University Physics III
University Physics III PHYS 262
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Resolving Power of a Diffraction Grating From our phasor discussion in relation to the multiple minima between adjacent maxima in a multislit diffraction pattern the phase difference A between adjacent minima is given by 27rN where N is the number of slits 27r lSt adjacent min WJ Resolving Power of a Diffraction Grating We also know how phase difference are related to the angular location 6 ds1nt9 0r 227Ids1nt9 27 xi xi Consider in nitesimal changes we have the following relationship 27rd cos 6 A M A6 Thus for the small change in phase Agz 27r N the angular position will change by 27 27rd cos 1 This gives the 12 width A6 or A6 N A Nd cos of the peak at order m Resolving Power of a Diffraction Grating We also need to know how small change in A will lead to change in 9 we can calculate this by taking the differential of the following relation dsin zm Taking the differential gives dcosQA6mAi or A6 Al dcosQ A reasonable criterion for resolvable peaks from two slightly different wavelengths is to have the peak of one to sit at least at or farther away from the 1St minimum of the other Resolving Power of a Diffraction Grating So we equate the two boxed quantities xi 2 m Al Ndcos dcos Canceling the common factor dcos6 we nally arrived at the desired relation So the resolving power of a diffraction grating increases by using a grading with a larger number N of lines measuring the spectra line at a higher order m Resolving Power again for Circular Apertures 639 is the angle between the center of the pattern and the first minimum image will not be sharp Copyright 7 2008 Pearson Edunanon incquot publishing as Pearson Addisvaeaey Because of di raction light spreads out after passing thru small apertures 9 this posses resolution limits to commonly used optical instruments such as microscopes and telescopes Resolving Power for Circular Apertures Consider two noncoherent point sources so that they don t interfere ie two distant stars angular separation 9 between the two stars We will observe two diffraction patterns on top star 2 lens of each others from telescope ID Resolving Power for Circular Apertures The overlap of the two diffraction pattern might prevent one om discerning the two sources of light A workable criterion is called the Rayleigh s Criterion which is similar in spirit to our discussion for the resolving power for the dif action grating The two diffraction pattern will be resolvable if the central max from one pattern is at least as far as the 1st min of the other image For circular aperture with diameter D the location of the its 1st order dif action minimum is xi Sln 91 1225 122 is a geometric factor Resolving Power for Circular Apertures The Limit of Resolution for a circular aperture is de ned as the smallest angular separation between two light sources that can be resolved according to the Rayleigh s Criterion and it is given by Resolving Power for Circular Apertures A optical device such as a telescope or microscope will have a high Resolving Power if its has a small Limit of Resolution so that nearby objects with a small angular separation can be resolved This gives the following ways to increase the Resolving Power increase the diameterD 9 use a bigger lenmirror in telescope decrease the wavelength A 9 use a shorter wavelength of light in chip production a 5min quotpomm b Medium upenum c Lurgc uperlmc Example 366 Resolving Power of a Camera Lens Given f50 mm I fnumber of 72 What 1s the m1n1mum d1stance between two Object distance 9 Om points on the object that one can resolve wavelength 500nm f number fD 9 D ff number 50mm2 25 mm Rayleigh s Criterion gives a 500x10 9m 6min s smamm 122 122 3 24x10 5 rad D 25x10 m Example 366 Resolving Power of a Camera Lens For a simple lens we know that the angular separation of two points on the object is given by 6 XLy s s y 9 separation of object points y 9 separation of the corresponding image points s 9 object distance s 9 image distance Physics 262266 George Mason University Prof Paul 80 Chapter 19 The 1St Law of Thermodynamics III Heat Work and Paths in Thermodynamic Processes III Internal Energy and the 1St Law of Thermodynamics III Kinds of Thermodynamic Processes More on Heat Capacities Adiabatic Processes Thermodynamic Systems ID II Thermodynamic Systems Any collection of objects considered as one system that may have potential to exchange energy with its surrounding II System States amp Thermodynamic Processes P gt State of a thermodynamic system is characterized by a set of marcoscopic variables P V T n and it can be Visualized as a point in the PV diagram state 1 gt A thermodynamic system changes om one state 1 to another state 2 through a thermodynamic process indicated by the blue curvepath in the PV diagram state 2 Thermodynamic Systems ID II Notes Thermodynamic states can only be speci ed if the system is in thermal equilibrium Every subparts of the system should have the same values of P T Vn etc A path for a thermodynamic process can only be represented in the PV diagram if the process is reversible quasistatic A quasistatic process can be thought of as a suf ciently slow still fast in macroscopic time process such that the system is approximately near equilibrium at each step Reversible vs NonReversible Processes Reversible Quasistatic State 1 NonReversible NonQuasistatic State 1 Insulation State 2 H c 7 rm same W initial amp r nal Vacuum states 5 0 1 f 20 L 20 L l 4 300 K l Breakable Gas at 300 K state 1 Small changes in forcing the piston can increasedecrease the volume reversibly state 2 partition p NO small changes can stop the gas in lling the container after the partition is broken Energy Transfer in a Thermodynamic System Heat Q A thermodynamic system can absorb or release heat during a thermodynamic process Work W Work is either done on or done by a system during a thermodynamic process Note Play attention to the sign conventions for Q and W Surroundings environment N b A Q 4gt 0 i System W A O E Avz 39 Heat is positive when Work is positive when it it enters the system is done by the system negative when it negative when it is done leaves the system an the system Surroundings environment x Q lt 0 x x W stem K V 0 7 V39 Work Done by a Gas System dx 7 r The system the gas 7 i exerts a force F pA 6 pA gt on its surrounding the K i A piston through a V distance dx Force that system exerts on piston The in nitesimal work done by the system dWis given by dWFdx PAdx PdV Work Done by a Gas For a nite change in the volume from V1 to V the total work done is given by integrating the differential a pV diagram for a system b pV diagram for a system undergoing an expansion with undergoing a compression with c pwdiaumm or a Syslsm c vanmg Pressure Varymg pressure undergoing an expansion with P I7 constant pressure pl p2 ICIO39Shhalldlll lg Il ldlC ICS negative k work p 2 1 P2 l7 I I I Work Area Work Area fvz I Wovrk Alea 1702 7 VI gt 0 gt z 2 WWW 0 fvlpdVlt0 I V V 0 V V2 0 V2 VI 0 V V2 Work Done by a Gas Case 1 isochoric processes Vis constant dVO Egg h quot 2 l A l I A r 1 Case 2 isobaric processes P is constant Case 3 isothermal processes T is constant Example 191 V2 Starting with W IPdV V1 Using the Ideal Gas law PVnRT Work Done by a Isothermal Process nRT dV V V2 We have W I V1 T is constant and we can pull it out of the integral together with the other two constants nR xed amount of gas For T constant we also have PIV1 PZV2 or g l Pi So we can also write W nRT1n 2 w Work Done is Path Dependent n m plthx m rhn39ru u m m h quotN Na quot lum W gsmnn numimc l m Marc 2 m quot u m quotquot quot quot 3 r a 0 quot c y l v 7 y Iquot Fm am MM mm mmquot quot mama nmnm mum u 5 m cmumnx olunm M m dcucusxng plcwsum quot39 uhm m y M vlunic Invrn quotquotm quot w mm Prxx urs39 5 i p7 m w Arrn quot T w o x v 0 v Work done by a ga s is dijfkrem depending on path taken even though the starting and ending states are the same for all these processes Heat AbsorbedRelease is also Path Dependent Process 1 isothermal expansion System does work on piston hot plate adds heat to system W gt 0 and Q gt 0 State 1 State 2 Vi Gas at 20LH 77 Ii 7 p L 15 Process 2 free expansion System does no work no heat enters 0139 leaves system W 0 and Q 0 State 1 State 2 Insulation Vacuum 20L Breakable Dartition Gas at 300 K Starting and ending states are the same for both processes Internal Energy U Internal Energy U the total amount of energy KE PE intrinsic to the system associated with all its microscopic components when Viewed in a reference frame at rest with the object system The KB and PE associated with a the bulk motion are NOT parts of the internal energy U of the system Internal Energy example KES translational rotational A bucket of water lt I Vibrational v PESC Within molecules between molecules rest on the table molecularVA bonds dipoledipole interactions A E Internal Energy for an Ideal Gas Recall in an Ideal Gas Gas molecules are 0 infinitely small No intermolecular forces We have explicitly calculated its Internal Energy U 9 3 translational degrees of freedom only 9 N molecules 9 each contributes 12 CT fl 9 h 315 Ideal Gas monoatomic The Internal Energy for an Ideal Gas U is a function of T ONLY Experimentally this is also shown to be the case for most diluted gases The 1St LaW of Thermodynamics We have seen that both heat and work can change the internal energy of the system In particular Heat Work 1 Q Piston pushing down on gas Heat enters the system Work done on the system U increases AU gt O U increases AU gt 0 The 1st Law of Thermodynamics ID Putting both of these mechanisms together we can combine them into one mathematical statement for the change in U in any processes 397 lSt Law of Therm This is a generalization of the principle of conservation of energy to include energy transfer through heat as well as through mechanical means work Important Property for AU Although both Q and Ware pathprocess dependent AU is independent of path U is a state variable and it depends on the initial and nal states only 5 1St Law of Thermodynamics AU can be positive negative and zero More heal is ttdded to system than system More heat ows out of system than work is does work Internal energy of system increases done Internal energy of system decreuses Surroundings environment Surroundings environment Q ISOJ quot W IOOJ Q ISOJ W 100 gt g x S stern 39 K jN 2 32393 AUQ WSOJ AUQ7W 750 Heat added to system equals work done by system Internnl energy of system unchanged Surroundings environment Q 1501 W 1501 System AUQW0 Tunneling Through a Barrier Consider the following potential barrier U x U0 0 s x s L U0 U x O elsewhere Ej2z m zaurt o A quantum particle with mass m and energy E is traveling from the left to the right 35 0 L Classical Expectation with E lt U0 In the region x lt 0 the particle is free but when it reaches x 0 the particle will hit a wall since its E is less than the potential at x 0 It will be re ected back and it could not penetrate the barrier l H Tunneling Through a Barrier