960, it was classified as Solution in search of a problem, and today laser technology is applied in many different areas. The number of applications of laser is enormous, and it is not possible to explain all of them her. In this course, the laser applications are divided into four groups as shown as below[1]. التطبيقات الصناعية .Industrial Applications .1 Medical .3 التطبيقات العلمية والقياسات .Metrological & Scientific Applications .2 Military.4التطبيقات الطبية .Applications التطبيقات العسكرية .Applications But at first we explain by details the material & laser parameters that effected on the laser applications. And our hope is that with time we will fill the missing information on most of the well known applications of lasers.

1.2 Material & Laser Parameters Material & Laser Parameters that affect laser processing are discussed in this section[2]. 1.2.1. Material Parameters 1. Reflectance (R): Is the ratio of the power reflected from a surface to the power incident on it[2].

n  n 2 R  ( 1 2 ) ...(1.1) n1  n2

Where: n1 & n2 are the refractive index of the substrate material and incident medium, respectively. The amount of light that absorbed by the metal equal to (1- R)% .

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Metals like copper & silver have high reflectance’s that increase with wavelength and, it is found that the reflectance of metals decreases as the temperature nears the melting point.

2. Absorption Coefficient (): Is the fractional loss of light power per unit distance for light traveling in a nonmetallic material. Beer’s (Lambert’s) law relates power to absorption by[2]:

Z P  Poe ...(1.2) Where:

P: is the power at depth (Z), Po: is the power entering the surface.

3. Heat Capacity & Specific Heat a- Heat Capacity (c): Is the amount of energy needed to rise the temperature of unit mass (sample) one Celsius degree (1oC). From this definition, we see that if energy Q produces a change T in the temperature of a sample, then[3]: Q = c T ...(1.3) b- Specific Heat (C): Is the heat capacity per unit mass. Thus, if energy Q transfers to a sample change by T, then the specific heat of the sample is [3]: Q C  (J/kg.oC) ...(1.4) mT

4. Thermal Conductivity (K): Is the heat flow per unit area per unit thermal gradient. And indicates to the units for K are (W/cm.Co)[2]. 5. Thermal Diffusivity (k): is a measured of how much temperature rise will be caused by a pulse of heat applied to the material. Units are generally (cm2/s). By using the laser, material melting depended on thermal flow of it. And this flow had a directly relation with thermal conductivity (K), and reversely with Specific Heat (C) and density of the material k = K/C. [2].

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6. Latent Heat (L):Refers to the amount of heat required to cause a change of phase of unit mass of material. Because this added or removed energy dose not result in a temperature change, the quantity L is called latent heat[3].

Lm: Is the Latent heat of fusion or the energy required to case melting of unit mass (J/kg).

LV: Is the latent heat of vaporization and has the same meaning and units as Lm. Latent heat of vaporization are much large than Latent heat of fusion.

7. Transformation Temperature (T): Refers to the melting temperature (Tm), [2] Vaporization temperature (TV) & other phase change temperature (TP) .

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1.2.2. Laser Parameters

Laser parameters relate to both the laser beam and to the laser power output as a function of time. Normally, lasers are either operated CW or pulsed. The parameters of the beam that affect processing are as follows[2]: 1. Wavelength: Is the distance over which the wave repeats itself and is represented by the Greek letter  (lambda). Each color of visible light has it’s own characteristic wavelength. 2. Focused Spot Size. 3. Mode Structure (CW or Pulse). The wavelength affects absorption and reflection characteristics, and spot size & mode structure affect average irradiance & irradiance distribution in the spot.

1.3 Controlling the beam after it is emitted out of the optical cavity A known rule in optics is that the product of the beam diameter (d) and its divergence angle () is a constant. Thus, when the beam divergence needs to be reduced, the beam diameter must expand. The following pages describe the beam expander[1].

1.3.1 Beam Expander: Optical device increasing beam diameter and reducing divergence. Result a smaller focused spot for more distance between lens and workpiece. We saw the basic equations describing the beam divergence[1]:

...(1.5)

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Increasing the beam diameter (d) causes a decrease in the beam divergence angle (). Beam expander is based on the telescope developed by Kepler in the 17 century (see figure 1.1).

