Calculate the intensity of a solid object made up of point light sources?

  • Thread starter ericwithakay
  • Start date
  • Tags
    Intensity
In summary, to calculate the intensity at a desired location from a light-bulb, it can be treated as a point source and a gaussian-sphere-type-of-thing can be used. However, for a solid object made up of "point" light sources, integration may be necessary to solve the problem. The differential element of the body needs to have a measure of luminosity, and the luminosity observed at a point decreases according to the inverse square of the distance. This results in a difficult integral to find the total luminosity at a specific location.
  • #1
ericwithakay
2
0
So I understand that to calculate the intensity a distance d away from a light-bulb you would treat the light-bulb as a point source and do a gaussian-sphere-type-of-thing to figure out the intensity at the desired location.

My problem involves a solid object made up of "point" light sources and I'm wondering whether I need integration to be able to solve the problem.

Any advice would be greatly appreciated.

Thanks

-E
 
Physics news on Phys.org
  • #2
Yeah, sounds like you'll need to integrate. I think you could get this with a little more work.
You want your differential element of your body, be it area, volume, what have you, to have some measure of luminosity. So say you have the unit circle that has luminosity m per unit area. Then you subdivide your circle into little differential elements dA. Then for each dA the luminosity you observe at point (p,q) is (m/[(x-p)^2+(y-q)^2])dA lumens from each dA or whatever measure of light intensity you are using (that's the Guassian sphere thing you mentioned). That is, the luminosity falls according to the inverse square of the distance. If you integrate over the area of the circle you get the total luminosity an observer at (p,q) gets. Not an easy integral though.
 

1. What is intensity and why is it important to calculate it?

Intensity is a measure of the strength or amount of something. In science, it is often used to describe the amount of energy or power in a given area or volume. Calculating intensity is important because it allows us to understand and analyze various phenomena, such as the intensity of light or sound, and how it affects our environment.

2. How do you calculate intensity?

To calculate intensity, you need to know the amount of energy or power in a given area or volume. This can be done by dividing the total energy or power by the area or volume. For example, to calculate the intensity of light, you would divide the total amount of light energy by the area it is spread over. The resulting unit is usually watts per square meter (W/m2) for light or watts per cubic meter (W/m3) for sound.

3. What are the units of intensity?

The units of intensity depend on the type of energy or power being measured. For light, it is usually measured in watts per square meter (W/m2), while for sound, it is measured in watts per cubic meter (W/m3). Other units of intensity include joules per square meter (J/m2) for radiation and watts per square centimeter (W/cm2) for laser beams.

4. How is intensity related to amplitude?

Intensity and amplitude are closely related. Amplitude is the measure of the maximum displacement of a wave from its equilibrium position, while intensity is the amount of energy or power carried by the wave. Generally, as the amplitude of a wave increases, so does the intensity. However, this relationship is not always linear and may depend on other factors such as distance.

5. Can intensity be measured directly?

In some cases, intensity can be measured directly using specialized instruments such as light meters or sound level meters. However, in many cases, intensity is calculated by measuring other quantities and using mathematical formulas. For example, the intensity of light can be calculated by measuring the energy of the light source and the distance from the source to the area being illuminated.

Similar threads

  • Classical Physics
2
Replies
61
Views
1K
Replies
3
Views
1K
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Other Physics Topics
Replies
2
Views
3K
Replies
6
Views
871
  • Introductory Physics Homework Help
Replies
1
Views
2K
Replies
1
Views
1K
Replies
1
Views
1K
Back
Top