Quantum Picture x lt O and x gt L free space 0 s x s L inside the barrier The wavefunction for a free E lt U0 9 wavefunenon is a particle with de nite E and decaying exponential e P is sinusoidal 6 or e39ikx exponential function within barrier A 39 x 0 L 1 4 I 1 I and sinusordal outs1de the barrier Tunneling Through a Barrier If the energy is high enough and the barrier is not too Wide so that the exponential decay does not signi cantly diminish the amplitude of the incidence wave then there is a nonzero probability that a quantum particle might penetrate the barrier reduced amplitude 9 reduced probability but not zero probability The transmission probability T can be solved om the Schrodinger equation by enforcing the boundary conditions 9 Application of Tunneling STM Scanning Tunneling Microscope STM x S urtace electrons i L Specimen this variations can be used to scan surface features In a SIM a extremely sharp conducting needle is brought very close to a surface than one wants to image When the needle is at a positive potential with respect to the surface electrons from the surface can tunnel through the surfacepotential energy barrier The tunneling current detected will depend sensitively on the width L of the surface gap and ii The Harmonic Oscillator Classically the harmonic oscillator can be envisioned as a mass m acted on by a conservative force F k x Hooke s Law Its associated potential energy is the familiar Ux l U k39 2 x 2 x For a classical particle with energy E I the particle will oscillate sinusoidally about x 0 with an amplitude A and angular frequency a ID The Harmonic Oscillator For the Quantum analysis we will use the same form of the potential energy for a quantum Harmonic Oscillator 2 2 JLMJrlkvxz d2 2 1 2m dxz 2 wxElx or W ml Boundary condition consideration Ux increases without bound as M gt 00 so that the wavefunction for particle with a given energy E must vanish at large x The wavefunctions for this ordinary differential equation with the boundary condition lx gt 0 as x gt ioo are called the Hermite functions WCX Ce mk39xzZh And we have the following quantized energies En njha n0l2 H The Harmonie Oscillator Hermite Functions 1 Enn hm n012 pm wow 2 1 1 i i 39 E5 I1 n I i 9 i0 7 5 w E2 3 n 3 E1 3 m 3 i 2 a i 0 x quotA note similar to previous note wavefunction penetration into examples the loweStE classically forbidden regions state is not zero E The Harmonic Oscillator Probability Distribution Function Wm Wm M u lIImlZ l l 1 l jl 4 1 l 1 120 1 l H I 1quot l I 1 4 1 1 l l l r A A 0 gtnn 444 l 0 A0 Classically the particle with energy E will slow down as it climbs up on both side of the potential hills and it will spend most of its time at x iA The blue curve depicts this classical behavior and the QM CM as the quantum number n increases g The Hatorn Important 13641013653 e39 does not exist in wellde ned circular orbits around the nucleus as in the Bohr s model e39 in a Hatom should be envisioned as a cloud or probability distribution function The size and shape of this cloud is described by the wavefunction for the H atom and it can be explicitly calculated om the Schrodinger equation 39quot V in 3D 4 239 I E The Hatom In the Schrodinger equation we have explicitly included the Coulomb potential energy term under which the electron interacts with the nucleus at the origin Eicclrnni charge we ill coordinale lamn 1 ez Nucleus 472398 Iquot chums r39i 0 at the origin is the radius in r x2 y2 z2 spherical coordinates Ur Electron Probability Distributions In 3D the probability in nding the electron in a given volume dVis given by llx y z2 dV A good way to visualize this 3D probability distribution is to consider a thin spherical shell with radius r and thickness dr as our choice for 61V dV 47zr2dr We denote the probability of nding the electron Within this thin radial shell using the radial probability distribution function Pr as Prdr Prdr l12 dV l12 47239r2dr ID Electron Probability Distributions Examples of the 3D probability distribution function 12 electron cloud l l l 1v L r r 47r 0h2 1 2 e 529X10 11m is the Bohr s radius which we have seen previously Quantum Number Recall that for a particle in a 1D box we have one quantum number for the total energy ofthe particle gt It arises from tting the wavefunction sin n 72xL Within a Z box of length L quantization In the H atom case we are in 3D the tting of the wavefunction in space Will result in 2 additional quantum numbers a total of 3 ID Quantum Numbers In the Hatom case we are in 3D the tting of the wavefunction in space Will result in 2 additional quantum numbers a total of 3 by this argument 1 n Principle Quantum Number related to the quantization of the main energy levels in the Hatom as in the Bohr s model E 136eV quot21 2 3 quot n 2 7 n The other two related to the quantization of the orbital angular momentum of the electron Only certain discrete values of the magnitude and components of the orbital angular momentum are permitted Quantum Numbers 2 l Orbital Quantum Number related to the quantization of the magnitude of the equots orbital angular momentum L ll1n l012n 1 note in Bohr s model each energy level n corresponds to a single value of angular momentum In this QM description for each energy level n there are n possible values for L 3 m Magnetic Quantum Number related to the quantization of the direction of the equots orbital angular momentum vector L2 mln ml 0i1i2il By convention we pick the zdirection be the relevant direction for this quantization Physically there are no preference in the zdirection The other two directions are not quantized ID Magnetic Quantum Number Illustrations showing the relation between L and L2 Zeeman Effect Experimentally it was found that under a magnetic eld the energy levels of the Hatom will split according to the magnetic quantum number m Semiclassical explanation No L B field A B 0 With 639 orbits around the nucleus and it forms a eld current loop L2 measures the orientation of L with respect to B and thus affects the energy level of the Hatom E Anomalous Zeeman Effect Zinc sharp triplet Zinc singlet Sodium principal doublet only one of three patterns shown 4N0 ivNo B cld B licld l A D A D1 A aWilh With B cld B cld lli lll ll lll Predicted Predicted Predicted Predicted normal splitting normal splitting normal splitting normal splitting Normal pattern A Experiment agrees wilh predicted quotnormalquot splitting canyngm 2005 newquot Educahun m puullihmg as Pearson Audisnnrwasley Anomalous patterns Experiment does not agree with predicted normal splitting Additional experiments shows that some of the Zeeman lines are further split Electron Spins In 1925 using again semiclassical model Samuel Goudsmidt and George Uhlenbech demonstrate that this fine structure splitting is due to the Spin angular momentum of the electron and this introduces the 4th quantum number 4 Spin Quantum Number The electron has another intrinsic physical characteristic akin to spin angular momentum associated with a spinning top This quantum characteristic did not come out from Schrodinger s original theory Its existence requires the consideration of relativistic quantum effects Dirac s Theory The direction of the spin angular momentum 52 of the electron is quantized S2 msh ms 2 i NIH Bopyrlgm zoos Peavsou EJuLaHun ma publlshmg as Pearson Addxaoanesley S sslh s m S NIH Wavefunction Labeling Scheme ID We have identi ed 4 separate quantum numbers for the Hatom n l m 1 ms For a given principal quantum number n the H atom has a given energy and there might be more than one distinct states with additional choices for the other three quantum numbers The fact that there are more than one distinct states for the same energy is call degeneracy And states with different orbital quantum numbers are labeled as Historically states with different principal quantum numbers are labeled as I 0 s subshell n21 Kshe 11 psubshell n22 Lshe 12 dsubshell n23 Mshe 13 fsubshell n 4 N She I 4 g subshell I 5 h subshell E Wavefunction Labeling Scheme Table 41 Quantum States of the Hydrogen Atom 1 l m Spectroscopic Notation Shell I 0 0 ls K 2 0 0 23 L 2 l 1 0 1 2p 395 0 0 3s 3 l l 0 1 3p M 3 2 2 lO 12 3d 0 0 4s N and so on Conwgmt mo PearsonEduhalanmpubllsmngasFearsonAemsoanesley ml and ms are not labeled by this scheme E Pauli Exclusion Principle In order to understand the full electronic structures of the all the elements beyond the simple singleelectron Hatom we need another quantum idea In 1925 Wolfgang Pauli proposed the Pauli s Exclusion Principle no two electrons can occupy the same quantummechanical state in a given system ie no two electrons in an atom can have the same set of value for all four quantum numbers n 1 m1 ms The Pauli s Exclusion Principle the set of the four quantum numbers give the complete prescription in identifying the ground state con guration of e39s for all elements in the Periodic Table Then all chemical properties for all atoms follow 9 Additional electrons cannot all crowded into the n 1 state due to the Pauli s Exclusion Principle and they must distribute to other higher levels according the ordering of the four quantum numbers Filling in the Ground State Example Hatom Z l one e39 L1th1um Z 3 three 639 E 1 lled 1 I free space m1 1 0 1 n1l0 1 0 quot223121 OOOOQO quot223120 E To n21evel HeliumZ2 two e39 n1zo TL E the lowest level is quot2110 fl nowfull Last electron must go to 112 0 level by Pauli s Exclusion Principle IE Filling in the Ground State Example Sodium Z 11 ml l 0 1 n3l0 LO quot227121 ill HLL E n2l0 IN nLlO iL Spectroscopic Notation in the Periodic Table Typically only the outer most shell including the subshells Within the outer most shell is labeled of e39 in that subshell H gt ls1 shell 11 value subshell I value He gt 152 8 electrons to fill 2 will ll K shell 0 252 2194 and 6 remaining will need to go to L shell Z 8 outer Shell two subshells sl 0 ml 2 0 9 only 2 max slots is n 2 10 and 1 1 p1 1 ml 2 l 01 9 6 max slots with s p 4 taken 1 Chapter 36 Diffraction El EIEIEIEI Diffraction and Huygens Principle Diffraction from a Single Slit Intensity in the SingleSlit Pattern DoubleSlit Diffraction Diffraction Grating XRay Diffraction Resolving Power l Diffraction from Sharp Edges Geometric optics predicts that this situation should produce a sharp boundary between illumination and solid shadow quot DOESN39T That s NOT what HAPPEN really happens Point source Geometric shadow Straightedge Screen Diffraction from Sharp Edges This is What actually happens in real experiments a i b 39 gPliolrigraph of a razor blade illuminated hy r monochromatic light from Ll point source a pinhole Notice the fringe mound l e blade oulline Enliu ged View ol me area oulside the geometric shadow of the blades edge Position of geuman39ic shadow a 2 Copyrigme 2005 Pearson Eaucaxm In publlshlng as Pearson A imanesley Interference fringes seen beyond geometric shadow D razors gap Diffraction and Huygen s Principle Consider a simpler case a single slit Slit is divided into many imaginary strips Slit width Each strip is a source of Huygens s wavelets