Figure 1.1: Beam expander based on the Kepler telescope[1].

The first positive lens has short focal length and small diameter, while the second positive lens has long focal length and large diameter. The distance between the lenses is exactly equal to the sum of the focal lengths of the two lenses. The laser beam enters the short focal length, and is focused to a real image at the focal point of the other lens. This image serves as a point source for the other lens. At the output of the second lens the beam has a larger radius, and a smaller divergence[1]. The relation between the beam diameters and the beam divergence angles is:

f1 d1  2   ...(1.6) f 2 d 2 1

f1: Focal length [m] of the input lens - ocular. f2 :Focal length [m] of the output lens - objective. d1 :Diameter of the input beam [m]. d2 :Diameter of the output beam [m]. 1:Divergence angle (Rad) of the beam at the input to the beam expander. 2 :Divergence angle (Rad) of the beam at the output to the beam expander.

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From equation (1.6) it is clear that the ratio between the beam diameters is directly related to

the ratio of the focal lengths of the lenses. In other ward the beam diameter d1 expanded to

f2 d2 by , and we can rewrite equation.(1.6) as below: f1

f2 d2  d1 ...(1.7) f1

f Where: 2 is the Expansion Factor (F ). f1

Example-1[1] The diameter of a beam emitted from He-Ne laser is 1.2 mm and its divergence angle is 1 mRad. A Kepler beam expander is used made of 2 positive lenses with focal lengths of 1 cm and 6 cm. Calculate: 1. The beam diameter at the output of the beam expander. 2. The beam divergence angle.

Solution 1. The beam diameter at the output of the beam expander is: -3 -3 d2 = d1*(f2/f1) = 1.2*10 m * 6 cm / 1 cm = 7.2*10 m 2. The divergence angle at the output of the beam expander (2): 2 = 1* (f1 / f2) = 1mRad * 1 / 6 = 0.17 mRad The beam expander caused a reduction of 6 times of the beam divergence (the ratio of the focal lengths of the lenses).

1.3.2 Beam Transport Beam transport is a technique that used for transporting laser beam from the laser to point where the beam is required is to send the beam down a flexible fiber- optical waveguide. Basically they consist of a small diameter (few hundred microns) glass or quartz fiber containing a core region where the refractive index is higher than the surrounding cladding, as shown in figure (1.2).

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A ray is able to travel down the core in a zig-zag fashion, undergoing total internal reflection at the core-cladding interface provide the angle it makes with the normal to the interface is greater than the critical angle[4].

Ray Path

Core Total internal reflection Cladding

Figure (1.2) Beam Transport down an optical fiber utilizing total internal reflection[4]

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Critical Angle & Total internal reflection When light crosses materials with different refractive indices, the light beam will be partially refracted at the boundary surface, and partially reflected. However, by increased the incident angle the refractive angle increased too, until refractive of light be parallel to the boundary surface, and at this point the incident angle called the critical angle (represent the boundary between two different phenomenon, the refraction & reflection). Then if the incident angle is bigger than the critical angle, all the incident light will be reflected and this is know total internal reflected, as shown as in figure (1.3)[5].

Where n1 < n2 n1 r 90

n2  c i r i

n2 Sin c = n1 Sin (90)

 Sin c = n1/n2 -1 c = Sin (n1/n2) Light Source

Figure (1.3) the Total Internal Reflection[5]

1.3.3 Beam Focusing A schematic diagram of the layout often used to direct the laser beam on to work piece is shown below[4]:

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Mirror Laser Beam Expander

2WL Focusing Lens

W0 Workpiece

Figure (1.4) Schematic layout of a laser beam delivery[4] system

Basically the beam is passed through a beam expander and then focused to a small spot on the workpiece using lens. The reason for incorporating a beam expander into the optical system is that w enables a smaller final spot to be obtained. The focused spot size ( o ) is given by[4]: f wo  ...(1.8) wL Where:

: Focus Spot Size (mm). f: The Focal Length of the Lens (mm).

 : Wavelength (µm). wL : The Beam Radius at the Final Focusing Lens (mm).

