One way to study this case is through the numerical analysis of diffraction, as described in my other answer to you.
You can also do this pretty much as you describe through Huygens's principle or as Feynman describes in his popular QED book. If you set up an equation to describe what you've said, you'll see that the amplitude at a point with transverse co-ordinate $X$ on a screen at an axial distance $d$ from the plane with the knife edge is:
$$\psi(X) \approx\int\limits_0^\infty\exp\left(i\,k\,\sqrt{(X-x)^2+d^2}\right)\,{\rm d}\,x\tag 1$$
where the line of sources runs from $x=0$ to $w$ (the width of the bright region), where we can take $w\to+\infty$ if we like. We have neglected the dependence of the magnitude of the contribution from each source on the distance $\sqrt{(X-x)^2+d^2}$. This is because we now invoke an idea from the method of stationary phase, whereby only contributions from the integrand in the neighbourhood of the point $x=X$ where the integrand's phase is stationary will be important. Thus for $x\approx 0$ we can assume $|X-x|\ll d$ and so:
$$\psi(X) \approx\int\limits_0^w\exp\left(i\,k\,\frac{(X-x)^2}{2\,d}\right)\,{\rm d}\,x\tag 2$$
an integral which can be done in closed form:
$$\begin{array}{lcl}\psi(X) &\approx& \sqrt{\frac{2\,d}{k}}\displaystyle \int\limits_{\sqrt{\frac{k}{2\,d}}(X-w)}^{\sqrt{\frac{k}{2\,d}} X} e^{i\,u^2}\,{\rm d}\,u \\
&=& \sqrt{\frac{d}{2\,k}} e^{i\frac{\pi}{4}} \sqrt{\pi} \left({\rm Erf}\left(e^{3\,i\frac{\pi}{4}}\sqrt{\frac{k}{2\,d}}(x-w)\right)-{\rm Erf}\left(e^{3\,i\frac{\pi}{4}}\sqrt{\frac{k}{2\,d}}\, x\right)\right) \\
&=& \sqrt{\frac{d}{2\,k}} \left(C\left(\sqrt{\frac{k}{2\,d}} X\right) + i\,S\left(\sqrt{\frac{k}{2\,d}}X\right) -\right.\\
& & \qquad\left.\left(C\left(\sqrt{\frac{k}{2\,d}}(X-w)\right) + i\,S\left(\sqrt{\frac{k}{2\,d}}(X-w)\right)\right)\right)\end{array}\tag 3$$
where:
$$\begin{array}{lcl}
C(s) &=& \displaystyle \int\limits_0^s\, \cos(u^2)\,{\rm d}\,u\\
S(s) &=& \displaystyle \int\limits_0^s\, \sin(u^2)\,{\rm d}\,u\\
\end{array}\tag 4$$
where $C(s)$ and $S(s)$ are called the Fresnel integrals.
If I plot the squared magnitude of this function (related to the Fresnel integrals) in normalised units when $k=d=1$ and $L\to\infty$ (noting $C(\infty)=S(\infty) = -1/2$) for $X\in[-10,20]$ I get the following plot:
which I believe is exactly your plot with a shrunken horizontal axis (yours is likely mine with the transformation $x_S = 2\,\pi\,x_R$ where $x_S$ is Satwik's $x$-co-ordinate and $x_R$ Rod's).
Footnote: One of the loveliest curves from eighteenth and nineteenth century mathematics is the Cornu Spiral, which is a special case of the Euler Spiral. $\psi(X)$ in (3) traces a path in the complex plane parametrised by $X$, which turns out to be the arc-length $s$ of the spiral path in $\mathbb{C}$ such that:
$$\begin{array}{lcl}x &=& {\rm Re}(\psi(s)) \propto C(s) + \frac{1}{2}\\
y &=& {\rm Im}(\psi(s)) \propto S(s) + \frac{1}{2}\end{array}\tag 4$$
and I plot the normalised and shifted path $z = C(s) + i\,S(s)$ I get the lovely spiral below. The curly bits spiral all the way in to $\pm(1+i)/2$ as $s\to\infty$. The shifting and then taking magnitude squared explains why the intensity plot above is not symmetric about $X=0$, oscillating as $X\to\infty$ and dwindling monotonically as $X\to-\infty$.
The link you gave set me searching for a more detailed answer, and I learnt an interesting fact:
Huygens' construction works in 1 and 3 dimensions, BUT NOT IN TWO!
The theory behind this is was first derived by Fresnel and later by Kirchoff - the math is explained in detail in this article. It all comes down to the fact that the wave equation for waves propagating from a point can be written as
$$\frac{\partial^2 \phi}{\partial r^2}-\frac{(n-1)(n-3)}{4r^2}\phi = \frac{\partial^2\phi}{\partial t^2}$$
where $n$ is the dimensionality. For $n=1$ or $n=3$ the second term on the left vanishes, and the expression becomes like that of a one-dimensional wave propagating outwards. For $n=2$, like for the ripples on a pond, there is in fact another term that results in waves "traveling backwards". In that case, the usual Huygens construction does not (quite) work.
Quite subtle stuff - and the fact that the mathematical treatment happened more than 100 years after Huygens' initial publication of his ideas (1679: publication of "Traité de la Lumière"; Fresnel was born in 1788) is something that I had not previously appreciated. The casual treatment it is given in French's book makes sense in the context of a big book covering a lot of ground quickly - but I appreciate you brought this question to my attention; it's more interesting than it looked at first sight.
Best Answer
What does an aperture do? It "applies" Huygens principle to every point within the aperture, and ignores those outside the aperture because they are blocked.
There are a couple of things going on when you consider a lens. Let's make sure we understand them.
An aperture produces a diffraction pattern in the space of diffraction angles. Recall from the simple derivation: the diffracted rays from every point in the aperture are parallel to each other. (The diagrams accompanying the discussion often have to be fudged so that parallel lines appear to converge, although some authors are careful to include a lens as described below.) The diffracted intensity is a function of diffraction angle. The diffracted field from the screen comprises sets of parallel rays, each set corresponding to a particular interference condition (max, min, or in between). Again, the diffraction pattern is in "angle space". In order to see the pattern you need to be far enough away from the aperture that the rays from different angles and different points in the aperture no longer cross each other causing a confused pattern. You need your viewing screen to be "at infinity".
Now consider what a lens does. Any set of parallel rays entering a lens will be focused to a single point in the focal plane.
Consider placing a lens after an opaque aperture. Any parallel rays entering the lens will be focused to a single spot in the focal plane. Hence all the rays in each set of parallel rays from the aperture will focus onto the same spot in the focal plane. We've created the usual "Fraunhofer pattern at the focal plane". Allow the aperture and lens to approach each other, and you end up with an aperture containing a lens, producing the usual Fraunhofer pattern on the focal plane.
To finally answer your question: remove the aperture and leave the lens. The rays that hit the lens behave as before, forming the diffraction pattern at the focal plane. The rays that fall outside of the lens are also diffracted, according to Huygens principle. But these do not pass through a lens. So the diffraction pattern from these rays stays in "angle space". They exist, and in fact some of those rays overlap the rays from the lens. But we don't see them because they are spread out and weak, and if you are close to the lens their rays cross in a confused manner. Note, however, that those rays outside of the lens do form a legitimate diffraction pattern, which could be viewed "at infinity". The pattern would be a dark spot surrounded by light, the complement of the pattern for an aperture. Look up Babinet's Principle for more details.