Plane waves incident on the slit Waves spread out om each strips as wavelets creating interference patterns beyond and around sharp edges The spreading out of waves thru small apertures or by sharp edges is called diffraction Similar to the twosource interference pattern these waves interfere as they spread out and create the diffraction pattern Fresnel amp Fraunhofer Diffraction b Fresnel near field diffraction C Fraunhofer farfield diffraction If the screen is close If the screen is distant H the rays from the the rays to P are different strips to a approximately parallel point P on the screen are not parallel I 5 Screen For simplicity we will consider Fraunhofer Diffraction from now on SingleSlit Diffraction Central Maximum 6 0 straight ahead All waves om each wavelets travel the same distance to the screen far away and they arrive in phase 9 constructive interference There will be a bright fringe in the middle at 6 0 Side note Poisson s Spot obstacle ball bearing SCI39CCII Wave spreading around from the top will travel the same distance as the wave spreading around from the bottom At the midpoint 6 0 these waves Will interfere constructively and create a bright spot although it is in the shadow region ID SingleSlit Diffraction Dark Fringes First Order Minimum 6gt 0 9 slightly above or below the central max Divide the wavelets into 2 groups J 1 top and bottom 1 If waves om the top group a2 ERAv 2 destructively interfere with waves a from the bottom group we will a sm6 have a dark spot on the screen at 6 2 39 a rZ rlzrz r12l2 a w r path dlff 35m 6 12 or SingleSlit Diffraction Dark Fringes Second Order Minimum a 4 sine 4 path diff WNH Divide the wavelets into 4 groups If waves from each adjacent groups destructively interfere we will have another dark spot on the screen at 6 I rzrlnr312 a gsin62i2 0T m SingleSlit Diffraction Dark Fringes For higher order minimum with larger angular distance 9 we can use the same argument by subdividing the slit into more groups 6 8 10 etc This leads to the following general formula for the dark fringes Note 1 m 0 is not the first minimum In fact it is the location for the central max 2 Secondary maximum occurs near 312 512 etc but not exactly ID Intensity in SingleSlit Pattern to observation point p on screen D1V1de the slit into N large of smaller str1ps Ay aN gt size of each small strip A Ay sin 6 gt small path diff bet adj strips If E 0 is the magnitude of the incoming wave then the E eld from each strip Will have a magnitude of AEE0N Consider the electric eld om each wavelets as a phasor AE the resultant electric eld E at p can be calculated as the phasor sum of all theAE39s Phase Difference from Path Difference For each pair of adjacent phasors there is a path difference Al and this path difference induce a phase difference A between these adjacent phasors M l 27 27 A Al A sum 27 2 2 A y Considering the phasor sum of all N phasors the total phase difference 8 is 27 A A 6 396 N 396 N xi ysm So the total phase d1fference 8 is a function of the angular location 6 2 NAysin627 asin6 E Summing Phasors to Calculate Ep Central Maximum A 0 3 0 straight ahead All phasors are in phase EPNAEE0 Douyugm 2a 2003 Pearson Euncahnn Inc pubhihmg as Pea rson Addisonrwesiey Summing Phasors to Calculate Ep Slightly away from Central Maximum A gt 0 3 gt 0 8 is the phase diff between the first and the last phasors ltE Summing Phasors to Calculate Ep First Order Minimum 6 N A 27z 1St minimum condition when last phasor s tip matches up exactly with the rst phasor s end EP0 Note 27 asin627z gt asin6xl same condition as previously derived E Intensity in SingleSlit Pattern For N gt 00 Ay gt dy we can nd an expression of the intensity I in terms of 9 C 1 g The polygon becomes 0 C is the center of the are 3 I8 3 an arc of a circle angle A and B are right angles 5 0 interior angle at D is 1800 8 E E Q B Q 4C I8 8 2 30 86 f For the c1rcu1ar sect1on ACB g5 radius gtlt angle arc lenghi g 5 0 q E Q P radius x 2 2 E0 6 has to be 1n rad1an B E A E0 radius 2 0 D g Intensity in SingleSlit Pattern Lastly from the right triangles sin 2 sing Ez 2 2 Eo gt ABEPEO With I proportional to E Lorentz Transformation derivation Assume the modi cation of the Galilean Transformation by a linear factor y x 7x 39 ut 39 Since S and S are in relative motion we should have a symmetric equation for x with u 9 u x39 7x ut Now set both S and S to coincide With each other at the origin at I l 0 and a light pulse is initiated at that time Lorentz Transformation derivation After a time I in S and a corresponding l in S x 2 ct in S 0 IS the same 1n both frames x39 2 ct39 1n S39 Substitute these into the previous equations we have ct 7ct39 ut39 70 ut39 ct39 7ct ut Substitute l om the bottom equation to the top equation we have 2 c 1 c Mange w 72 czu2 m 1 7 W the desired 7factor CZ y202u2 Lorentz Transformation derivation One can also try to eliminate the x variable in the original two equations x39yx ut xyx39ut39 x39 yx39 ut39 ut x39 yzx39 yzutL yut yut yzut39 72 lx39 2 1 2 tyt39y x39yt39y 2 If 7 7 u Consider the yfactor on the RHS of the expression 1 c 2 c2 y 9 l uzc2 cz u2 C 15 Lorentz Transformation derivation More algebra the desired Lorentz Transform in time E Lorentz Transformation Reduction back to Galilean Transformation in the regime u ltlt 6 For Zltlt1 gt y21 C l uzc2 x39 E lx ut x ut u Galilean Transformation t39 1 t Zx t Einstein Relativity is more general and it reduces to previous results Galilean in the u ltlt 6 limit Intervals between Two Events We have de ned Lorentz Transformation for an event P Now we want to extend it to a spacetime interval Ax Al Events 1 and 2 will have different coordinates for different observers O and 0 2 AxiAx39 and AtiAt39 Ax x2 x1 At t2 t1 In general Ax39zx39Z x391At39t392 t391 Intervals between Two Events By direct substitution we have the following transform for intervals between two events Ax39 y Ax uAt 7 At39yAt Ax dx39 ydx udt u for differential changes dt39 9dt 2dx 0 Note One can show that the combination a s2 olx2 a z2 a s392 a x39Z a t392 is the same invariant for all inertial observation Time Dilation Revisit In S frame consider two clicks 1 amp 2 of a clock stationary in S these two events occur at the same place x 1 x 2 Ax 0 but at different time t 391 7t t39z At39 72 0 According to S frame t2 t1 2 At 7At39 yAt39 Since the two clicks occurs stationary in S At is the proper time Ato At At yAtO So the time interval measurement in S frame is time dilated Simultaneity Revisit In S frame let consider two events 1 amp 2 at two different locations 6 391 7t x39z Ax39 7t 0 happening at the same time so that Ax39 OAt39O Then in S frame these two events will be separated by the following time interval u 5 t1At7MgAx39j gt A y 2Ax39 0 Thus event 1 and 2 are not happening at the same time simultaneous in S frame Length Construction Revisit Consider a ruler in S frame so that its length between the two end points is given by Ax L0 the proper length An observer in S frame locates the ends of this ruler at the same time according to hisher own clock At 0 Let say that 0 measure a length of Ax L in S frame From Lorentz Transformation we have Ax397Ax7WAx L07L This is the length contraction formula ID Lorentz Velocity Transformation In S frame let say that we have an object moving in the xdirection with speed 39 dx39 Note V x g u 9 relative speed between S and S Lorentz Transform gives dxv 2 dx udt Vx 9 Speed Of Object u dt397dt C 2dx dx v 7 dx udt 1011 3 dt12dx1dt Fi C 02 dt Lorentz Velocity Transformation In S frame the velocity of the object is measured as This then gives Lorentz Velocity Transformation Slow relative speed u ltlt 0 ucltltl szc V u 39 c u Cl uc V E x 2 u sz z zc 39 x 1 0 x 1ic 1uC 2 Galilean Velocity Transform c c is the same in all frames Object moving at the speed of light Lorentz Velocity Transformation If the object has velocity components in y and 2 directions vy amp vz how would these components transform u is in x direction only W Vg Hw quotg yw u uw uw wm w n 7E 62 l 7E 62le y 39 71 1 Vx Smularly for the 2 component Inverse Lorentz Velocity Transform From the principle of relativity there should be no physical distinction for the two inertial observers in relative motion So the Lorentz Velocity Transform equation and its inverse transform should have the same form but with u 69 u for the inverse transform of v in term of v E Example 378 s 39I I r u 09006 a What is the probe s veloc1ty ux 0950 relative to earth w v 11795 b What is the scoutship s A velocity relative to spaceship Scoutship UT Spaceship Robot space probe Setup Two frames S Earth 8 9 Spaceship u 09000 a In S frame the probe moves at V39pmbe 07000 vxu 0700009000 209820 Vx luv39x02 107000900 E Example 378 s V u 09006 Ux 0950C gt v 0700c39 e gt 9 Scoutship 7 Spaceship Robot space A probe a The scoutship s speed is given in Sframe with respect to Earth v 09500 scout The scoutship s speed with respect Sframe is given by the inverse transform v vx u 0950c o900c x 1 chc2 1 09500900 03450 9H Relativistic Doppler Effect Moving source emits waves Source emits Position of first wave of frequencyfb First wave second wave crest at the instant that the crest emitted here crest here second crest is emitted W xi Stationary observer K39 d t etects waves 0t i i I 7 I gt frequencyfgtjbi F uT gtilt A fl Stanley 5 CT 4 nonwgmanna Peason Education the publishing as PeasonAuuisnanesley Statement of the problem A source of light is moving at constant speed it toward a stationary observer Stanley The source emits EM waves withfO 1 To in its rest ame What is the frequency measured by Stanley Relativistic Doppler Effect From Stanley point of view A cT uT c uT dist between wave crests Then since ft f c we have f c c uT As we have seen time intervals are measured differently by different observers T0 is measured in the rest frame of the source so that it is the proper measurement and Tis not and in fact T is dilated ie T T T7To c 0 l uZc2 cz u2 l cz u2 l cz u2 This ives g T c T f0 Calorimetric calculations dQ mch or dQ nCdT speci c heat Q mTL latent heat General Formulas applicable to ALL processes in an Ideal Gas PV nRT dU dQ dW CV R monoatomic dWPdV CPCVR dU nCVdT Speci c Processes Isothermal Tconstant 9 dT 0 9 dU 0 9 dQdW f f V W deV j dt nRTln f x x V V Adiabatic dQ0 9 dU dW PV7 const TVy391 const Isobaric Pconstant a W PdVPVf 44 Isochroic Vconstant 9 dV0 9 dW0 9 dUdQ Entropy General Reversible Processes dSdQT 9 AS nCVlnTfZnRh1VfK Non Reversible Processes one cannot write dSdQ T for the process But one can use a surrogate reversible process with the same initial and nal states to calculate AS 2quot 1 Law of Thermodynamics ASAS 202 system AS gt0 environment Physics 262 examl 3 25 pts 500g of water at 150 C is being heated in an experiment The nal product is 500g of steam at 100 0C LV 226102 Jg 0C cwmr 4186 Jg 0C Calculate the entropy change for the water in this heating process fdQ f In general we have AS IdS For the case of a calorimetric change we have x f T dQ mch Thus we have AS J m mcln Using the above equation we can calculate the entropy change