Another aspect of interest is the so-called Depth of Focus (Z) of the beam, which is the distance we can move the workpiece a way from the minimum beam radius and still have an acceptably small spot. ....(1.9)

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1 2 w 2  w(z)   2 Z  o   1   w   o  

Example-2[4]: It is required to focus the output of an Nd:YAG laser (=1.06µm) down to a spot of (50µm) radius. Given that the beam has a radius at the final focusing lens of 1mm). Calculate the focal length of the lens required, what will be the resulting depth of focus w(z) (Z)? Take = 1.1. wo Solution: f w w 1103  50106 w  o L o  f   6 148mm wL  1.0610

6 2 1  (50 10 ) 2  1.1 12 =3.39mm 1.06 106

Example-3[4]: If the spot radius for He-Ne laser (12.5 cm), & focal length lens is (1.5 cm) Calculate the focused spot size at (=632.8 nm).

f 9 2 w  (632.810 )(1.510 ) Solution: o  2  24nm wL  (12.510 )

1.4 Optical Processes When a laser beam hits matter, 4 processes can take place[1]: 1. Reflection - according to the law of reflection: The reflected angle is equal to the incidence angle. 2. Scattering- laser energy is scattered to all directions. 3. Transmission- laser beam pass through the material. 4. Absorption- laser beam is absorbed by the material.

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Only the last process (Absorption) can transfer energy to the material & causing a rise in temperature or chemical reaction.

1.5 Energy Balance Approximation Simple energy balance approximations frequently provide reasonable ball-park for many applications. For that reason, some discussion of this approach is presented here. In the figure (1.5) a schematic representation of laser beam focused onto the surface of a workpiece. If it is assumed that the material is heated to a depth (Z) with cross section area (a2), then the energy (U) required to bring the material to temperature (T) is given by[2]:

2 U  (CT  Lm  LV )a Z ...(1.10) Where (T) is used for temperature to avoid confusion with time(t) & represents the change in temperature of the part. It has been assumed in equation (1.7) that (  &C) are independent of temperature & are the same for liquid & solid.

Figure (1.5) Laser beam focused on a workpiece[2]

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Questions of Chapter One Q1.1 What are material & laser parameters that affect laser processing?

Q1.2:In each case, choose the best option: 1. Beam expander is an optical device increasing beam diameter and reducing divergence. Result a smaller focused spot for (a) small distance between collimated lens and workpiece. (b)more distance between collimated lens and workpiece. (c) more distance between laser and workpiece. (d) small distance between beam expander and collimated lens.

2. Critical angle represent the boundary between two different phenomenon, the: (a) scattering & reflection. (c) absorption & reflection. (b) refraction & transmission. (d) refraction & reflection.

3. The quantity L is called latent heat because this added or removed energy (a) result in a temperature change (b) result in a phase change (c) dose not result in a phase change (d) dose not result in a temperature change

Q1.3: Explain in briefly form the reasons for Incorporating a beam expander into the optical focusing system. Answer: The reason for incorporating a beam expander into the optical system is that enables a smaller final spot to be obtained.

Q1.4:Consider CO2 laser beam (=10.6µm) which has a radius of (5mm), and which is focused using a lens of (100mm) focal length. Calculate the minimum focused beam diameter & calculate depth of focus. w Answer ( o = 67m, Z = 0.6mm)

Q1.5:The diameter of a beam emitted from He-Ne laser is (1.2mm). A Kepler beam expander is used made of two lenses with focal lengths of (2cm) & (10cm). Then the beam output from this expander inter to focusing lens f =1.5cm. Calculate the focused spot size (Wo). Answer ( = 10m)

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Industrial Applications 2.1 Introduction

Industry accepted the laser as a tool soon after the laser was invented in 1965. The purpose of this chapter is to discuss the techniques used in material processing with lasers and to point out some of the advantages & disadvantages. The primary

concern is with heat treatment, drilling, cutting & welding.

