Lect33_[Compatibility_Mode]

# Lect33_[Compatibility_Mode] - Physics 344 Foundations of...

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Physics 344 Foundations of 21 st Century Physics: Relativity, Quantum Mechanics and heir Applications Their Applications Instructor: Dr. Mark Haugan Office: PHYS 282 [email protected] TA: Dan Hartzler Office: PHYS 7 [email protected] Grader: Shuo Liu Office: PHYS 283 [email protected] Office Hours: If you have questions, just email us to make an ppointment. e enjoy talking about physics! appointment. We enjoy talking about physics! Notices: Problem Set 11 is due next Monday before the beginning of our last lecture session before Thanksgiving. idterm II at 8:00pm Thursday December 2 in MSEE B012 Midterm II at 8:00pm, Thursday, December 2 in MSEE B012

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This week’s problem set includes two problems that ask you to use the real-valued and complex-valued representations of electromagnetic fields in a resonant cavity to make predictions about the probable outcomes of photon detection experiments. As you will recall, you had a problem last week that asked you to determine the α m and β m coefficients that allow you to construct such representations as superpositions of modal fields each with its own frequency. For the y -polarized elds that we have been discussing, fields that we have been discussing, 0 2 sin( ) ( ,0) a mm y kxE x d x a β= and 0 2 cos( ) a z kxcB x d x a α = Question 1: Which of the α m coefficients do you expect to have the largest magnitude for the field that has E y ( x , t =0) = 0 and cB ( x , t =0) = (1-2* x ) V/m in a cavity 1 m z long? A) α 1 B) α 2 C) α 3 D) α 4 ) none of these E) none of these
Answer: Here are plots showing cB z ( x , t =0) and cos( π x/a )c o s ( 2 π x/a ) cos(3 π x/a ) So, plots of the products of these functions (the integrands determining α m ) are Clearly, the integrand of the first of these and, therefore, α 1 has the largest magnitude. Note, too, just as α 2 = 0 we find that α m = 0 for all even m because of symmetry about the cavity’s center.

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In lectures 29 and 30 we used the real- and complex-valued representations of single-mode cavity fields to determine their cycle-averaged energy density nd of two- ode cavity fields to determine their cycle- veraged energy () 22 2 00 (,) 44 nn n uxt E ε α β <> = = + 0 ( , ) 2 cos(( ) )cos(( ) ( )) 4 nm m n n m n m E E EE k k x t ωω δδ = + + + and of two mode cavity fields to determine their cycle averaged energy density We realized that because photon detectors cannot resolve optical frequency oscillations and measure u(x,t), it is cycle-averaged energy densities like these that determine where and when we are likely to detect photons.
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Lect33_[Compatibility_Mode] - Physics 344 Foundations of...

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