in heating the water from 150 C to 100 C f m c dT T 1 1n 50g4186JgK1n M 15027315 541J K At the phase change point the temperature remains at 100 C and the entropy change is given by f dQ mLV ASatzrrhzat I 1 I T T 50g226gtlt102Jg 37315K 414JK So the net entropy change is AS AS AS mam Emprchangz latentrhzat 468JK 78 Physics 262 examl 4 25 pts One mole of an ideal monoatomic gas is taken through the cycle shown below An ideal gas initially at PA100kPa and VA250L state A doubles its volume VB500L by an adiabatic process Then the gas is isothermally compressed to state C Vc250L and the cycle is completed by an isochoric constant volume process to bring the gas back to state A Calculate a the net work done by the gas in one cycle b the heat absorbed by the gas c the heat expelled by the gas d Is this cycle a heat engine or a heat pump e Calculate the efficiency or coefficient of performance for this cycle 1L 10393 m3 V liters First we use the gas law to find the PV and T at states A B and C Qnmn 10X105Nm225gtlt10393m3 8315JmolK TA 30066K A TA PBVBy PAVi V 7 1 53 B PBPA A lX105Pa 03l498gtlt105Pa VB 2 TB1894K nR To TB 1894K B gt C isothermal C VC 25L given nRTC P 062996 X 105 Pa C C a AgtB This is an adiabatic expansion and Q AB0 AUAB WAB quotCV TB TA 3 E8315JmolK1894K 3007K WA 1389ld l389ld 78 Physics 262 examl W AB l 3 89k B gtC Isothermal compression AUBC 0 constant temperature QBC W3C nRTBln C1mol8315JmolK1894Kln12 B 1092k CgtA Isochroic WCA0 3 AUCA QCA Ean TC 138 Thus W r12 WAR WBC 138911 l092c 0297kJ net work is done BY gas b Heat absorbed by the system includes all positive Qs and the system absorbs heat only during the isochoric process CgtA so that Qabosmbzd QCA 139 c Heat expelled by the system includes all negative Qs and the system releases heat during only the isothermal compression B gtC so that Qrzlzmzd Q30 109kJ d Since net work is being done by the gas and net heat is being absorbed by the gas during the cycle this cycle is functioning as a heat engine e The ef ciency of this heat engine is W n22 029 214 138911 e Qabsmbzd 88 Symmetry of the Lensmaker s Equation Because the Lensmaker s equation is symmetric with respect to the sign convention for Ra and Rb the left and right focal lengths are the same irrespective of the values of R and Rb Example Application of the Lensmaker s Equation 111 Ill Ra l00cm Ca same side as outgoing light Rb 150cm Ca same side as outgoing light 1331 f 909cm The lens is converging Symmetry of the Lensmaker s iEqua on Now we ip our lens around so that the refracting surface with Rb Will be on the incoming light side 1 1 1 1 f Rb Ra R 1 1 gt a 2 133 1 Ca i 15 10 p Cb 033 i i OOII Rb 15 10 30 f909cm Ra Rb 150cm Ca same side as incoming light l00cm Ca same side as incoming light Same as previous calculation Ray Tracing Methods for Lenses a Converging lens Q A D39 2 3 PI P Fl Q 6 Parallel incident ray refracts to pass through second focal point F2 Ray through center of lens does not deviate appreciably Ray through the first focal point F1 emerges parallel to the axis Copyright 2005 Pearson Educahon um publishing as Pearson Audisnnrweslay E Lateral Magni cation 0f Lenses Q A A r s and s39 are both positive the image is inverted y y Q7 K f f X S39 f 9 K s s 39 Copyright 2003 Pearson Educalmn Inc pubhshmg as Pearson Adaisnnrwesiey 7 Rays Tracing Methods for Lenses 6 b Diverging lens Q g V 3 2 9 P 52 P 39F1 Parallel incident ray appears after refraction to have come from the second focal point F 2 Ray through center of lens does not deviate appreciably Ray aimed at the first focal point F 1 emerges parallel to the ax1s Admmwmey lgg Obj ectImage Relations Thin Lenses Thin Lens A let b FuKwun Hwan a Object 0 is outside focal point image I is real w b Object 0 is closer to focal point image I is real and farther away 1 0S3sz F1 lgg Obj ectImage Relations Thin Lenses C Object 0 is even closer to focal point image I is real and even farther away 1 0 2 F2 F1 I 2 d Object 0 is at focal point image I is at infinity K1 0 2 F 2 1 t r 2 h Obj ectImage Relations Thin Lenses e Object 0 is inside focal point image I is Virtual and larger than object 7 l I N x 1 0 2 F 2 F1 l r nmmnmmannu DAanA mmmvm um mmm mn a Dearth AdNrrmJNn nu 2 f A Virtual object 0 light rays are converging on lens Example 3411 Compound Lenses k120 cm gtlt 24o cm elem cm 120 cm l Kaoeegoel lt60gt 60gt cm CH1 cm Example 3411 1 L 12 SI39 1 39 S1 S1 fl s1 39 2 24cm For the 1St Lens This image om the 1St lens is on the light incoming side of lens 2 so that s2 2 36cm s139 12cm i Li A LA 1 12 1 s2 s239 f2 12 s239 6 s239 12 12 S1 39 2 120m nal image is 12 cm on the outgoing side of lens 2 The combined lateral magni cation is the product om both lenses s1 39 52 39 24cm 12cm mt m m 0t 1 2 S1 2 upright and real 5 12cm 12cm 2 Fresnel Lens Fresnel Lens for the lighthouse at Jones Point in Old Town Alexandria Use in headlights and lighthouse lamps to save weight D immel lens Format s Principle Pierre de Fermat 16011665 A general mathematical principle that can be used to analyze light path When a light ray travels between two points its path is the one that requires the least time Application 1 uniform material 11 or v is the same everywherel Between any two points the least time requires the shortest distance in an uniform medium Light will travel in a straight line in an uniform medium a P Fermat s Principle Pierre de Fermat 16011665 Application 2 Snell s Law Within 111 and 112 light travels d P 6 1n stra1ght11nes and total tlme F of travel om p to q is a d x 71 61 lt7 n lt gt 2 x 62 b V1 V2 r2 92 quota2x2 b2d X2 3961 6111 6712 Note With two different speeds the fastest way to get from p to q is not necessary a straight line Fermat s Principle Application to Snell s Law x 2 lat2 c2 le a c2 6101 6102 Find the value ofx the crossing point such that the total travel time is minimized 20 a alj2xmaljWAzo E XRay Production xRays are produced when rapidly Electrons are emitted thermionically from the heated cathode and are accelerated toward the movmg electrons that have been r I i i anode when they strike it x rays ate pIOdLILELI accelerated through a potential difference 103 to 106 V strike a Heated metal target cathode Anode Power XRay emission is the inverse of supply the photoelectric effect for heater Photoelectric hf K of e XRay prod K ofe 9 hf Accelerating Voltage V Two Processes ID 1 Independent of target material bremsstrahlung braking radiation 9 gives maximum f energy or minimum ft max KE of accelerated e 2 Dependent of target material electrons With suf cient KE can excite atoms in the target material When they decay back to their ground state characteristic spectrum of XRays will be emitted The energy levels involved in these transition are typically separated by hundreds or thousands of eVs rather than a few eVs as is typical for visible photons Both of these cannot be explained by classical physics Compton Scattering In 1923 Arthur H Compton provided an additional direct conformation on the quantum nature of Xrays Scattered Detector photons X ray source V a a I d l NOTE Energy of Incident photon gtgt binding photons quot Target energy of e39 in graphite anyvgmfiim Pearson Eu Damn MC 9 blsn mg 3 Fee son mason we lay Xrays of well de ned xi are made to fall on a graphite target For various scattering angle intensities of scattered Xrays are measured as a function of the wavelength E Compton Scattering Experimental Observation The scattered Xrays have intensity peaks at two wavelengths xlo and xl A z Lio is called the Compton Shift Intensity per unit wavelength Photons scattered from tightly bound electrons undergo a neg ligible wave length shift 3 Photons scattered from loosely bound electrons undergo a wavelength shift given by Eq 3 23 I39 Compton Scattering But most importantly A is found to be a function of the observation angle Classical prediction Electrons in graphite absorbs Xrays and reemit them back If the electron is stationary then the reemitted But electrons are moving before amp after scattering so that 2 will be Dopper s shifted depending on equots velocity ll Compton Scattering Classical prediction Since diff electrons will have diff velocity the intensity profile for the scattered Xrays is expected to be a single peak with a spread around xlo But the actual experiment gives two peaks with a Compton Shift which depends on Classical physics can t explain this E Compton Scattering Quantum Explanation Compton and his coworkers showed that if we take the Xrays as particles With E hf pl hf c and treat the scattering process as a billiardlike collision then the observation of Aft s dependence on can be explained 9E Compton Scattering a Before collision The target electron b Aflequot COW 10 The angle belwee he is an 3 directions of the scattered photon and the incident photon is d Incident photon Target electron wavelength A at rest momentum 17 wavelength A I momentum F Vec7 4 4 6 Recall Ephoton hf 2 6 Recoilng elgctrogx Scattered photon 2 Eelectron 77W HITIQt rl entt lrn Pe a h a a lehgt m Z Pelectron 7mv E Compton Scattering Conservation of Total Relativistic Energy before after Note high energy collision 9 e s speed 2 he 2 might be relativistic 7 me 39 7771C Conservation of Relativistic Momentum before after h h x dz r cos mvcos A 1 7 y dzr Oz smq ymvsm knowns 9 ft given 45 observation angle 3 eqs with 3 unknowns 6 xl v Compton Scattering Dividing 0 throughout the energy conservation equation and de ne the following shorthand ght and g ht the three eqs can be written as 12 9 gg392 2g g39mc wjmzcz m v 1 gmc g39 7mc 2 gg39cos 7mvcos6 3 Og39sin 7mvsin6 g g39 mc2 gZmzc2 g g392 2g g39mcmzc2 gZmzc2 1 g g 2 2g g mc 72 1mzc2 2 1v2cz 1mzc2 2 2 l VZC2 1 v262 Compton Scattering 22 32 9 g g39cosg 2 72m2V2 cos2 9 G g392 sin2 5 yZmzv2 sin2 9 g2 2gg39cos g392 cos2 5 sin2 5 72m2V2 cos2 9 sin2 9 22 2 2 39cos 39sz v g gg 15 g 1v2cz Notice that the RHS is the same as the one from the energy conservation equation Combining them gives g g3922g g39mcg2 2gg39cos g392 f Zgg39Zg g39mc Zgg39cos gg39l cos g g39mc Compton Scattering Substituting the shorthand ght and g hl back 2 1 cos mc A W Compton Shift Equation The factor hmc has the units of length and 9f22426m is called the Compton Wavelength Compton Scattering The Compton Shift equation gives the exact prediction for the observed wavelength shift in the Compton scattering experiment The particle description for photon is correct ID Blackbody Radiation Continuous Spectra In contrast to line spectra emitted from excited atoms in a diluted gas radiations from many thermally excited atoms in a solid will in general emit a continuous spectrum with characteristics depending to its temperature Typically a good light absorber Will also be a good emitter so that a perfect emitter is typically called a blackbody Radiations emitted from a blackbody Will be characteristic of that particular system only since all other strayed radiations from the surrounding Will be absorbed by the system and not re ected back out The continuous spectrum emitted by such a body is called blackbody radiation An example of a blackbody