2.2 Laser Drilling One of the very first industrial uses of the laser was reported in 1965 when a diamond die was drilling using pulse ruby laser. A hole 4.7mm in diameter & 2mm deep was made in about 15 minutes; using a mechanical process this had previously taken 24 hours[4]. When the laser beam is focused on the material, the speed of penetration will be[2,4]: H VP  ………………...... (2.1) (CTV  LV ) Where: 2 VP : Penetration Speed (mm/s). H: Heat flow or Intensity (w/m ).  : Density of Material (kg/m3). C: Specific heat capacity (Jkg-1.k-1). -1 TV: Boiling point (k). LV: Latent heat of vaporization (J.kg ). And the maximum depth of penetration (d)[4]: d VP  ………………...... (2.2) t  Where: d: Hole Depth t : Pulse Duration

d VP t ………………...... (2.3)

H t d  ………………...... (2.4) (CTV  LV )

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Example-4[4] Suppose a heat pulse with (H = 1011 Wm-2) & pulse duration (500µs) is incident on to a copper surface. Calculate the hole depth[4]. 6 C = 385 LV =4.75x10 ρ = 8960 TV = 2855 Solution:

11 6 H t 10 50010 d   6 = 0.95mm (CTV  LV ) 8960(385 2855  4.710 )

Example-5[4]: It is required to drill (1mm) diameter holes in a Nickel sheet (1mm) thickness. Using pulsed Nd:YAG laser with a (5kW) peak power output. Estimate the pulse duration for length required[4]. 6 C = 444 LV =6.47x10 ρ = 8900 TV = 3110 Solution:

3 P P 510 2 H   2   6.36 109 (w/m ) A r 1 3 2  ( 10 ) 2

d(CTV  LV )  t  H

1103 8900(4443110  6.47106 )   0.011S 6.36109

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Example-6[4]

Holes are to be drilled in (0.5mm) thick nickel with a radius of (0.13mm) using a pulsed Nd-YAG laser with a (4kW) peak power output. Use energy balance to estimate the energy needed and the pulse length[2]. 3 C = 0.44 J/g.k  = 8.9 g/cm Lm = 298 J/g

LV = 6303 J/g TV = 3187 k R = 60% Solution:

2 U  (CT  Lm  LV )a Z = (0.44 x 3187 + 298 + 6303) 8.9 x  x (0.013)2 x 0.05 = 1.89 Joules Energy Needed = U x 100/100-R = 1.89 x 2.5 = 4.725 Joules Pulse Length = U/P = 4.725/4000 = 1 ms

Question: Proof that Penetration speed (VP) equal to: H VP  (CTV  LV ) Answer:

2 U  (CT  Lm  LV )a Z

U P Z 2 P  & H  &VP  & A  a t A t

2 U  (CT  Lm  LV )a Z %t

2 P  (CT  Lm  LV )a VP %A

H  (CT  Lm  LV )VP

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2.3 Laser Cutting

Industrial laser cutting is done with CW or pulsed CO2 & high-repetition pulsed Nd-YAG lasers. The processes is a gas-assist technique in which, under pressure, forces molten from the Kerf , Oxygen is used with oxidizable material to increase cutting speed[2].

Figure 2.1 Laser cutting Advantage[2] سهولة األتمتة ) تتشغي اللهها أتووهتغيغه ًا( .Ease of automation .1 صشر حلم المسهحة المتعرضة للحرارة .(Small Heat Affect Zone (HAZ .2 القطع ضغق أبهلغ الدقة .Narrow & high-precision Kerf .3 القطع بهللغزر سرع ون الطرق األخرى .Frequently higher speed than other methods .4

When the laser beam is focused on the material, the speed of cutting will be[2,4]: dH VC  ………………...... (2.5) Z(CTV  LV ) Where:

VC : Cutting Speed (mm/s). d: Focused beam diameter (mm). H: Heat flow or Intensity (w/m2). Z: Cutting depth (mm) or thickness.  : Density (kg/m3). C: Specific heat capacity (Jkg-1.k-1). -1 TV: Boiling point (k). LV: Latent heat of vaporization (J.kg ).

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Most materials can be readily cut using a CO2 laser with the exception of those such as brass, copper & aluminum which have high reflectance at 10.6 m. However, since the reflectances are much lower at 1.06 m Nd:YAG lasers can be used instead[4].

