That's a question solved by looking at physical units.

The irradiance (watt per square metre) on the whole object determines it's illumination, this unit varies with distance between object and light because the "subtended surface" diminishes by 1/r² with the distance.
(The radiant flux (W) of the light being constant)

For simplicity let's imagine a case where the light is at the camera position and we are looking at a disc. The radiance of the disk is the emitted light by its surface considering our vision angle : in watt per steradian per square metre.

The radiance is necessarily less than its irradiance if the material is diffuse. Why is that ? because the camera only subtends a tiny angle of the total re-emission directions; while the disc re-emits its energy at 2π steradian (hemisphere).

So the radiance seen by the camera is irradiance / 2π. As you can see it does not depend on the distance between the object and the camera. Now, radiance is a unit per square meter, which means it defines "light intensity" per area, so when discretized, it means that it is the pixel value :)

I hope I'm right, this is always confusing.

That's a question solved by looking at physical units.

The irradiance (watt per square metre) on the whole object determines it's illumination, this unit varies with distance between object and light because the "subtended surface" diminishes by 1/r² with the distance.
(The radiant flux (W) of the light being constant)

For simplicity let's imagine a case where the light is at the camera position and we are looking at a white disc. The radiance of the disk is the emitted light by its surface considering our vision angle : in watt per steradian per square metre.

The radiance is necessarily less than its irradiance if the material is diffuse. Why is that ? because the camera only subtends a tiny angle of the total re-emission directions; while the disc re-emits its energy at 2π steradian (hemisphere).

So the radiance seen by the camera is irradiance / 2π. As you can see it does not depend on the distance between the object and the camera. Now, radiance is a unit per square meter, which means it defines "light intensity" per area, so when discretized, it means that it is the pixel value :)

I hope I'm right, this is always confusing.

That's a question solved by looking at physical units.

The irradiance (watt per square metre) on the whole object determines it's illumination, this unit varies with distance between object and light because the "subtended surface" diminishes by 1/r² with the distance.
(The radiant flux (W) of the light being constant)

For simplicity let's imagine a case where the light is at the camera position and we are looking at a disc. The radiance of the disk is the emitted light by its surface considering our vision angle : in watt per steradian per square metre.

The radiance is necessarily less than its irradiance if the material is diffuse. Why is that ? because the camera only subtends a tiny angle of the total re-emission directions; while the disc re-emits its energy at 2π steradian (hemisphere).

So the radiance seen by the camera is irradiance / 2π. As you can see it does not depend on the distance between the object and the camera. Now, radiance is a unit per square meter, which means it defines "light intensity" per area, so when discretized it means that is the pixel value :)

I hope I'm right, this is always confusing.

That's a question solved by looking at physical units.

The irradiance (watt per square metre) on the whole object determines it's illumination, this unit varies with distance between object and light because the "subtended surface" diminishes by 1/r² with the distance.
(The radiant flux (W) of the light being constant)

For simplicity let's imagine a case where the light is at the camera position and we are looking at a disc. The radiance of the disk is the emitted light by its surface considering our vision angle : in watt per steradian per square metre.

The radiance is necessarily less than its irradiance if the material is diffuse. Why is that ? because the camera only subtends a tiny angle of the total re-emission directions; while the disc re-emits its energy at 2π steradian (hemisphere).

So the radiance seen by the camera is irradiance / 2π. As you can see it does not depend on the distance between the object and the camera. Now, radiance is a unit per square meter, which means it defines "light intensity" per area, so when discretized, it means that it is the pixel value :)

I hope I'm right, this is always confusing.