opticsradiometryvisible-light
In Radiometry, radiance (L) is defined as flux density per unit area per unit solid angle. If we move further along a ray, away from a point, shouldn't the radiance decrease? I am unable to grasp an intuitive reason for the statement:
Radiance remain constant along rays of light through empty space.
Any explanation in this regard would be highly appreciated.
There is a limit to how small you can focus an ideal single-mode laser beam. The product of the divergence half-angle $\Theta$ and the radius $w_0$ of the beam at its waist (narrowest point) is constant for any given beam. (This quantity is called the beam parameter product, and is related to the $M^2$ beam quality measure you may have heard of.) For an ideal Gaussian ("diffraction-limited") beam, it is:
$$\Theta w_0 = \lambda/\pi$$
So, to answer what I interpret as your main question:
Let's say that I have a laser beam of some given power that starts with some diameter $D_0$ at the point of emission and increases to $D_f$ at some distance $r$ away. Would this be sufficient information to imply a limit to the power per unit area (W/m^2) that could be obtained through focusing and what would that be?
The answer is no.
The parameters you have given are sufficient for calculating $\Theta$, but only if $r$ is large enough so that the points at which you measure the diameter are in each other's far field.
You would also need to know the beam radius at the waist, so you could calculate the beam parameter product. Then, to get the minimum spot size, you would need to refocus the beam so that it is maximally convergent. The absolute limit is the fictitious divergence half-angle of $\pi/2$, or 90 degrees, although in practice the theory breaks down for half-angles of more than 30 degrees (this number is from Wikipedia) since the paraxial approximation stops being valid. For an ideal beam at this impossible opening half-angle, this gives you a minimum waist radius of $2\lambda/\pi^2$. So yes, it does depend on the wavelength.
What lens characteristics and approaches would someone look for in order to do this with a laser pointer?
You need a lens with a very short focal length. This gives you the largest convergence. Note that the more convergent the beam, and the smaller the waist size, the smaller the Rayleigh range is. That is, the beam radius will get very small, but it won't stay very small, it'll get bigger very quickly as you move away from the focus. (The Rayleigh range is the distance over which the beam radius increases by $\sqrt{2}$.
In addition, thinking of a Gaussian beam as being "straight" is not quite correct. There is always a waist, always a Rayleigh range less than infinity, and always a nonzero divergence angle.
EDIT
Also, it is important to realize that there is no difference between an unfocused and a focused Gaussian beam. Refocusing a Gaussian beam with a lens just moves and resizes the waist.
The aperture size of the laser is not the same as the waist size. If the beam is more or less collimated, then the aperture will still be larger, because the waist radius is usually defined in terms of the radius at which the intensity drops to $1/e^2$ of its peak value. If the beam is cut off by an aperture at that radius, then even if it were close to diffraction-limited, it certainly wouldn't be anymore. So, apertures are always larger.
The waist is the thinnest point of the beam. Usually this point is inside the laser cavity, or outside the laser if there are focusing optics involved, which there often are. So still, the answer to your question is no. You are not missing the definition of $\lambda$; rather, you are comparing your minimum waist radius to the value of $2\lambda/\pi^2$ that I said was "impossible". I called it impossible, because to make a beam converging that strongly, you would need a lens with a focal length of zero!
Let's try a more realistic example with some numbers. Take your red laser pointer with $\lambda$ = 671 nm. Laser pointer beams are often crappy, but not so crappy as you might think, if they are single-mode. Let's assume that this particular laser pointer has an $M^2$ ("beam quality parameter", which is the beam parameter product divided by the ideal beam parameter product of $\lambda/\pi$) of 1.5. A quick Google search didn't give me typical $M^2$s of red laser pointers, but this doesn't seem to me to be too much off the mark.
Note that if you know the $M^2$ and measure the divergence of a beam, then you can calculate the waist radius. We are going to do that now. Suppose the laser pointer beam is nearly collimated: you measure a divergence of 0.3 milliradians, about 0.017 degrees. Then the waist size is
$$ w_0 = \frac{M^2 \lambda} {\pi\Theta} = \frac{1.5 \times 671 \times 10^{-9}} {\pi \times 3 \times 10^{-4}} \approx 1\,\text{mm}. $$
In this case, they probably designed the laser pointer with an aperture radius of 2 or 3 mm.
Now suppose you focus your collimated beam with a 1 cm focal length positive lens, which is quite a strong lens. The beam's new waist will be at the lens's focal length. That means you can calculate the divergence half-angle: it is the smaller acute angle of a right triangle with legs 1 mm and 10 mm. So,
$$\tan\Theta = 1/10,$$
or $\Theta\approx$ 6 degrees. Applying the formula once more to calculate the waist yields a waist radius of 3.2 microns, which is quite small indeed.
A "safe" laser pointer might have a power of 1 mW. The peak intensity is equal to $2P/\pi w_0^2$, so before the lens the peak intensity is about 600 W/m^2. After the lens it is about 100000 times larger.
So, to summarize:
When your book says energy it should say radiant intensity. I didn't read Lambert's Photometria myself, but multiple sources say that this is how Lambert defined his law. A lambertian surface follows Lambert's cosine law, so for this surface we have: $$I_\theta=I_n \cos\theta$$
Radiance's definition can be written as: $$L=\frac{\partial I}{\partial A\cos\theta}$$
From this definition we can say that for $\theta=0$, then $\cos\theta=1,$, so: $$L_n=\frac{\partial I_n}{\partial A}$$ The radiance for some random $\theta$ for a Lambertian surface would be: $$L_\theta=\frac{\partial I_\theta}{\partial A\cos\theta}=\frac{\partial I_n\cos\theta}{\partial A\cos\theta}=\frac{\partial I_n}{\partial A}$$
This last relation only works for lambertian surfaces since I used the Lambert's law in it. As you can see, if you apply Lambert's law, the definition of radiance doesn't depend on $\theta$ anymore.
To answer your second question, every surface that subtends the same solid angle receives the same light (radiant flux) from the radiant area ($\partial A$). This means that every unit of the fortshortened surface of this receiving surface receives the same radiant flux. Therefore it is not about the distance to the source, but about the solid angle it subtends. If you only think about human eyes as sensors, then it has to do with distance, because the size of the eye is kind of the same for everyone and it doesn't depend on distance, so you have to be at the same distance to subtend the same solid angle, but you shouldn't think of radiance as being a function of distance, but of solid angle.
Best Answer
Imagine you start right next to a star. As you move away from the star, the intensity of the light $I$ (in $W/m^2$) goes down depending on distance $r$ following an inverse-square law:
$$I\ \alpha\ \frac{1}{r^2}$$
but the solid angle $\Omega$ (in $sr$) also decreases in the same proportion:
$$\Omega\ \alpha\ \frac{1}{r^2}$$
Therefore since radiance $L$ (in $W/m^2\cdot sr^{-1}$) is intensity per unit solid angle:
$$L\ \alpha\ \frac{I}{\Omega}$$
the $r^2$ cancels out and radiance does not depend on distance.