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I am a bit confused on what the area A means in the equation for intensity. I know that the area here uses the surface area of the sphere equation and the book I use says that A is the area of the surface, but what exactly is this surface referring to. For example, for the intensity of the Sun on Earth you use the distance between the Earth and the Sun as r for the area. How would this change, if at all say for a point source like a laser. Wouldn't you need to know the dimensions of the laser beam to work out the intensity? Furthermore, say if a beam of light from either the Sun or a flashlight was shinning on a piece of paper, would A be the cross-sectional area of the paper. Could someone please elaborate, thanks!

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For light emitted from the Sun, we use a spherical surface to calculate the intensity because we assume light is emitted equally in all directions. The surface of the sphere represents the set of points that light passes through after a given propagation distance (or equivalently, after a given time). For a laser beam, you need to know the beam dimensions. Most lasers produce Gaussian beams, so the "beam waist" is typically quoted as the beam spot diameter at the focal length. Given the beam waist, it is easy to then find the beam spot area at any distance away from the focal length. If a beam of light is shining on a piece of paper, it is a similar story to the laser beam. You need to know the area of the beam. However, this is not well-defined since the shape of the beam from incoherent sources can differ greatly. Usually, taking the full-width-at-half-maximum point of the intensity distribution of the beam spot is sufficient for estimating the effective area of the beam spot. If you used the area of the paper instead of the area of the beam spot, then you are calculating the average intensity across the entire sheet of paper. This has little physical significance, since almost all the energy is concentrated in the beam spot, so looking at an area much larger than the beam spot is often redundant.

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  • $\begingroup$ I understand the concept for a beam of a laser, but then why wouldn't you need to know the dimensions of the beam from the Sun, or is that what 4pir(squared) is, if so how? $\endgroup$ – Albert Aug 9 at 8:49
  • $\begingroup$ The light from the Sun is emitted equally in all directions, so the "beam spot" (if you want to call it that in this case) is the entire surface of the sphere. $\endgroup$ – Feel My Black Hole Aug 9 at 8:50
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what exactly is this surface referring to.

The light is striking or passing through some surface, and you want to know the intensity of the light at that surface.

You could pick any surface you like and measure (or calculate) the intensity of light on or at that surface.

How would this change, if at all say for a point source like a laser. Wouldn't you need to know the dimensions of the laser beam to work out the intensity?

Yes. Furthermore, the intensity of the beam is not uniform. It's (often) brighter in the center, and dimmer near the edges. So you need to decide which part of the beam you want to know the intensity of.

You could consider, for example, an infinitesimally small surface at the center of the beam, if you want to know the peak intensity of the beam.

You could consider a 1 cm square surface centered in the beam if you want to know the average intensity over that 1 cm square (perhaps because you have a 1 cm square detector you're going to use to measure the beam power).

Or whatever other surface is relevant to the problem you're trying to solve.

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  • $\begingroup$ Should we use a differential surface dA instead of A in the above definition? $\endgroup$ – ado sar Aug 16 at 19:51

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