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6.013/ESD.013  Electromagnetics & Applctions

Fall 2007

Instructors: David H Staelin, Erich P Ippen, Mehmet F Yanik

TAs: Theodore Golfinopoulos, Alan Lee, Aristeidis Karalis, Johnna Powell

Lecture:  TR1  (32-141)        

Announcements

Practice Final Exam Solutions Posted

Hello, All,

    The solutions to the practice final exam have been posted in the "Materials" section under the "General" heading.

    Also, we'll be holding another tutorial this Sunday from 2:00 PM to 4:00 in 36-112, as well as on Monday and Tuesday in 26-328 from 4:30 PM to 6:30 PM.

Good luck,

Ted

Announced on 14 December 2007  4:41  p.m. by Theodore Golfinopoulos

Review tutorial Friday (tomorrow) 1-2pm at 26-328

Hello all.

We will have a review tutorial tomorrow Friday 1-2pm at 26-328. The focus will be problem solving, by examining problems from past and sample exams and more, and highlighting the  absolutely basic and fundamental things you should really know to avoid big mistakes.

Aristeidis

Announced on 13 December 2007  6:13  p.m. by Aristeidis Karalis

Corrections on Solutions of PS#12

Problem 12.1(c):

kx = k1*cos(85) <=> 2π/λx = 2π/λo*sqrt(ε1)*cos(85) <=> λx=λo/[sqrt(ε1)*cos(85)]=5.737*λo     =>     D~0.6*λx=1.721*λo     [and not D~4*λo]

Problem 12.3(c):

Inside the laser cavity, the fields have the following profile: E=Eo*sin(km*z), H=Ho*cos(km*z), where    Eo/Ho=η=sqrt(μ/ε)    and   from the boundary conditions:

km*D=[ωm/c]*D=[2π/λom*co/c]*D=m*π,

where   c=1/sqrt(ε*μ) and co=λο*ω/2π=1/sqrt(εο*μο).

To find the total energy stored in the laser cavity for the mth mode:

Wm = 1/2*μ*Η(z,t)^2 = 1/2*μ*Ho^2*cos(km*z)^2*cos(ωm*t)^2   =>

<Wm> = 1/4*μ*Ho^2*cos(km*z)^2   =>   <wm> = D/8*μ*Ho^2   =>

<w>=2<wm> = D/4*μ*Ho^2

Since the standing wave is made out of two counterpropagating waves, each one of these ways, and in particular the forward propagating one that leads to small transmission through one of the mirrors, has half the amplitude of the total standing wave. Therefore, the power transmitted is:

<P> = T * 1/2*η*(Ho/2)^2 = T*1/8*η*Ηο^2

Therefore:

Qc = ωm*<w>/<P> = ωm * (D/4*μ*Ho^2) / (T*1/8*η*Ηο^2) = 2*m*π*c*μ/η/Τ = 2*m*π/Τ,   since c*μ/η = c=1/sqrt(ε*μ) *μ/ sqrt(μ/ε) = 1, so finally:

Qc = 2*m*π/Τ = 2*π/T * (2D/λom*co/c) = 2*π/0.01 * (2*10^-3/10^-6*sqrt(2)) = 1.8*10^6

Announced on 13 December 2007  2:47  p.m. by Aristeidis Karalis

Final Exam is Cumulative!!!

Many of you have asked about this!  The final exam is cumulative.

Announced on 12 December 2007  6:31  p.m. by Theodore Golfinopoulos

Additional TA Review Sessions; Exam Preparation Materials

Hello, All,

     We have scheduled four additional review sessions, and are working on a fifth.  So far, we've planned to run interactive-style tutorials on

Friday, 12/14, 2:00 PM-4:00 PM in 26-328
Sunday, 12/16, 2:00 PM-4:00 PM in 36-112
Monday, 12/17, 4:30 PM-6:30 PM in 26-328
Tuesday, 12/18, 4:30 PM-6:30 PM in 26-328

We will likely have an additional hour of problem-solving review on Friday from 1:00 PM to 2:00 PM, where we'll go over a few problems from the sample final (so it would be good if you had worked these problems, yourselves, beforehand).  Please check the announcements on the Stellar site over the next few days for announcements regarding this and other topics (including corrections to homework solutions, etc.).
     Lastly, a sample final exam has been posted to the Materials Section, along with the actual formula sheet that will be included in your final exam packet (solutions will be posted soon).  And you can also download a final exam from 2005 from OpenCourseware.

Good luck, and please contact us if you have any questions.

Take care,

The TA's

Announced on 12 December 2007  4:50  p.m. by Theodore Golfinopoulos

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