How is radiance conserved? - Power/intensity misconception

Before getting to the main question I'd like to clear some pretty basic concepts regarding energy/power/intensity as my physics is pretty rusty.

According to wiki

Now suppose we have a surface consisting of 10 sources. Each source is emitting a radiation per second having energy of 200 joules.

1) Is the resultant flux 200 * 10 = 2000 J/s or same as 200?

2) If it's not 200, I am confusing this with something else which I forgot. If anybody can remind me I'd be grateful. In that case increasing the number of instances (like sensors here) didn't affect the original thing under consideration (energy maybe), just it's intensity i.e. the number of photons/particles arriving at the surface.

After that's clear let's come to the main question. According to wiki.

In radiometry, radiance is the radiant flux emitted, reflected, transmitted or received by a given surface, per unit solid angle per unit projected area.

Consider the picture here, According to the radiance in-variance property, the radiance at $dA1$ (the point at which solid angle rays meet) is same as that at $dA2$ in the 3rd row of the picture. To prove that almost every derivation does this as the first step or any intermediary step

$d^2\phi_1 = d^2\phi_2$

If we suppose as the left surface being a light source and the right one as a reciever and the source is radiating energy in a broad cone like in a flash light. Then radiant flux represents the total energy leaving the surface per second. This includes energy radiated not present in the cone drawn in the picture.

As shown in the picture. This means that the total radiant flux arriving at reciever $R$ is less than what was emitted by source $S$. There lies my confusion.

Edit - If you think it's pretty common sense to even state your thoughts DO STATE IT because that's what I'm thinking I am missing some basic concepts here.

• Yes, the flux at 2 is less than the flux at 1. But in the derivation of the radiance theorem, only the radiation that goes through both surfaces is considered. The radiation that misses surface 2 is not of any concern to the radiance theorem. – garyp Feb 24 '18 at 20:32
• @garyp - don't understand, if flux isn't equal, then how are we assuming it's equal. – gallickgunner Feb 24 '18 at 20:34
• Wait, if you mean to say that radiant flux in the theorem doesn't mean for the entire surface but only for the part of the surface that matches with the other surface? then it wouldn't be radiant flux anymore, it'd be radiant intensity i.e. radiant flux per solid angle? – gallickgunner Feb 24 '18 at 20:42
• Close, you have to include the surface area difference. So it's not radiant intensity, rather it's radiance. – garyp Feb 24 '18 at 20:47
• Yes but radiance and radiant flux are 2 different things. My confusion is, the radiant flux is just the energy emitted from a surface, its not bounded by solid angle or surface area. If the flux emitted from source is greater than that at the receiver, how did we assume flux at source equals flux at receiver. – gallickgunner Feb 25 '18 at 7:07