Question: Proof that cutting speed (VC) equal to: dH VC  Z(CTV  LV )

Answer: By assuming that the cutting is a limited number of drilling, from the figure

below we get: L L: Cutting length Z d: Drilling or focused diameter

td: Drilling time

tC: Cutting time Speed = displacement / time d n: number of holes

tC  ntd ……………………………………(1) L VC  ……………………………………(2) tC L From (1 & 2) td  ……………………………………(3) nVC Z VP  ……………………………………(4) td L V From (3 & 4) V  P ……………………………………(5) C n Z Cutting length = n× Drilling diameter ……………………………………(6) L = nd

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dV From (5 & 6) V  P ……………………………………(7) C Z H Penetration Speed VP  ……………………………………(8) (CTV  LV ) dH From (7 & 8) VC  ***The end*** Z(CTV  LV )

Example-7: As an example of metal for which we have the required thermal constant: 6 C = 435 LV =6.8x10 ρ = 7870 TV = 316 And we suppose the laser beam to have a power of (1 Kw) and to be focused down to a spot diameter of (0.25 mm). Find the cutting speed at thickness of (2.5 mm). Solution: P P 1103 H     21010 2 2 0.25 (w/m ) A r  ( 103 )2 2 H 21010 VP   6  366 (mm/s) (CTV  LV ) 7870(435316  6.810 ) dV 0.25310 V  P   36.6 (mm/s) C Z 2.5

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Example-8: Estimated the diameter of the focused spot size required to achieve the cutting speed (97 mm/s), for aluminum sheet at thickness (1.3 mm) & power output (2Kw) with thermal constants: 6 C = 903 LV =10.9x10 ρ = 2710 TV = 2720 Solution:

P P P H   2  VP  2 …….(1) A r r (CTV  LV )

dV ZV ZV V  P  V  C & d  2r  V  C ……(2) C Z P d P 2r

ZVC P From 1 & 2  2 2r r (CTV  LV )

2P 2  2 103  r   3 3 6 ZVC (CTV  LV ) 1.310  97 10   2710(903 2720 10.9 10 ) r = 0.278 mm d = 2r = 2 x 0.278 = 0.57

Summary 1.Laser cutting & drilling can be performed on[4]: a. Metals. b. Ceramics c. Plastics d. Cloths e. Glass when it’s surface is coated with a radiation absorbing material such as carbon. 2. A long focal length lens is used in order to produce: a. Narrow Kerfs. b. Reduce the tendency for the cut. 3. To correct or avoid the (tendency of wide Kerf), a gas-jet assisted laser beam can be used, the gas being: a. Either an inert gas, such as helium or argon, and employed when the material is prone to burn or oxidize such (cloths or plastics). b. Or a reactive gas such oxygen, used to obtain exothermic reaction with metals to produce a clean cut & rapid rate of cutting or drilling. 4. For drilling or cutting: a. Ruby laser b. Nd-YAG laser c. Carbon dioxide laser Can be used either in the pulsing mode or in the CW.

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2.4 Laser Welding In the basic welding process two metals (which may be the same or dissimilar), are placed in contact and the region round the contact heated until the material melt and fuse together. The laser welding divided here into two types[4]: a. Microwelding b. Deep penetration welding

a. Microwelding: The ability to focus a laser beam down to an area only a few microns across and the ease with which it can be directed and controlled has naturally led to the use of lasers in welding and soldering minute metal contact such as are found in electronics circuits.

b. Deep penetration welding: With the advent of CO2 lasers with 1kW & higher CW power output it was discovered that the phenomenon of keyholing occurs. In this phenomenon a hole is produced in the material, allowing the beam to penetrate deeply into material. Figure (2.2) illustrates this process when a CW is being used.

Figure (2.2) formation of a “keyhole” during high power laser welding[4]

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Figure 2.3 Laser Welding

Advantages of Laser Welding Among the advantages of laser welding are the following[2,4,6]: 1. Minimum heat input, which results in very little distortion. 2. Small heat affected zone (HAZ) because of the rapid cooling. 3. Narrow, generally cosmetically good weld bead. 4. High strength welds. 5. Easily automated process that can produce very precisely located welds. 6. Weld some metals difficult to weld by other techniques especially dissimilar. 7. Weld in areas difficult to reach with some other techniques. processing can be economical.

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Example-9[4]: By using an “Energy Balance” argument for laser welding, Show that the welding depth (d) may be written[4]: 2P d   rVm (CTm  Lm ) Where:

P: Laser power CW.  m : Welding Speed. r: The focus beam radius.