is a cavity with a small opening Dicavit v Blackbody Radiation The total intensity of emitted light by a blackbody at absolute temperature T is given the Stefan Boltzmann Law I 0T4 where ais call the Stefan Boltzmann constant and it has the following value a 567gtlt10 8 4 m2K Blackbody Radiation The intensity is not unifonnly distributed over all wavelengths The intensity distribution for a given range of wavelength is called the spectral emittance 12 As the temperature increases the peak of the spectral eluiltance curve becomes Experimental observed spectral emittance for higher and shifts to shorter wavelengths different T s A 10 Wm 2000 K Note as expected I 1102de There is also the experimentally obtained Wien displacement law mT29OXIO3mK 7 A rum Dashed blue lines are values ot39Am in AS T M dominant A moves to Shorter 3830 for each temperature a m g Blackbody Radiation As we have seen ametallic cavity with a small opening is agood candidate for a hlachhody Most importantly one can derive the spectral emittance fora hlackhody using such acavity Model An unit cube metallic cavity lled with EM waves forming standing waves V normal modes with nodes at the walls For a given range of wavelengths am all the spectral emittance can be calculated as the combined energy from all the allowed normal modes within agiven wavelength range Blackbody Radiation ID This is given by loud1 NM E 0 where N1d1 276 cm A is the of EM modes standing waves allowed in the unit cube within the wavelength range 1 x1 cm and E is the average energy per normal mode standing wave within the cavity As we will see the different between the classical Rayleigh prediction incorrect and the quantum Planck prediction depends on how E is calculated Blackbody Radiation Distribution of energy states The energy of each standing wave normal modes is distributed according to the Boltzmann distribution iEkT PEdEe T dE Where k is the Boltzmann s constant and T is the absolute temperature recall the special case for molecular speeds the Maxwell s distribution of molecular speeds Huygens Principle Christiaan Huygens 1629195 The Huygens Principle can be used to predict the spreading of light wave It is a geometrical construction using every point on a wave front as the source of secondary wavelets that spread out in all directions with a speed equal to the speed of propagation of the wave B Secondary Application to the Law of Re ection wavelets Successive positions of a plane wave AA as it is re ected from a plane surface Huygens Principle Applies to Re ection I direction of 9r re ected ray Since 2 OP 4Q AP 900 gt AAQO A0PA and 9 6 A0 A0 Huygens Principle Application to Refraction a Successive positions of a plane wave AA as it is refracted by a plane surface I nb gt nu vb lt va nu C B K x P A qu 0 6a Note Wave slow down in material b U bf C B Material a S Material 1 A Huygens Principle Application to Refraction B Ut l opyngm c Materlal a 02mm MW 55 B Ubt Material 7 Mammary In time t the wavelet at Q travels to O and the wavelet at A travels to B 0 t 3 Q va ABzvbt Using AAQO and AOBA W3 haVe V a sin 9a 2 A0 sin 9 2 Vi A0 Substituting Va 2 cna and vb cnb and dividing Sln a cna nb 3 n sin6 n sin6 s1n6b cnb n a a b b Snell s Law a Chapter 34 Geometric Optics II Re ection amp Refraction at a Plane Surface II Re ection amp Refraction at a Spherical Surface II Thin Lenses II Optical Instruments E Images Formed by Flat Mirrors Object Image Rays tracing to find image Q 1 V Q 6 Originate rays from a point on an 2 I object 3 y 6 E 6 Follow thru their re ections P 6 V P Image is located at where E E E E re ected rays converge to or S S seem to diverge from follow ray 1 and 2 Lateral magni cation rmrv De nitions Real and Virtual Images ID II Image can be real or Virtual Real Image rays actually rst converge then diverge from the image point Image real Virtual Image rays do not actually pass thru the image point but they appear to be diverging from t E Images Form by Flat Mirrors Object Image Q 1 V Q 0 2 y 0 9 y P 6 V P S gtlt s Since APQV AP39Q39V congruent S 392 S for at mirrors y y m 1 Sign Rules ID 1 2 3 Object Distance s is if the object is on the same side as the incoming light for both re ecting and refracting surfaces and s is otherwise Image Distance s is if the image is on the same side as the outgoing light and is otherwise Obj ectImage Height y y is if the image object is erect or upright It is if it is inverted Sign Rules graphs a Plane mlrmr Object distance s is positive because the object is on the some side as the incoming light In both of these specific cases Image distance 5 is negative because the image is NOT on the some side as the outgoing light Obje t distance s is positive because the object is on the same Side is the incoming Image distance 5 is negative because the image is NOT on the some side as the outgoing light b Plane refracting interface Images by Multiple Re ectionsRefractions When there are multiple re ecting and0r refracting surfaces image formed by previous surface serves as object for the next surface Image of object P Image of image PI formed by mirr0r 1 formed by mirror 2 P1 P Mirror 1 7 Image of object P formed by mirror 2 Mirror 2 ig Re ection at a Spherical Surface Sign Rule 4 for the radius of curvature of a spherical surface COHCEIVG Center of curvature on same side as outgoing light R is positive Outgoing aThe radlus of curvature R is when the center of COHVGX curvature C is on the same s1de as the outgomg llght R lt 0 C concave and otherwise P Center of curvature not on same side as outgoing Outgoing light R is negative CV is called the optical axis Re ection at a Spherical Surface a Construction for finding the position P of an image formed by a concave spherical mirror From APBC 05 9 130th From ACBP 39 15 lt9 8 object V Eliminating 9 between these 2 Eqs Center of curvature Vertex a 2 s and s are both positive IF s39 gt I l yrgmoaaospearsonEuunaunnmcpuuusmngaspe n r Re ection at a Spherical Surface a Construction for finding the position P of an image formed by a concave spherical mirror B Relating the angles to the h physical distances we have 5 For a spherical mirror Point 0 object R 6 A a A b P39 P Center of curvature h V tanaz tang h 5 5 s39 6 Vertex Optic axis s and s are both positive IF Eel I l Cmyrgm 2005 Pearson Education me nubllshmg as Pierson Addisonrwesley h tan 2 R g 2nd Law Disorder amp Available Energy In a natural process a block sliding to a stop e g v J gt Macroscopic Mechanical Energy KB of the block is converted into the Internal Energy through heat absorbed as a result of friction slightly warmer due to friction gt At the end the coordinated motion of the block is converted into the KB amp PE of the slightly more agitated random motions of the molecules in the block 2nd Law Disorder amp Available Energy gt Since it is very unlikely that one can coordinate all these randomly moving molecules in a concerted fashion one typically cannot convert the internal energy of a system completely back to macroscopic mechanical energy gt However this does not mean that internal energy is not accessible An Heat Engine is exactly the machine that can perform this conversion but only partially The 2nd Law of Thermodynamics is basically a statement limiting the availability of internal energy for useful mechanical work ID The Carnot Cycle The Most Ef cient Heat Engine I A reversible cycle described by Sadi Carnot in 1824 I It provides a theoretical limit to the thermal ef ciency of any heat engine I The Carnot cycle consists of Two reversible isothermal processes Two reversible adiabatic processes An Ideal Gas as the working substance Steps of the Carnot Cycle Details on the Carnot Cycle The isothermal expansion a9 9 and compression 09d 0 T is constant and UT is a function of T only an Ideal Gas AU Isothermal V IQHI Wab nRTH 1n a9 isothermal expansion IQCIZVVCdZI ZRTCh l g nRTcln 10 KgtVd V0 Vd c9d isothermal compression Details on the Carnot Cycle The adiabatic expansion 990 and compression d a Qadiabatic O de nition From Section 198 we learned that TV71 const THVJ I W4 796 adiabatic expansion THVay l TchH 6196 adiabatic compression Dividing these two equations gives VLE 7am for adiabatic processes Ef ciency of the Carnot Cycle IQHI From de nition 6 2 Using our results for QC and QH from the isothermal processes nRTC1nVCVd TC1nVCVd nRTH 1nVVa TH 1nVbVa From the adiabatic processes we have Vb Va 2 V0 Vd t I I In 1 gt Carnot CYCle H Tmust be in K Efficiency of the Carnot Cycle ID 361 Carnot Cycle General Comments gt Higher efficiency if either T C is lower andor T H is higher gt For any realistic thermal process since the cold reservoir is far away from absolute zero T C will strictly gt O gt Thus realistic e is strictly less than 1 No 100 ef cient heat engine gt Realistic heat engines must take in energy from the high T reservoir for the work that it produces AND some heat energy must be released back to the lower T reservoir KelvinPlanck s Statement Refrigerators Refrigerators are basically heat engine running in reverse gt Heat from inside the refrigerator cold T reservoir is absorbed and released into the room high T reservoir with the input of mechanical work Expansion valv m Refrigerator Evapnralor 1 Condenser Low pressure lLigh I pressure Inside DI39 refrigerator c Inside of refrigerator at temperature Tc Compressor WM Compressor Refrigerators From 1St Law AUcycle O chi IQH W QH chW Note we have put in the explicit signs according to our sign convention V QC absorbed 9 positive V QH released 9 negative V W work is done by refrigerator 9 negative Coefficient of Performance for a Refrigerator W7 Q z you my its m A perfect Refrigerator chl KZIQC W IQHHch A perfect refrigerator means K 00 This means that IQHQc 039quot lWlO No mechanical work Wis needed to transfer heat from the cold reservoir to the hot reservoir Workless refrigerator The Clausius s statement of the 2nd Law does not allow this gt Kris IE Entropy We have seen that the 2 Law of Thermodynamics is a statement on nature s preferential direction for systems to move toward the state of disorder And Entropy is a quantitative measure of disorder Let consider an isothermal expansion of an ideal gas Intuitively as the gas expands into a bigger volume the degree of randomness for the system increases since molecules now have l Gasat r 100 K more choices in position for them to move 50 L 20 L about One can associate the increase in 7 iii 79797 randomness to the ratio 300Kquot 9300K AV dV 39 i 39 i 0 V Entropy Since this is an isothermal process we have the following relation from the 1st Law nRT dVLdQ d dWPdV dV gt Q V V nRT We introduce a new macroscopic variable S entropy and de ne as the in nitesimal entropy change for a reversible process at temperature T For any nite reversible process the total entropy change AS is SJK l Entropy another derivation Recall om a Carnot Cycle we have derived the following relationship QcTc IQHI chI QHVTH Q 3W6 0 Formally we can rewrite this as We have absorbed the explicit sign into the Q variable 2 ZTO cycle where Q represents the