How to Evaluate Power Ratio in Higher Order Hermite-Gaussian Beams?

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SUMMARY

The discussion focuses on evaluating the power ratio in Hermite-Gaussian beams of order (1,0) by integrating the power contained within a circle of radius W(z) against the total power. The integral is expressed as P = ∫IdA, with the user initially struggling with the integration in Cartesian coordinates due to the presence of additional variables. The user successfully resolves the issue by converting to polar coordinates, using the transformation x = r cos(θ), which simplifies the evaluation of the integral.

PREREQUISITES
  • Understanding of Hermite-Gaussian beam theory
  • Knowledge of integral calculus, specifically double integrals
  • Familiarity with polar coordinate transformations
  • Basic concepts of beam optics and power distribution
NEXT STEPS
  • Study the properties of Hermite-Gaussian beams in detail
  • Learn about the application of polar coordinates in multivariable calculus
  • Explore advanced integration techniques for evaluating power ratios in optical systems
  • Investigate the implications of higher-order modes in beam propagation
USEFUL FOR

Optical engineers, physicists, and students studying beam optics or laser technology who seek to understand the power distribution in higher-order Hermite-Gaussian beams.

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Homework Statement


Evaluate the ratio of the power contained within a circle of radius W(z) in the transverse plane to the total power in the Hermite-Gaussian beams of order (1,0)


Homework Equations


P=\intIdA


The Attempt at a Solution


I have determined the ratio for the Gaussian beam but the higher order modes have a extra 'x' or 'y' in the integral but I am integrating w.r.t r (r2=x2+y2). So how can I evaluate the integral?
 
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Why don't you integrate in Decartes coordinates x and y? Because, if I understood correctly, then the integral is can be represented as
\int^{W(z)}_{0} x*e^{-x^2}dx \int^{\sqrt{W(z)^2-x^2}}_{0}y*e^{-y^2}dy
this is only a quarter of the full energy of course.
 
Last edited:
I don't understand what is Decartes coordinates but I managed to solve it. Its simply x=rcos\vartheta. Have no idea why I didn't think of that when I posted this. Thanks
 

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