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E = A cos (kx - w t + f 0 )

“Amplitude” “Phase”. E = A cos (kx - w t + f 0 ). Or “phase” and “inititial phase”. E = A cos (kx - w t + f 0 ). E = A cos (kx - w t + f 0 ). In this class, we will be adding up a lot of oscillatory electric fields!. E = A 1 cos ( w t + f 1 ).

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E = A cos (kx - w t + f 0 )

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  1. “Amplitude” “Phase” E = A cos (kx - wt + f0)

  2. Or “phase” and “inititial phase” E = A cos (kx - wt + f0)

  3. E = A cos (kx - wt + f0) In this class, we will be adding up a lot of oscillatory electric fields! E = A1 cos (wt + f1) + A2 cos (wt + f2) + A3 cos (wt + f3) + A4 cos (wt + f4) + A5 cos (wt + f5) Solve for Atotal And ftotal. Good luck! + A6 cos (wt + f6) + A7 cos (wt + f7) + A8 cos (wt + f8) = Atotal cos (wt + ftotal)

  4. eif=cosf + i sin f |M| cos(wt+f) = Re ( |M| ei(wt +f) ) = Re ( |M| eif eiwt ) = Re ( A eiwt ) = Re ( A eiwt ) A=|M| e if

  5. E = A1 cos (wt + f1) + A2 cos (wt + f2) + A3 cos (wt + f3) + A4 cos (wt + f4) + A5 cos (wt + f5) + A6 cos (wt + f6) Take these real magnitudes and phases, Covert them into complex amplitudes… + A7 cos (wt + f7) + A8 cos (wt + f8) = C1 e iwt + C2 e iwt + C3 e iwt +… CN e iwt + = Ctotal e iwt With Ctotal = C1 + C2 + ….CN You’ve solved for Atotal and ftotal. Good job! Note, we’ve “forgotten” to say “real part of”

  6. Im(A) Im(A) Re(A) Re(A) “Graphical sums” A2 A3 E= Re (A1 e iwt + A2 e iwt A3 e iwt ) A1 At E(t=0) Three different At give same E(t=0)?

  7. Im(A) Im(iA) Re(A)=E(t=0) Re(iA) “Graphical sums” Three different At give same E(t=0)? E= Re (A e iwt ) Yes, but they correspond to very different E(wt=p/2)

  8. “k” Plane waves, sure, but spherical waves are solutions, too. k

  9. X CX=C1 + C2… Ci = (Ai/ri)eikri

  10. 6 5 …. Clicker Question 4 Q. Imagine a large number N of identical light sources, each small compared to a wavelength of light, each equidistant from the detector at X, and each emitting single-frequency E&M spherical wave with same phase and initial amplitude. If the average intensity at X from one of these waves is I0, what is intensity of Total field from all N sources? 3 2 1 X • On average, zero. • On average NI0 • On average N2I0 • Depends. (on what?) • Can’t tell from what we know.

  11. 6 5 …. Clicker Question 4 Q. Imagine a large number N of identical light sources, each small compared to a wavelength of light, each equidistant from the detector at X, and each emitting single-frequency E&M spherical wave with same phase and initial amplitude. If the average intensity at X from one of these waves is I0, what is intensity of Total field from all N sources? 3 2 1 X Answer: C. Electric fields all add up in phase to give a phaser longer by N, So intensity, goes as field squared, is greater by factor of N2 E field N times longer

  12. 6 5 …. Clicker Question 4 Q. Same as previous question, except now the initial phase of each source is random, in general, different from the other sources. If the average intensity at X from one of these waves is I0, what is intensity of total field from all N sources? 3 2 1 X • On average, zero. • On average NI0 • On average N2I0 • Depends. (on what?) • Can’t tell from what we know.

  13. 6 5 …. Clicker Question 4 Q. Same as previous question, except now the initial phase of each source is random, in general, different from the other sources. If the average intensity at X from one of these waves is I0, what is intensity of total field from all N sources? 3 2 1 X B. Random walk of N steps takes E-field N1/2 steps from origin, on average, so average intensity goes as (N1/2 )2. • On average, zero. • On average NI0 • On average N2I0 • Depends. (on what?) • Can’t tell from what we know.

  14. …. Clicker Question 1 2 3 4 Q. The four emitters are far enough away from X that the path length from each to X is the same to within much less than a wavelength. But each adjacent emitter is ¼ wavelength close in path to Y. The initial phase of each emitter is the same. Which is true? In units if I0, the average intensity.. • …at X is 16, at Y is 0 • …at X is 4, at Y is 4 • …at X is 4, at Y is 41/2 • Depends. (on what?) • Can’t tell from what we know. X Y

  15. …. Clicker Question 1 2 3 4 Q. The four emitters are far enough away from X that the path length from each to X is the same to within much less than a wavelength. But each adjacent emitter is ¼ wavelength close in path to Y. The initial phase of each emitter is the same. Which is true? In units if I0, the average intensity.. Answer: • …at X is 16, at Y is 0 Field increased by 4 at X Field adds to zero at Y X Y

  16. P Clicker Question X M 1 A B S w/ mysterious M 200 l - w/o mysterious M 100 l - x Q. Light travels from source 1 to detector at B, passing through plane P. The total path length S from 1 to B depends on the position X at which the path pass through P, and it also depends on whether or not mysterious thing M is inserted in the path. The two curves above show how the optional presence of M modifies the path length. Assuming M does not absorb any light, but just speeds it up or slows in down, is the intensity at B… A. … higher with M in place. B. …lower with M in place. C. ...the same.

  17. P Clicker Question X M 1 A B S w/ mysterious M 200 l - w/o mysterious M 100 l - x Q. Light travels from source 1 to detector at B, passing through plane P.The total path length S from 1 to B depends on the position X at which the path pass through P, and it also depends on whether or not mysterious thing M is inserted in the path. The two curves above show how the optional presence of M modifies the path length. Assuming M does not absorb any light, but just speeds it up or slows in down, is the intensity at B… Without M With M Answer A, … higher with M in place..

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