In the situation you are describing, the only thing one can take away is that the room you are in is brighter than it would be if the bright room were not there. It does not mean that your room has to be as bright as the bright room, just that your room is brighter than it would be otherwise.
For your new situation with the street lamp:
If you move very far away, but still within sight of the lamp, the light from the lamp or nearby bright objects will still reach your dark room. Your dark room will not be as dark as it was before, even if you judge it to be just as dark using only your eye. Since light is reaching the dark room, that light will cause the room to be a bit brighter.
Using the idea of intensity of light:
An intensity of zero at some point in space means there is no electromagnetic radiation; there is no light. If you stand in a place where the intensity of visible light is actually equal to zero, there is no light there. You won't be able to see anything with your eyes or any other visible light detector.
Make sure you're not confusing zero intensity with a very small intensity. These are very different ideas.
It might help if you take a particle-view of light. Think about the photons leaving the lamp and entering your eye. If they enter your eye, and you move away, they're going to hit the walls of the room instead.
Unfortunately, I think you are speaking about what people commonly say is "Huygen's Principle", "In order to explain waves diffraction, it says that every point in a wave front behaves as a source, so the next wave front is the sum of all secondary waves produced by these points.", but this is not actually what Huygen's principle says.
Huygen's principle has to do with the propagation of light, which is electromagnetic waves, governed by Maxwell's equations. It can be shown that upon decoupling Maxwell's equations, one obtains spacetime wave equations of the form:
$u_{,t,t} = c^2 \left(u_{x,x} + u_{y,y} + u_{z,z}\right)$, (commas indicate partial derivatives) subject to the boundary conditions:
$u(\mathbf{x},0) = u(\mathbf{x}), \quad u_{,t}(\mathbf{x},0) = \psi(\mathbf{x})$.
The solution is given by D'Alembert's formula, but in the context of space-time wave equations, is known as Kirchhoff's formula or the Poisson formula, but it is the generalization of the Huygen-Fresnel equation, and is given by:
$$u(\mathbf{x},t_{0}) = \frac{1}{4\pi c^2 t_{0}} \iint_{S} \psi(\mathbf{x})dS + \left[\frac{1}{4 \pi c^2 t_{0}} \iint_{S} \phi(\mathbf{x}) dS\right]_{,t_{0}}.$$
You see from the solution that the point of Huygen's principle is to ensure causality of wave propagation. That is, as can be seen from the solution that $u(\mathbf{x}_{0},t_{0})$ depends on the boundary conditions on the spherical surface $S = \{ |\mathbf{x}-\mathbf{x}_{0}| = c t_{0} \}$, but not on the values inside the sphere! That is, the boundary conditions influence the solution only on the spherical surface $S$ of the light cone that is produced from this point.
This is precisely Huygen's principle: Any solution of the spacetime wave equation travels at exactly the speed of light $c$. So, as you can see Huygen's principle is independent of any specific slit/aperture configuration, it will apply in any situation where you can set up such boundary conditions for the spacetime wave equation!
Best Answer
The bending of waves around corners is known as “diffraction,” and its natural length scale is the wavelength of the diffracted wave. So if you want to block the sound from a speaker playing a middle C, with a wavelength in air of about a meter, then you need an obstacle which is many meters across. (A building is a good size.) But to block visible light, with sub-micron wavelength, a millimeter-scale object is a sufficiently enormous obstacle.