First of all, I'd like to discuss 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. Therefore, when you make a straight wave pass through a little aperture, it spreads out. But the problem is that this spreading or diffraction depends on the size of the aperture compared to the wavelenght. Although I'm sure you've seen this many times, I'll put a video:


Here you can see that when the aperture is large enough, there is almost no diffraction. But doesn't this contradicts Huygen's principle? I mean, it should spread out anyway. If every point in the aperture is a wave source, the oscilation should reach any point beyound the aperture. It is as though diffraction only happened in small holes, not in large ones nor in corners. Where am I wrong? I've read explanations for this effect with light, but they use Quantum Electrodynamics, and water waves are not quantum, right?

The second doubt I have is about seeing with light. How does light hitting a surface reflects its shape, so that it cannot reflect the shape of sufficiently small objects, such as atoms?

Thank you

  • $\begingroup$ Hi,Miguel Bolín ,welcome to physics stack exchange and please try to ask only one question in one post if possible. $\endgroup$ – Paul Jul 11 '15 at 15:53
  • $\begingroup$ That one can't determine the position of atoms with high resolution (even with atomic resolution) with light is a 19th century approximation that uses the limitations of the human eye to detect small changes in light intensity to derive a VISUAL resolution limit. With today's optics, sensors and computing we can build super-resolution microscopy systems just fine, since we are not limited by our eyes anymore. $\endgroup$ – CuriousOne Jul 11 '15 at 18:52
  • $\begingroup$ @Paul I'm sorry $\endgroup$ – MBolin Jul 11 '15 at 23:28
  • $\begingroup$ @CuriousOne I'm sorry, but I think you're not right. Isn't it a light limitation, instead of a human's? I thought the only way to "see" an atom is using the scanning tunneling microscope, which uses electrons instead of light $\endgroup$ – MBolin Jul 11 '15 at 23:36
  • $\begingroup$ Look up super-resolution microscopy and try to understand the principles behind the different versions in use. Quantum mechanical uncertainty limits the detection of an atom's position to the order of its De Broglie-length, not to the wavelength of the radiation that is used to make the measurement. This is frequently misunderstood. $\endgroup$ – CuriousOne Jul 12 '15 at 19:11

First off I think I should sort out a misconception about Huygens Principle. You can apply this principle efficiently if you have a slit, which is equal or smaller than the wavelength you are considering. If on the other hand the slit is substantially larger than the wave length, you should consider multiple Huygens sources.

Take a look at this animation from wikipedia. As you can read in the description of the animation, the wavelength of the waves are equal to the width of the slit and you see a nice demonstration of Huygens Principle.

However as the slit gets wider and wider the Huygens Principle breaks down and you have to consider multiple Huygens sources as it is illustrated in this diagram from wikipedia:

enter image description here

You can immediately see that when you make the slit larger, the diffraction effect becomes less pronounced.

Your second question is explained in this answer, I suggest you should take a look at it.

  • $\begingroup$ regarding Huygen's principle, I know that you have to consider different Hougen's sources, but the picture you put doesn't fit the animation. In it, the wave only affects a sort of triangular zone, while in your picture it affects every point beyond the barrier. $\endgroup$ – MBolin Jul 11 '15 at 23:26
  • $\begingroup$ And regarding the second question, I know about tunneling, I know you can "see" atoms with electric current, but my question was why not with light. $\endgroup$ – MBolin Jul 11 '15 at 23:27
  • $\begingroup$ Note that the animation and the picture are referring to two different setups, whence it's to be excepted that they look different. Regarding the answer to which I linked in my answer, it clearly explains why you cannot see the visible range of light. Read in particular the second paragraph. $\endgroup$ – Gonenc Jul 11 '15 at 23:34
  • $\begingroup$ This is the second paragraph: $\endgroup$ – MBolin Jul 11 '15 at 23:38
  • $\begingroup$ "In microscopy, there is a rule of thumb that the smallest things you can distinguish with a perfectly engineered microscope have to have a size about half the wavelength of the light you're shining at it. The more exact version of this is known as the Abbé difraction limit. Visible light has wavelength of about 400-700 nanometres. This is of course about 4000-7000 times as much as the diameter of the atom, so there is indeed no way we can see an atom with a (diffraction) microscope using light. $\endgroup$ – MBolin Jul 11 '15 at 23:38

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!

  • $\begingroup$ Then there is no Huygen's principle for waves in general, only for light? $\endgroup$ – MBolin Jul 11 '15 at 23:53
  • 1
    $\begingroup$ @Miguel Bolín : No,Huygens principle works for any wave,its a property of wave . $\endgroup$ – Paul Jul 12 '15 at 2:41
  • $\begingroup$ @Paul , Dr. Ikjyot Singh Kohli said: "Huygen's principle has to do with the propagation of light, which is electromagnetic waves, governed by Maxwell's equations" $\endgroup$ – MBolin Jul 12 '15 at 15:19
  • $\begingroup$ @MiguelBolín Hi. I was just giving that as a specific example. However, Paul is quite correct, as Huygen's principle works for any wave. Although, I cited Maxwell's equations above, you will note the derivation was done for a general wave equation which indeed shows that Huygen's principle applies for any wave. $\endgroup$ – Dr. Ikjyot Singh Kohli Jul 12 '15 at 15:38

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