heat absorbedreleased along the isotherm at temp T Entropy another derivation Any reversible cycles can be approximated as a series of Carnot cycles l p Reversible cyclic 1 process for an ideal gas 39 Approximating the path of the cyclic process by a series Isotherms Entropy another derivation This suggests that the following generalization to be true for any reversible processes Where er is the in nitesimal heat absorbedreleased by the system at an in nitesimal reversible step at temp T j denotes the integration evaluated over one complete cycle cycle Entropy another derivation We have seen this property previously 95 dU 0 Changes in the internal energy U over a cycle Closed cycle is zero This is a consequence of the fact that U is a state variable and dU for any processes depends on the initial and nal states only Thus the result I dQ 0 indicates that there is another state variable S such that We and This new state variable S is called the entropy of the system ID r x E 2nd Law AS gt O amp Carnot Theorem For any reversible heat engines ASengme O ASW QC QH 211d Law TC TH Astor ASengme A env 2 0 C H WZQHQC Cold reservoir at temperature TC 2nd Law AS gt O amp Carnot Theorem Consider the ef ciency of the heat engine e l QC lQHI Applying the inequality from previous slide 6 1 IQCI s 1 T C IQHI TH T C Recall that the ef c1ency of a Carnot Cycle 1s given by 60mm 2 l H This gives our desired result Carnot engine is the most ef cient Internal Combustion Engine The Otto Cycle A fuel vapor can be compressed then detonated to rebound the cylinder doing useful work Intake thch Exhaust vulvc open c osetl l Cylinder Piston Crankshaft a Connecting l n l l i 39t39 PNUH tum us do n L qutng u 39onlprtssion slrukc lntukc till a closes Imnl ucntnn in ll1lr lquot mixluie l complexme n gttmlinc ui uuxunu cnlcis through Intttke vtllw piston mums up Both valves closed Spark plugY fires Ignition Spark plug tgnilcgt nllxtute Pomr stroke Hul htunt tl mixture expnnds pitslting piston tlown lntukc vulvc Exhaust valve Clnscd upon L 7 x l A A Exhttusl stroke FAllillN zllh opum pistun moves up smellingt nhutist unl lc39th ing L39ylintlui tetttly tut next ittltlke stroke The Otto Cycle Otto cycle 39M H I I II M I P t i c Heating at constant m z 39 volume fuel combustion iquot i rquot QH Cmdcr 7 1 Adiabatic expansion I I gt t L runkshuli anL Ilnncclmg 39 x t f gt 1 K I intake a b b c c d exhaust 39 ch For the two constant Vprocesses 2 and 4 I exhaust 01 0 V rV V we can calculate Adiabatic compression QH AUZ nCV gt O compression stroke Cooling at constant volume cooling of exhaust gases QC AU4 nCV T T07 lt O r is the compression ratio 10 to 13 The Otto Cycle Applying the de nition of efficiency C T T elgln Vquot d1Tquot Tquot QH quotMTG72 TcTb Now we can utilize the two adiabatic processes 1 and 3 Tallfl YEW 1 and TCVCH Tdde l TarT71 TbW 1 TC 17 1 TdrVy1 TQM 1 Tb To Td l ii The Otto Cycle Substituting T b and T 0 into the ef ciency equation we have e17 Td1 T6721 zyi Tc Tb Ta1397 1 71 7 1 TaTdy 1 gt l ill lir quoti ll zipT751quot Using a typical value for the compression ratio r 8 and 7 140 gives Note This is a theoretical value 6 056 or 56 Realistic gasoline engine typically has e 35 E The Diesel Cycle Diesel cycle Key difference 19 Fuel ignition heating at constant I No fuel in cylinder at the beginning of the QH Pl cssl39l elluel COFl bUS 101 l Th t k 1 b 7 39C 1s a Significant difference between compreSSIOn S 10 6 process 39 the Diesel and Otto C cles II Fuel is injected only moments before ignition in the power stroke process 3 I No fuel until the end of the adiabatic compression can avoid preignition III Compression ratio r value can be higher 15 Adiabatic expansion power stroke chl to 20 I V El Higher temperature can be reached during 0 V quotV the adiabatic compression 6 Adiabatic compression compression stroke II Higher e and no need for spark plugs 3 Cooling at constant volume cooling ol exhaust gases D steam eug me Chapter 33 The Nature and Propagation of Light III The nature of light Re ection and Refraction III Total internal re ection Dispersion Polarization HEIDI Huygens principle The Nature of Light Wavelengths in In I 021 l 0 N Frequencies in Hz mm IOIS Ion mm Visible light m quot m m m The wuve is traveling in the V positive 39 direc1i0n the same a A as he du39egllon of E X B vmuzr BLHF RED GRANGE vmnw GREEN E hat Light is a propagating electromagnetic waves The Nature of Light But as we will learn later light can also be characterized as discrete packets of energy called photons Also as we will learn later light travels in vacuum at the same speed c 299792458 x lOgms and it is the universal speed limit for all physical processes The Nature of Light ID El Index of Refraction n In materials light interacts with atomsmolecules and travels slower than it can in vacuum e g 3 vwater E Z c The optical property of transparent materials is called the Index of Refraction r27 Table 331 Since v material lt 6 always n gt1 3 Index of Refraction n Table 331 Index of Refraction for Yellow Sodium Light AU 589 nm Index of Substance Refraction n Solids lee 130 309 Fluorite CaFl 434 Polystyrene 49 Rock salt NaCl 544 Quartz SiOz 544 Zircon ZrOl SiOZ 923 Diamond C 2417 Fabulitc SrTiOA 2409 Rutilc TiOl 262 Glasses typical values Crown 52 Light lint 58 Medium lint 62 Dense int 66 Lanthanum int 80 Liquids ill 20 C Methanol CHlOl l 329 Water H30 333 Ethanol C21150H 36 Carbon tetrachloride CCI4 460 Turpentine 1472 Glycerine 473 Benzene 50 Carbon Llisul de CS 628 Index of Refraction and Wave Aspects of Light lt gt 7L medium a 7L medium b l a b Oscillating e Source of EM wave 0 0 na l lb Va Vb Note frequency of the EM wave is dictated by the oscillations of the charge and the timing of this oscillation can Z change for an observer in either medium a or b f does not change across media Index of Refraction and Wave Aspects of Light Recall for a EM wave we have V f 1 So in the two medium we have v f xi and vb f 1 Dividing these two equations we have Va 1a 1a 61161 1a nb Vb 1b 1b 6721 1b quota So the wavelength of a light changes in different medium With one of the medium being a vacuum we have Sources of Light El Thermal Radiation EampM radiation from accelerated charges due to thermal motion Mixture of many xt s a Bluer With higher T Sources of Light El Electrical Discharges thru Ionized Gases will study this later Lights are in a selected set of wavelengths e g sodium vapor lamps street lamps neon signs uorescent lamps IE Mean Free Path for Gas Molecules Assumptions 0 Molecule with nite radius r note pointlike particles do not collide The number density per unit vol is given by N V 0 Molecules move at an average speed v For simplicity let assume that the red molecule is the only one which is moving and to the right with speed v Then within a time interval dt the number of molecules that it might collide will be given by spatial volume indicated blue 47er vdtNV number density Thus the number of collisionsunit time is 47rr2vN dt V Mean Free Path for Gas Molecules ID To take into account that other molecules besides the red one are also moving the estimated collision rate should be higher and it can be shown that this mean collision rate will be larger by JE So dn 47rxEr2vN dz V The average time between collisions mean free time is then the reciprocal of this value V t mm 47zxEr2vN And the mean free path will be lvt V mean 4 er 0r we used PV 2 NkT ig Heat Capacities of Gases at constant V 6 TH Using the KineticMolecular model one can m calculate heat capacities for an Idea Gas U D For pointlike molecules molecular energy V V consists only of translational kinetic energy Ktr I And we know that Ktr is directly proportional to T B When an infinitesimal amount of heat dQ enters the gas dT increases and thr increases accordingly 0119ngde 0ranT Heat Capacities of Gases El From de nition of molar heat Table 18 Molar Heat Capacities capacity we also know 0f 33595 Type of Gas Gas CVJmol K d Q nCvdT Monatomic He 1247 I From energy conservation Aquot 1247 requiring leclKtr gives dWO Diatomic H2 2042 N2 2076 nCvdTganT 02 2110 I jl CD 2085 gt i l Polyatomic C02 2846 III Monoatomic Ideal Gas so 3139 matches well with prediction H25 2595 Heat Capacities diatomic Tr slalional motion The molecule Rotational motion The molecule roml h v gt x gt 39 my may he Llescnhed aboul I15 comer of mass This molecule has V39 mtquotde muvtmquot The n etlutl mu 1m Incuy Components nl39 Independent uxcs oi muumn quot5 lh mgh ll mle Lu quotn uL y a P39 11439 Independent AV axes of mania mum39s mass as being localed ul its nucleus W quotquot WW A diatomic molecule can absorb energy in its translation and also in its rotation and in the Vibrations of its molecular structure Equipartition of Energy ID This principle states that each degree of freedom separate mechanisms in storing energy will contribute 12 H to the total average energy count per molecule El Monoatomic 3 translational dofs 9 3 12 H This give Km 32 NkT same as before El Diatomic without vibration 3 trans dofs 2 rotational dofs This give Km 52 NkT or 52 nRT Again consider an infinitesimal energy change we have nCVdT zgchT and this gives CV 52 R liq Heat Capacities real gases I At low T only the 3 C BeluwSU KH AppI cuuhlc Apprccmblc molecules undergo I39olalional molion th39LIlionul I IIolion activated 7 JL I EL ESQZI KZSim X l g s ii lii gza quot W I At higher T additional 3R rotational dofs are 5 1 quot 2 activated 77777777777777 m I At higher T still R Translation vibrational dofs might m I I I I l I I THQ 0 25 50 00 250 500 000 2500 5000 0000 also get activated Heat capacity for 21 H2 gas m Heat Capacities of Ideal Solids El Atoms are connected together by springs El Assume harmonic motions for these springs Hooke s law filly till 1 El For each spatial direction quot v m quot there are two dofs A 39 Vibrational KE Vibrational PE and we have 3 dims 39 El For the entire solid we have 32 kT KE 32 kT 2 PE 9 E 3kT tot gt CV3R llllljlll Dulong Petit Prediction CV 7R2 Dulong and Petit prediction Lead Aluminum 3R 5R2 Diamond 2R 3R2 R Rz I I I I TK 0 200 400 600 800 1000 w Waves and Wave Fronts y Expanding wave fronts in 3D Propagating waves are usually visualized as a sequence of expanding wave fronts in space Point sound source producing spheiical sound waves alternating compressions the wave has the same pha51c relat10n and rarefactions of air e g crests troughs Wave front the locus of points Where l7 lE Waves Wave Fronts and Rays Expanding wave fronts Rays are always perpendicular to the wave fronts When wave fronts are sphericaL the rays radiate from the center of the sphere Rays Source Wave fronts When wave fronts are planar source is suf ciently far away rays are perpendicular to the wave fronts and parallel to each other Rays Point sound source producing spherical sound waves alternating compressions and rarefactions of air Wave fronts A ray indicates the direction of travel of the wave The Study of Light Optics II Geometric or Rays Optics In most daily situations light rays travel in a straight line in a uniform medium At the boundary