Power through a circular aperture

Can you explain that a bit more?In summary, the conversation discusses determining the ratio of power contained within a circle of radius w0 to the total power in Gaussian beams of different orders. It also mentions comparing the ratio to the total power for specific Gaussian beams and the use of spatial filtering in selecting the principal mode from a multimode beam. The conversation also touches on the relationship between power and intensity, as well as the use of cylindrical coordinates in formulating the integrals for the problem.
  • #1
thewall12
10
0

Homework Statement



Determine the ratio of the power contained within a circle of radius of w0 (in the transverse plane at z = 0) to the total power in the Gaussian beams of orders (0,0), (1,0), (0,1) and (1,1). Then compare the ratio of the power contained within a circle of radius w0/3 to the total power for the (0,0) and (1,1) Gaussian beams. (This transmission difference is the principle behind spatial filtering that is used to select the principal mode from a multimode beam.) You may find it easier to use cylindrical coordinates for this problem.


Homework Equations





The Attempt at a Solution



I've attached the solutions. The solutions are for #4. I'm not exactly sure how power and electric field are proportional. Can anybody explain that? Also, I don't understand how the integrals for TEM 01/10 and TEM11 were formulated.
 

Attachments

  • Chem 485 PS4 solutions.pdf
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  • #2
Do you know the formula for the intensity distribution of a gaussian beam? (remind yourself of it here: http://en.wikipedia.org/wiki/Gaussian_beam)

Notice (also confirmed on that page) that intensity is proportional to power and you should be able to convince yourself that the relationship quoted in the solutions is indeed correct.

It may be helpful to post your own attempts at formulating the integrals you mentioned, so people can help you understand them.
 
  • #3
Thanks, L-x. I can see the relationship now. I'm still a little lost on the integrals, though. I know that since the beam's optical axis is z, the cylindrical coordinates (r, theta, z) just become a function of r and theta. I understand the bounds on the integrals and the E^2 portion, but not how you have r, r^3, and r^5.
 

1. What is power through a circular aperture?

Power through a circular aperture refers to the amount of light or energy that passes through a circular opening or hole. It is commonly used in optics and engineering to measure the intensity of light or energy that is transmitted through a circular aperture.

2. How is power through a circular aperture calculated?

The power through a circular aperture can be calculated by multiplying the area of the aperture by the intensity of the light or energy passing through it. This can be represented by the equation P = IA, where P is the power, I is the intensity, and A is the area of the aperture.

3. What factors affect the power through a circular aperture?

The power through a circular aperture is affected by several factors, including the size of the aperture, the intensity of the light or energy source, and the distance between the aperture and the source. Additionally, the wavelength of the light or energy and any obstructions or imperfections in the aperture can also impact the power.

4. How is power through a circular aperture used in practical applications?

Power through a circular aperture is used in various practical applications, such as in cameras to control the amount of light that enters the lens, in laser cutting machines to control the intensity of the laser beam, and in telescopes to measure the brightness of stars and other celestial objects.

5. What are some common units of measurement for power through a circular aperture?

The units of measurement for power through a circular aperture vary depending on the type of light or energy being measured. For example, in optics, it is commonly measured in watts (W), while in radio frequency engineering, it is measured in decibels (dB). Other units such as lumens, lux, and candela can also be used in certain applications.

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