 : Density. C: Specific Heat. Lm : Latent Heat of Melting. Tm : Melting Point.

Use this model to estimate the welding speed possible when welding two (1mm) thick iron plates together using (2kW) laser. Assume that the laser output is focused down to a spot of (1mm) diameter, assume surface reflectance of (0.5) Take:

5 C = 449 = 7870 Tm = 1810 Lm = 2.7x 10

Solution: 2 U  (CTm  Lm )  r Z Energy Balance 2 P  (CTm  Lm )r vdm

P DVdm DP Vdm  2 Vm   2 r (CTm  Lm ) Z Zr (CTm  Lm ) Where: D=2r Diameter or spot size, Z=d (depth or thickness). 2rP 2P Vm  2  d  dr (CTm  Lm )  rVm (CTm  Lm ) 2P(1 R)  m   (CTm  Lm )  r d

2(2103 ) 0.5  149mm.s 1   (0.5103 )(7870)(1103 )(4491810  2.7105 )

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Example-10[2]

Consider the weld of steel with a CO2 laser capable of providing a power of (2.5 kW) at the workpiece. A weld depth of (2mm) is desired with a weld width of (1mm). Using the energy balance model & assuming (50%) reflectivity, estimate the energy that is required to bring the material to Tm & the pulse length. According to the uniform irradiance model estimate the time required to reach Tb at the surface[2].  = 7.87 g/cm3 C = 0.46 J/g.Co K = 0.21 cm2/s o o o Lm = 272 J/g Tm = 1547 C Tb = 2752 C k = 0.75 W/cm.C Solution: 2 -1 2 U  (CTm  Lm )  r Z = (0.46 × 1547 + 272) 7.87 ×  × (1×10 /2) ×2 × 10-1 = 12.16 J Note: U should be doubled (to compensate for heat conduction losses)[2]  U = 24.32 J U total = 24.32/(1-0.5) = 48.64J P = U/t  t = U/P = 48.64 / 2.5x103 = 19.5 ms

2.5 Applications for Surface Treatment [6] Lasers have been used in a number of ways to modify the properties of surfaces, especially the surfaces of metals. Most often, the objective of the processing has been to harden the surface in order to provide increased wear resistance. In some cases, the goal has been to provide improved resistance to corrosion.

Laser applications in surface treatment have been dominated by the CO2 laser, usually operating at multikilowatt levels, although there have been demonstrations of surface treatment using Nd:YAG, carbon monoxide, and excimer lasers. We shall

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describe several different approaches to surface modification with lasers, including heat treating, glazing, surface alloying, and cladding.

2.5.1 Surface Hardening Laser surface hardening (heat treatment) is a process whereby a defocused

beam (generally from a 1kW CW or higher power CO2 laser) is scanned across a hardenable material to raise the temperature near the surface above the transformation temperature. Normally the cooling rate due to self-quenching by heat condition into the bulk material is sufficiently high to guarantee hardening.

2.5.2 Re-melting (Glazing) Thin layer of the material surface is melted, and on solidification, a structure of fine-grains is produced by the fast quench rate. This structure gives glassy appearance with some desirable characteristics. This technique is applied to metals & Ceramics.

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2.5.3 Alloying This involves spreading metal powder over the surface then laser is passed so that surface melting occurs which allows for the metal powder to be alloyed with the base metal, so that modified surface characteristics are produced.

2.5.4 Cladding[7] 1.Definition: Laser cladding is a melting process in which the laser beam is used to fuse an alloy addition onto a substrate. The alloy may be introduced into the beam–material interaction zone in various ways, either during or prior to processing[7].

2. The goal Laser cladding, provided wide range from cladding alloys for use. Because of the most substrates that tolerate laser melting are generally suitable for cladding: carbon– manganese and stainless steels, and alloys based on aluminum, titanium, magnesium, nickel and copper. Popular cladding alloys are based on cobalt, iron and nickel. 3. Laser Source

- CO2 lasers have traditionally been used for cladding because for many years they were the only sources that could provide the power density required for melting. - As multikilowatt Nd:YAG lasers with fibreoptic beam delivery became more widely available. Because: 1. The potential for lower cost. 2. Flexible coating of three-dimensional components was realized. 3. The improved absorption of the shorter wavelength Nd:YAG laser beam by metallic alloy additions provides a means of increasing the process efficiency and increasing coverage rates.