between two materials air amp glass a ray s direction might change 1 Wave characteristics of light are not important Geometric Optics is the study of the propagation of light with the assumption that rays are straight lines in a xed direction through an uniform medium The Study of Light Optics ID Lgtgtl II Condition for Rays Optics W Relevant system size gtgt wavelength mirror L In this approximation wave characteristic of light is not important and rays model of light gives accurate predictions Visible light ft 500 nm ltlt L 9 Rays Optics works well with typical optical instruments mirror lens cameras telescopes II Physical Wave Optics Ch3536 The study of light when wave properties of light are important diffraction and interference L z 1 Re ection and Refraction incident ray normal re ected ray 6a 9r p N boundaryinterface a1r material a na between two media 39 l b glass materla nb refracted ray J 6b When light hits a boundary typically a part of it Will be re ected amp a part of it Will be refracted Re ection III Specular Re ection Cl Re ection from a smooth surface a Re ected rays are parallel El Diffused Re ection D Re ection from a rough surface n Re ected rays are in various directions a Specular re ection l b Diffuse re ection g Law of Re ection incident ray normal re ected ray material a na mateiial b 11 914K f s axe elf E Law of Re ection general interface incident ray re ected ray normal ea 9 A 1 tangent plane i Retrore ector Example 333 Mirror 2 7 Mirror 1 180 201 Re ected ray goes back in the same direction as incident ray independent of incident angel 6 D comer min39ot IE Law of Refraction Snell s Law incident ay no rmal H Nilquot ea b material a na mammal b nb refracted ray 6 l wfw r ie uf quot A mm Qty 3971 Snell s Law 3 cases A ray oriented along the normal does not bend regardless of the materials 0a 0b Incident 7 r IR f rt d N 0U a1 Re ected e me 6 Recall t n C V Dpvrexioll A W entering 3 material 0f 1W8 index Of A ray entering a material of smaller index 39C aCtiO bends WWW the quot0mm of refraction bends away om the normal Material 1 Material b I 39 lt Inc1dent Inc1dent quotb quota nb gt n 4 I t N0rr2a1 w Re ected Reflected Mgtgrial a Refrac ted Matenal b 7 Total Internal Re ection gt Light moves from a medium with a larger n to one with a smaller n gt As the angle of incidence becomes more and more acute the light ceases to be transmitted only re ected gt0 crit At the critical angle of incidence 0cm the angle of refraction 0 2 900 Any ray with 6 gt 6cm shows a total internal re ection gt Incident laser beams Refracted at interface Total internal re ection Two mirrors at different angles Total Internal Re ect II Critical Angle 9 is determined by the borderline ease ray 3 na sin 60m nb sin 900 from Snell s Law my 3 39 bit sum 3 quot 39 only valld for nagtnb 5 Photon Absorption In general a photon emitted when an excited atom makes a transition from a higher level to a lower level can also be absorbed by a similar atom that is initially in the lower level f m Ef An atom is raised from an initial level i to a higher W energy final level fby absorbing a photon with hf Ef E1 energy equal to E Ei A E 1 U i gg Absorption Spectrum When a white light with continuous spectrum pass through a gas photons with particular wavelengths corresponding to these transitions can be absorbed These will result in dark lines in the absorption spectrum of the transmitted light Astronomers can analyze the atomic components in interstellar gases by studying these absorption spectra Example 384 Emission and Absorption Spectra Hypothetical atom with 33900 6V three energy levels 200 eV a 7 can be emitted when l 300 CV 300 CV this atom is excited 100 eV I A 100 eV 100 6V b 7 can be absorbed Ground V when this atom is initially level I I I in the ground state EmlSSlon Absorptlon Wtransitions transitions Example 384 a Three possible transitions AE 16V 26V and 3eV red downward arrows he 4136x10 15evs300x108ms 1240nm 620nm or 414nm AE leV 2eV or 3eV b Two possible transitions AE 16V and 36V blue upward arrows he 4136x1015evs300x108ms l l240nm or 414nm AE leV or 3eV 414 nm 620 11m 1240 um I arson Euunahon me Unuhshmg 35 Pearson Admsuurwesley cowngme gone P2 ig The Nuclear Atom a bit of history What were known by 1910 0 1897 J J Thomson discovered e and measured em ratio 1909 Millikan completed the measurement of the electron charge e 0 Size of atom is on the order of 103910 m 0 Almost all mass of an atom is associated with the charge But the mass and charge distributions inside the atom were not known The leading assumption was by J J Thomson and an atom was modeled as a sphere with positive charge and the electrons were thought to be embedded within it like raisins in a muf n Ernest Rutherford designed the rst experiment to probe the interior structure of an atom in 19101911 together with Hans Geiger and Ernest Marsden ig Rutherford Scattering Experiments Alpha particles high energy charged helium Source of nuclei It can travels alp al several centimeter par 10 es through air and O lmm through solid matter Target Thin gold sliver JZinc sulfide or copper fOIIS scintillation screens The alpha particle is about 7300 times heavier than an electron So by momentum considerations alone it Will have only minimum interactions with the much lighter electron Only the positive charge Within the atom s majority mass can produce signi cant de ections in scattering the alpha particles Rutherford Scattering Experiments Thomson s mode of the atom An alpha BUT there were large scattering angles particle is scattered through only a small angle N 18009 almost backward Observed Indicating that Thomson s model was not correct Rutherford s model of the atom An alpha particle can be scattered through a Positive charge inside atom is distributed throughout the volume so not drawn to scale that electric eld inside the atom is n K expected to be small and the force a If ONUC1 US that it exerts on the a will also be I l small 9 de ection angles are t X X expected to be small a 5 H liq Rutherford Scattering Experiments These large angle de ections can only be possible if the positive charge within an atom is concentrated in a small and dense space called the nucleus The following is a computer simulation of the de ection angles with various nucleus size done much later The experimental results agreed with simulation data using small 103915 m nucleus sizes a A gold nucleus with iudius 7 n x 0 H m gives iai39geenngle scattering b A nucleus with IO times the radius oi39lhc nucleus in a shows In lui ge suulc scuttcring The nucleus occupies only about 103912 of the total volume but it contains 9995 of the total mass liq The Bohr Model It is a mixture of classical and new quantum ideas in trying to theoretically calculate the energy levels of a hydrogen or hydrogenlike atom Assumptions of the model 1 e39 moves in circular orbits around a proton under the in uence of Coulomb Force Law Only certain circular orbits are allowed angular momentum of e39 around nucleus must be multiples of h h 27 Photons are only emitted when e39 jumps from a higher energy y orbit to a lower one The proton is assumed to be stationary Proton Me Electron The electron revolves in 1 circle of radius r with speed u m 7e The electrostatic attraction provides the needed centripetal acceleration Bohr s Model general comments Bohr s Model is not a fully quantum mechanical theory It contains both classical ideas electrons in circular orbits around the nucleus not correct and new quantum ideas the idea of quantized light amp discrete energy levels correct Nonetheless it is a successful theory since it provided accurate predictions on the observed atomic spectra lines a V Conceptual problems with the classical circular orbits idea According to classical EM an accelerating electron in circular orbit around the nucleus will radiate EM radiations Thus its energy will decrease continuously and spiral rapidly toward the nucleus The radiated frequency will also depend on the frequency of revolution and since its orbit is expected to decay continuously the radiated spectrum will also be continuous with a mixture of frequencies Bohr s Model general comments The correct quantum picture is that 9The electrons are in stable energy levels orbits in the atom 9The atom only radiates a photon When an electron makes a transition om a higher energy level to a lower one hf El E f The classical picture of an electron revolving around a nucleus is not correct As we Will learn later the quantum picture of an electron is more like a cloud With a given distribution function around the nucleus Bohr found that these stable energy levels are quantized With discrete levels characterized by its angular momentum 1 x 2 4 n is called the principal quantum number for the energy levels ID Bohr s Model mathematical details From a classical calculation for a Hydrogen atom the electron is bounded to the nucleus by the Coulomb s Force 162 47590 r2 For an electron in circular orbit around a nucleus with radius r we have 1 e mv2 4723980 r1 162 4780 mv2 Fma gt Solving for r we have P 11 From the angular momentum quantlzatlon we have V mr I l g Bohr s Model mathematical details Substituting this expression for v into the radius equation we have 1 62 mr 2 rm 2 472290 m nh e2 mrf quot 47rgo 11th orbital radii for a Bohr atom 312 w and Wm 5139 orbital speeds for a Bohr atom Bohr s Model mathematical details The smallest radius for the Bohr atom corresponds to n 1 and this minimum radius for the Bohr atom is called the Bohr s radius 2 ax 7412 quotfob With this fundamental length scale for an atom the other radii can be written as r aOnZ n 123 Now consider the total energy for an e in an orbit at r En KnUn Bohr s Model mathematical details ID l e2 mv2 l e2 Recall from F ma we have 2 gt mv2 4723980 r r 4723980 r So as in any circular motion under 11quot2 type of force we have K U 2 This gives the total energy for an e in an orbit at r l 62 62 me2 1 En 2 2 8750 r 872290 47rgoh n A me 1 ri 13146ie V 27 23 39 39 energy levels of aBohr s atom Note Energy levels are quantized as a consequence of the angular momentum quantization ID Bohr s Model notes The ground state energy of the Hatom is given by E1 136 eV when n 1 0 Comparing the expression for Hatom s energy with the empirical formula derived by Balmer we can derive an explicit expression for the Rydberg Constant me4 me4 7 Agree well with previous hCR T R 2 3 1097 X 10 171 1 experimental value using 880 h 880 h C wavelength measurements 0 The energy required to remove an electron completely is given by the transition from n 1t0 n 2 00 and it is called the ionization energy Ew E1136eV 139 onizatz39 on The Ideal Gas Law graphical View PVT relationship in the Ideal Gas Law can be Visualize graphically as a surface in 3D 7727wa1177 Constant pressure isobars Constant volume isochors Constant Ternperal ure isothenns l g PV Diagrams Each curve represents pressure as 1 function of El Shoes Of the volume for an ideal gas at a single temperature P V 1 preVIOus surface F01 eacll CtuvepVIs 42011 le and Is directly