Light gets dimmer as distance to it increases, since the amount of energy per unit area decreases.
somaniarushi
One way I felt very comfortable understanding this was through the lens of the inverse square law: https://en.wikipedia.org/wiki/Inverse-square_law. The law says that any physical quantity's magnitude is inversely related to the square of the distance from the physical quantity's source. It is very fascinating to me that intensity of irradiance fits into this wider law — wondering if there's a reason for that!
CarneAsadaFry
@somaniarushi The reason that the inverse square law holds (and applies to irradiance) is actually due to conservation of energy (as @phoebeli23 mentioned). You can imagine drawing concentric spheres around the point source. As light travels away from the source, the total energy at each point of time is conserved, and so the total intensity going through each concentric sphere must be the same. Since the surface area of the spheres increases with r^2 as we travel farther from the source, the intensity must decrease with r^2 so the total intensity stays constant.
Js2604
It was mentioned in lecture by Professor O'Brien that in the case of someone standing close (relative to the size of the light) to a very large tube light, the falloff in irradiance would be proportional to 1/r instead of 1/r^2. Is this 1/r factor because the footprint of the light emitted by this light would end up being more rectangular than perfectly circular as a result of the shape and width of the light?
StephenYangjz
I think this formula here is rather intuitive. The conservation of energy states that the amount of light energy is unchanged, but the area that the lights spans gets larger proportional to r^2. This means that the intensity of the light per unit area shrinks to 1/r^2. However, this is rather only an approximation as the footprint of the light may change.
shreyaskompalli
It's pretty interesting though that the intensity is proportional to r^2 and not r -- intuitively, you'd think that the intensity would decrease linearly with the distance from the source. This is actually what I naively thought, especially when I pictured light as a single ray in my head. However, when thinking of light energy/flux as a wave, it makes a lot more sense that the intensity is proportional to the area that it covers.
ksaralle
For the comment above: thank you for providing an intuitive interpretation of the irradiance falloff! i was confused about why it is proportional to r^2 until I was reminded that the sense of illuminense depends on how the area that is covered. in this sense, the distance from light source itself cannot not fully defines irradiance
cchendyc
Is this Falloff just referring to the inverse proportional character of distance r and the radiance intensity E? since clearly if we are far away from the source, we are gonna receive lesser of the light
Light gets dimmer as distance to it increases, since the amount of energy per unit area decreases.
One way I felt very comfortable understanding this was through the lens of the inverse square law: https://en.wikipedia.org/wiki/Inverse-square_law. The law says that any physical quantity's magnitude is inversely related to the square of the distance from the physical quantity's source. It is very fascinating to me that intensity of irradiance fits into this wider law — wondering if there's a reason for that!
@somaniarushi The reason that the inverse square law holds (and applies to irradiance) is actually due to conservation of energy (as @phoebeli23 mentioned). You can imagine drawing concentric spheres around the point source. As light travels away from the source, the total energy at each point of time is conserved, and so the total intensity going through each concentric sphere must be the same. Since the surface area of the spheres increases with r^2 as we travel farther from the source, the intensity must decrease with r^2 so the total intensity stays constant.
It was mentioned in lecture by Professor O'Brien that in the case of someone standing close (relative to the size of the light) to a very large tube light, the falloff in irradiance would be proportional to 1/r instead of 1/r^2. Is this 1/r factor because the footprint of the light emitted by this light would end up being more rectangular than perfectly circular as a result of the shape and width of the light?
I think this formula here is rather intuitive. The conservation of energy states that the amount of light energy is unchanged, but the area that the lights spans gets larger proportional to r^2. This means that the intensity of the light per unit area shrinks to 1/r^2. However, this is rather only an approximation as the footprint of the light may change.
It's pretty interesting though that the intensity is proportional to r^2 and not r -- intuitively, you'd think that the intensity would decrease linearly with the distance from the source. This is actually what I naively thought, especially when I pictured light as a single ray in my head. However, when thinking of light energy/flux as a wave, it makes a lot more sense that the intensity is proportional to the area that it covers.
For the comment above: thank you for providing an intuitive interpretation of the irradiance falloff! i was confused about why it is proportional to r^2 until I was reminded that the sense of illuminense depends on how the area that is covered. in this sense, the distance from light source itself cannot not fully defines irradiance
Is this Falloff just referring to the inverse proportional character of distance r and the radiance intensity E? since clearly if we are far away from the source, we are gonna receive lesser of the light