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- The absorptivity of powder additions to a multikilowatt diode laser beam is about

twice that of a CO2 laser beam, and the rectangular cross-section of the beam is an ideal shape for high coverage rates. The properties of clads produced by

using diode lasers are similar to those produced by CO2 and Nd:YAG lasers.

4. Power For practical laser beam cladding a power density of about 100Wmm−2 and a beam interaction time of about 1 second are used. About 2kW is the normal minimum laser beam power needed for cladding, since insufficient power will result in limited melting of the alloy addition, whereas too much power causes excessive melting of the substrate and dilution of the clad.

5. Advantages 1. High quality (The molten clad solidifies rapidly, forming a strong metallurgical bond with the substrate). 2. High speed, 3. Easily automated.

2.5.5 Annealing One of the main areas of semiconductor processing is the annealing of ion implant damage. Annealing (Re-crystallization) may be made by heating to about 1000oC for 30 minute. Laser annealing is quite fast than the classical methods (heating in a furnace).

2.6 Micromaching[4] There are several areas in which the ability of laser to selectively vaporize small areas of material is useful. One such example is in the laser trimming of

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resistors. Resistors are made by a wide variety of techniques but often consist of a thin film of conductive material deposited on an insulating substrate between two electrodes Figure (2.4-a). As manufactured the films may not have exactly the required resistance and so require some kind of trimming. Whit a laser this may be achieved in several ways. For example, the resistance may be increased by selectively removing material from the film either by drilling holes or by cutting slots in the film figure (2.4-b-c).

Conductive material Electrodes

(a) (b) (c)

Figure (2.4) Use of laser for resistor trimming: (a) shows the basic resistor structure; (b) the resistance is increased by drilling holes in the conductive material; (c) a slot is cut[4]

2.7 Laser marking Is a process whereby serial numbers or other identification include logos, is placed on parts by evaporating a small amount of material with a pulsed CO2 or Nd- YAG laser[2]. 1. High speed, 2. Easily automated. 3. No mechanical contact with the workpiece.

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2.8 Laser Scribing 1.Definition: Scribing is a process for making a groove or line or line of holes either fully penetrating, or not, but sufficient to weaken the structure so that it can be mechanically broken[8].

2.The goal The usual method of scribing a wafer is by the use of diamond scriber; however a diamond scriber produced residue on the wafer that may lodge between microcircuit components, for the later reason laser scribing is found.[2] The quality, particularly for silicon chips and alumina substrates, it is measured by the lack of debris and low heat affected zone. Thus low energy or high power density pulses are used to remove the material principally as vapor..

3.Advantages 1.Clean (lack of debris ) 2. Fast. 3.Accurate. 4. Safe (low heat affected zone). 5. No mechanical contact with the workpiece.

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Questions of

Q2.1:In each case, choose the best option: 1. In cutting an drilling by laser a long focal length lens is used in order to produce (a) faster processes. (b)Narrow kerfs & reduce the tendency. (c) very clean kerf.

2. The goal for exchanging laser scriber instead of diamond scriber is (a) diamond scriber requested a high capital. (b) diamond scriber produced residue on the wafer that may lodge between microcircuit components. (c) laser scriber produced residue on the wafer that may lodge between microcircuit components. (d) laser scriber can be easily automated. .

3. The goal from used laser in cladding processes is (a) The laser provided thin layer from cladding alloys for use. (b) The laser provided fat layer from cladding alloys for use. (c) The laser provided narrow range from cladding alloys for use. (d) The laser provided wide range from cladding alloys for use.

4. In surface hardening laser is scanned across a hardenable material to raise the temperature near the surface above the (a) latent heat. (b) vaporization temperature. (c) melting temperature. (d) freezing temperature.