proportional to T Boyle s law ngtngtngtn El Evolution of a gas at constant Twill move along these curves called isotherms Gives P vs Vat a various T P nRT KineticMolecular Model of an Ideal Gas ID Macroscopic description Microscopic description of gases of gas molecules P K T lt gt V p F KE Ideal Gas Law Kinetic Theory NeWtOH S Eq5 I An example of a successful theoretical linkage between the micro and macro descriptions of an ideal gas I Explicit expressions of P amp T in terms of microscopic quantities Kinetic Theory assumptions ID I III A very large N of identical molecules each with mass m in a container with volume V Molecules behaves as pomlparlicles Molecule sizes ltlt avg separate bet particles amp dim of container Molecules moves according to Newton s laws and they move randomly with equal probability in all directions and with a xed distribution fv histogram of speed v m Fractions of molecules moving at a given speed range stay the same speed v Molecules interact with each others only thru elastic collisions and the container walls are perfectly rigid and in nitely massive 9 both KB and momentum are conserved applet i Kinetic Theory model Left Wall L before collision lt gt vi 123 II Vx f 0 A lt I I H x H l Vf y after collls1on Idea Gas in a box with VAL Vx 0 Velocity component parallel to the wall y component does not change 0 Velocity component perpendicular to the wall xcomponent reverses direction 0 Speed U does not change Conyngm 2003 Pearson Education inc publishing as Pearson AdcwsorvAWesley Pressure Exerted by an Ideal Gas ID Pressure on left wall due to molecular collisions 1 Momentum change in Xdir by a molecular moving to the left at Vii Amv Pf Pl 2 mvx mvx 2mvx 2 Duration At that this molecule takes to collide with the same wall again diluted gas 2L At 3 vx Left Wall Pressure Exerted by an Ideal Gas 3 Force exerted by this molecule on the left wall F Amv 2mvx mv At 2Lvx L 4 With N molecules total force on wall in At Pressure of an Ideal Gas 5 Random direction assumption VXZLWZ Vy2av V22aw xyz are the same Since v2 vx2 vy2 V22 9 We haVe V2aV 3Vx2av V2 L 3 Finally the pressure on the wall is 0V This gives Fm F wz lev2 tot A AL3 3 V 61V 61V Pressure of an Ideal Gas ID Rewriting we have avg KE per molecule 2 1 PV NEmv2avj NKEW I This tells us that P inside the container quot is proportional to the of molecules N quot is proportional to the avg KB of molecules These are microscopic properties of the gas g Molecular Interpretation of Temperature From before P177 gaging From Idea Gas Law P VNkT Recall k Boltzmann Constant For both of these to be true we need to have Note Ktr in your book is the lolalKE for all molecules in volume V 1w welcome Temperature is a direct measure of the average translational KB of the molecules in an ideal gas 9 Distribution of Molecular Speeds II Within an ideal gas molecules moves with a 39 si of s eeds 11Ver W p Fraction of molecules I A mathematical rigorous way with Speeds from 01 to v2 to describe this statistically is thru a distribution function 17v II Properties fv0v0 0 0102 UA 39 ule fay 9 0 v 9 large Fractlon of molec s wrth speeds greater than vA fv largest at midrange E MaxwellBoltzmann Distribution f U 44 026W II vdv gives the probability of finding molecules with speed in flU range vvdv II Averages with respect to the distribution of molecular speeds can be calculated using v 00 l Vavivfvdv avg of v 0 As temperature increases 39 the curve flattens 2 oo 2 39 the maximum shifts to higher speeds 2 V V 2 JV fvdv avg 0f V2 Dispersion Index of refraction n I The index of refraction 17 n is usually a property of the medium but equally important it also varies With the 16 wavelength A in vacuum of the light dispersion I n typically decreases With increasing 1 1 4 400 00 600 700 Wavelength in vacuum nm ID Physical Observable Consequence of Dispersion III The Visible Spectrum of White Light According to Snell s Law angle of refraction depends on n n I 39gt angle of refraction I white light c V 6a quot s1n 9a 2 n s1n 9M and n s1n 9m a1r T glass y fixed n gt n ebl 1nc1dent E d1rection 91mm lt 9m e of whlte r light g5 Dispersion by a Prism White V Deviation of light yellow light f j imil Douy gl c 2003 Pearson Euucanun me pubhshmg as Pearson Addlsmlrweiey Dispersion in a Rainbow Forming a rainbow The sun in this illustration is directly behind A dOUble rambow the observer at P The rays of sunlight that form the The two refractions primary rainbow refract into the z disperse the colors droplets undergo inter rial reflectlon and I refract out Water droplets in cloud To the down sun point Angles are exaggerated for clarity Only a primary ObseTVer rainbow is shown at P x i Copyrighl 2003 Pearson Education ine pubhshing as Pearson AddisonWesley Capyrighlm 2003 Pearson Education Inc publishing as Pearson Addisonrwesiey Polarization For a transverse wave on a string the direction of the wave s displacement gives the polarization of the wave a Transverse wave linearly polarized in the b Transverse wave linearly polarized in the y direction zdirection y ii A Linearly Polarized EM Wave Similarly for an electromagnetic wave the direction of the electric eld vector 171x z gives the polarization of the wave The wave is traveling in the V positive 39directionthe same as the direction ol39E X B gt An transverse electromagnetic wave with a E polarization in the ydirection Ext coskx wt l3xt Bm coskx at E E A polarized wave in a well de ned direction is called a linearly polarized wave Polarization by Filters A nonlinearly polarized wave on a string can be polarized by a slot barrier Tbe mfmw mmasam m mg km pas ngn yxnnponenwlx a zedinthe ydirection y Bwloti 0L lA x V U V Z Polaroid Filter for an EM Wave Filter only partially absorbs vertically polarized component of light InCidenF Polarizing unpolarized axi S 39 light rF 4gt Polaroid filter Filter almost completely Transmitted light is absorbs horizontally linearly polarized in polarized component of the vertical direction light Dapyrlgm 2005 Pearson Educsunn ma publishing as Pearson Addismvwesley E The Action of a Polarizing Filter Pulmm l Unpolarized incident light will be linearly polarized parallel to the polarizing axis after transmission hmnccll palmwing We can analyze the intensity of the quotquotquot ii 1 23 WW I st or me I transmit light passing thru the second polarizer an analyzer u npulurizcd EH Fem 4 Only E will be transmitted Wm H Thu IIHCIMII Io llgltl mu m 1 5m u muuumll lien Vin lnlhcrmk I me 1 Em EH Ecos mumu minmnmmlcnhL ml ymmllclAmlVi rpululicnlun Icvu ll vl n m pnlmwmg 11MI m uualmt E The Action of a Polarizing Filter hhlhcmglclvclwcrnllvcpnlnnmg Slnce Intensrty 1 1s proportlonal to E9 quotW my I Pnlal izer 11161 1 c n l unpolumcu liqu 2 tram max COS Malus s Law Transmitted intensity of linearly polarized light through a polarizer leluucll 391 lw IlllL lhll In hgm lrnm IlIC umlym I11IIHMI um when gt 7 1 AI nllm Anglm I 1 le m Mum mm cmlmmlcnk 1 mm 1 mumI m pcllmlrliculluv I39cwpccln39rl m lhc 1Yllll39llll1gll m m Allyer For an upolarized light E is in all directions Transmi 7c 271 l 2 trans maxICOS 2max 0 l llomcell i 39I39ln llllclhll ul my tmmnnlml llghl h Ilu39 mmc Iquot All nnumuum m Ilh palmmin mm nuqu u39l 39l39 39u lI Polarmng I I l l ulp lumr Ilm nu u 11mm mmmn n mu m Inculcnl Inmmh m p D p xlanzex 9H Polarization by Re ection then the reflected light is 0 poltu ietl perpendicular to the plane of incidence D If tmp olnrized ljvht is incident at the N ormal Plane of l incidence 6 Altcrnntivcly il llt39tpt tltu lZELl light is incident on the reflecting surface at an angle other than 0P he rcl39lcctcd light l partially 39 pnluriicd e rViym t It7 and the transmitted light is mrriuIy polarized parallel to the plane of incidence At a special angle polarizing angle or Brewster s angle 81 the electric field component parallel to the plane of incidence will not be re ected This occurs when a Polarization by Re ection From Snell s law we have na s1n 6p 2 nb s1n 6b Then using the condition for 9p 6b 7z39 2 617 na sin 6p 2 113 sin7z2 6p This gives i la Sin 6p I lb COS 6p Brewster s Law Component perpendicular Reflected to plane of page ray Normal When light strikes a surface at the polarizing angle the re ected and refracted rays are perpendicular to each other and h tan 0p I Copyright 2mm Pearson Education um publishing as Pearson AddisoaneMey Another Thin Film Example non 12 nai lt I l lm lt 71ng air refractive coating on lens beating glass wave 1 reflected from top interface of the coating n gt n 3 1800 or 7239 phase shift coating 117 wave 2 re ected from bottom interface of the coating n gt n 3 1800 or 7239 phase shift glam coating Interference from 21 Thin Film ID Since both wave 1 and 2 suffers the same phase shift upon re ection the net phase difference will be om the path difference 21 only So we have the standard condition net phase diff due to path diff only There is one more consideration the path difference is accumulated in a so that the relevant wavelength should be ft 111 coating medium with negating Constructive 2t mg 7 m 0172 znwatmgf quot11 W 012 39 quot Note In addition to wavelength modi cation D 1 the RHS dependence are CSthtlve39 2t m 3 in m 0 1 2 39 quot switched with respect to 1 the air gap case 2nwa ngt m 3 1 m 01 2 m Thin and Thick Films a Light re ecting from a thin film b Light re ecting from a thick film Z The waves re ected from 391 the two surfaces are from V Bursts0f light a different bursts and are not coherent few urn long a The waves reflected g J from the two surfaces 9 h fare part of the same 3 burst and are coherent Thick film 1 Thin film Interference effects are dif cult to observe Interference effects can be observed Eiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiigg Example 356 Thin Film h150 Plastic h00200 mm n14o FHhT lass 9 h WWWWWMWWWWWWV Question a Will there be a bright or dark fringe at the point of contact b What is the distance x to the next bright fringe IE Example 3 56 Since n plastic lt quotsilicone and quotsilicone lt quotglass K0 hm Silicone grease Y 1 Both wave 1 and 2 suffer a phase shift 0 b I i 00200 W So at the point of contact 10 FHm glass the re ected wave 1 and 2 9 71396um will arrive at the eyes in phase bright fringe To find the location of the next constructive interference we use 2nsiliconet 1 0 m E Example 3 56 From the two similar triangles we have x 7 0150 Plastic I Eh 00200 mm 0 140 Substitute tinto the previous eq Flint glass 21500th anmg 7 10 I a 100 500 x Mquot m m 0833mm 2n h 215000200mm coating 51 Newton 3 Rings F I k rmgcgtnnp ac u n helwecn lens and muster b Newton s rings circular interference fringes Muster Lens being tested a A convex lens in contact wilh a glass plane Canyrlgm a 2an Pearson Edumtmn W publishing as Peavsnn Addlsonrwesley When Viewed in monochromic light the interference pattern is a set of concentric rings called the Newton s rings Since each fringe corresponds to a path difference A the lack of symmetry of these rings can be used to check for precision in lens making
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