No, because all the off-rectilinear parts are destructively interfered away.
Huygen's principle is often stated differently than how Huygens stated it. He said that if you draw a sphere at every point of the wavefront, the future wavefront is the envelope of the spheres so drawn, and only one of the envelopes (the one that keeps going forward). So from your sphere, you are supposed to imagine drawing a bunch of other spheres, and then looking at the outermost limit of where they reach. This is a larger sphere, of radius ct, and this is the new wavefront.
The modern version of Huygen's principle (I learned from this site) is the fact that the Green's function of the wave equation in 3+1 dimension is a delta-function on the light cone. This is a way of saying that the full propagation of the wave is by superposing the circles from propagating each point on the previous front. The superposition though is cancelling away from the region where the nearby circles have a mutual tangent, and this is the envelope criterion.
This is wrong. A wide region (width $\gg\lambda$) of many, evenly distributed in-phase spherical wave emitters, like the Huygens equivalent of a laser beam's cross section, behaves like a phased antenna array, with the result that destructive interference between the emitters cancels radiation deviating significantly from the beam and reinforces radiation along the beam. So you do indeed get arbitrarily narrow beams from Huygens's theory.
What you might be thinking is that this theory does not cope well with a unidirectionally propagating beam. If we replace the laser beam's cross section in-phase with spherical emitters, a narrow beam is radiated both forwards and backwards (a phased antenna array does this, for example). This is where one must call on an obliquity factor to restore unidirectionality. Fresnel-Huygens theory multiplies the spherical radiation pattern by $\frac{1}{2}(1+\cos\theta)$, where $\theta$ is the angle between the direction of propagation and the line joining the centre of the Huygens wavefront and the point in question.
If you look at the Huygens-Fresnel diffraction integral in the Wikipedia link above, then consider the beam cross-section $B$ in the $x-y$ plane and then also a point $P$ with co-ordinates $(x, y, z)$ in the farfield (i.e. so that $R = \sqrt{z^2+y^2+z^2} \gg w$, where $w$ is the maximum width of the beam cross section, then a Huygens superposition of spherical emitters spread over $B$ will beget a disturbance at $P$ of:
$$\int_B \frac{\exp\left(i\,k\,\sqrt{\left(x-x^\prime\right)^2+\left(y-y^\prime\right)^2+z^2}\right)}{\sqrt{\left(x-x^\prime\right)^2+\left(y-y^\prime\right)^2+z^2}}\,h\left(x^\prime,y^\prime\right)\,\mathrm{d}x^\prime\mathrm{d}y^\prime\approx $$
$$\frac{e^{i\,k\,R}}{R} \int_B \exp\left(-i \frac{k}{R}\left(x\,x^\prime+y\,y^\prime\right)\right)\,h\left(x^\prime,y^\prime\right)\,\mathrm{d}x^\prime\mathrm{d}y^\prime$$
where $h\left(x^\prime,y^\prime\right)$ represents any beam apodisation and aberration (intensity variation and phase, respectively) and I have simply pulled the denominator out of the integral: this is justified because, as a proportion of itself, $\sqrt{\left(x-x^\prime\right)^2+\left(y-y^\prime\right)^2+z^2}$ does not vary much for $\left(x^\prime,y^\prime\right)\in B$ as $R\rightarrow\infty$, but, as a number of wavelengths it varies a great deal, so we must keep the numerator. Thus we see two results:
- The expression is symmetric in $z$, i.e. the same disturbance propagates backwards as it does forwards, so that the beam is not unidirectionally propagating. This is why Fresnel introduced his obliquity factor $K(\chi)$ to remove the backwards propagating wave;
- The farfield beam cross section is given by the Fourier transform of the beam cross section at $B$, but with a scale factor $\frac{k}{R}$ in the Fourier transform's argument, i.e. once we get a reasonable distance from the cross section at $z=0$, the "shape" of the projected beam does not change - it is always equal to a Fourier transform of the beam at $z=0$, but the same shape is dilated by a factor proportional to $R$.
This diffraction scheme is called Fraunhoffer diffraction. So the diffracted beam can be made tightly focussed in the farfield if the cross section at $z=0$ is wide. By dint of the Fourier transform, there is an inverse proportionality between the width of the beam at $z=0$ and that in the farfield. This is exactly the behavior of a high quality laser beam - you will always get a dilation proportional to $R$ - the only way to avoid this is to have a plane wave of infinite extent.
To get some intuition for the result, witness that the above theory gives, for the diffraction of the Gaussian beam defined by:
$$h\left(x^\prime,y^\prime\right) = \exp\left(-\frac{{x^\prime}^2+{y^\prime}^2}{2\sigma^2}\right)$$
the diffracted result at the point $P=(x,y,z)$:
$$\frac{e^{i\,k\,R}}{R} \int_B \exp\left(-i \frac{k}{R}\left(x\,x^\prime+y\,y^\prime\right)\right)\,h\left(x^\prime,y^\prime\right)\,\mathrm{d}x^\prime\mathrm{d}y^\prime = \frac{\sigma^2}{R} \exp\left(-\frac{k^2 \sigma ^2 \left(x^2+y^2\right)}{2 R^2}\right)$$
Best Answer
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!