5. For practical laser beam cladding a power density of about (a) 10Wmm−2 & a beam interaction time of about 1 second (b) 1000Wmm−2 & a beam interaction time of about 1 second (c) 100Wmm−2 & a beam interaction time of about 1 second

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(d) 100Wmm−2 & a beam interaction time of about 10 second

Q2.2: Proof that Penetration speed (VC) equal to: dH VC  Z(CTV  LV )

Q2.3:Estimate the hole diameter in a Nickel sheet (1mm) thickness. Using pulse Nd:YAG with a (5kW) peak power output. And pulse duration (0.11s). 6 C = 444 LV =6.47x10 ρ = 8900 TV = 3110 Answer: (d = 1 mm)

Q2.4: Determine the peak power required to drilling a sheet of Nickel with (1mm) thickness, if the pulse duration is (0.11s), and the diameter of hole (1mm) 6 C = 444 LV =6.47x10 ρ = 8900 TV = 3110 Answer: (P = 5 kW)

Q2.5: Estimate the cutting speed for Aluminum by used laser beam at power (3Kw) & focused spot (0.57 mm), if the aluminum has thickness (13 mm). 6 C = 903 LV =10.9x10 ρ = 2710 TV = 2720

Answer: (VC = 14 mm/s)

Q2.6: Estimate the size of the focused spot diameter required to achieve the cutting rates quoted for a metal cutting speed equal to (32.3 mm/s), thickness (2.5 mm) & power output (1kW) with thermal constant:

6 C = 435 LV =6.8x10 ρ = 7870 TV = 316

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Answer: (d = 0.25 mm)

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Metrological & Scientific Applications 3.1 Introduction

There are many types of metrological & scientific applications in use today such as (distance, velocity, pollution) measurements. The purpose of this chapter is to discuss the some of these techniques. And all systems that use lasers for

metrological & scientific applications.

3.2 Optical Alignment The simplest applications of lasers such as the HeNe laser is in producing a visible line which can be used for[4,9]: 1. Positioning an object. 2. Surveying, guidance of equipment in construction. 3. Aiming other lasers or optical instruments. The use of a beam of light for alignment, often called optical tooling. Advantages of optical alignment: 1. The optical alignment are fast compared to mechanical alignment. 2. Very easy to used. 3. Often required one operator. The accuracy of laser alignment is limited by the divergence of the beam. And the divergence can be reduced by expanding the beam.

3.3 Laser Depth Sounder(LDS)[9] A Nd-YAG laser system is being used to sound & map the sea floor by Royal Australian Navy. Both the primary radiation of 1.06 m & the SHG or frequency- doubled 0.532 m wavelengths are being directed to the sea surface from an airplane.

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The 1.06 m infrared radiation IR is mostly reflected from the surface of the water, while the green 0.532 m traverses the water & is reflected from the sea floor back to the receiver of the sounder. The delay between the two beams determines the depth of the sea or ocean floor. It is reported that the depths to 30m have been thus measured with a resolution of one meter. The principle of measurement of the sea floor are dependence on the calculate the trip time of the laser pulse, and this is similar to that of a rangefinder, with the exception that here there are two laser beams project to the target, where one beam reflects from the surface of the water & the other reflects from the sea floor as shown as in Fig.3.1.

 0.532m 1.06m

Water

Figure 3.1 Laser depth Sounder The computer of the laser depth sounder subtracts the distance of plane-to- water surface from the total distance of plan-to-sea floor, and the remainder is the depth of the sea floor from the surface of the water. Putting this relation in a simplified equation form: D  F  S ………………...... (3.1)

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Where: D- the depth of the sea floor from the surface of the water. F- the distance of plan-to-sea floor. S- the distance of plane-to-water surface. However, the distance F & S are equal to:

CTF F  ………………...... (3.2) 2

CTS S  ………………...... (3.3) 2 Where: C- 3x108 m/s

TF- time of plan-to-sea floor pulse trip.

TS- time of plane-to-water surface pulse trip. In Eqa.3.2 & Eqa.3.3 we divided the trip time on two to obtained the leaving time of laser pulse. Combining the Eqa.3.2 with Eqa.3.3 & substituting them in Eqa.3.1 we have:

CTF CTS D   ………………...... (3.4) 2 2 C D  (TF TS ) ………………...... (3.5) 2

Advantages 1. High resolution about one meter. 2. Faster than conventional methods. 3. Safe at the shallow levels.