I'm a high school student. I still don't understand the reflection, diffraction and refraction of light. It seems to me that in concept of quantum physics, reflection is just a process of object absorbing and reemiting light at a particular angle. What really is diffraction? Is it the behavior of light propagating over an obstacle? How big is the obstacle? I remember in the double slit experiment that if the slit is too big, the light will just go through it. The width of the slit must be smaller than the wavelength? Why wavelength? It seems to me that, in a simple model of wave propagating forward, the amplitude will affect whether it can go through the slit or not.

  • $\begingroup$ Want to correct the fact that if the slit is to big one see fringes on both sides of the slit. Then ever you have an edge you get diffraction. $\endgroup$ Commented Jan 20, 2015 at 6:48
  • $\begingroup$ Do you understand the classical concept of diffraction, with pond waves for example. Without a classical understandment quantum physics will only make things worse. $\endgroup$
    – jinawee
    Commented Jan 20, 2015 at 10:14
  • $\begingroup$ I'm voting to close this question as off-topic because the user shows a too low level in physics. Also, there is a lot of material in Internet and Wikipedia that explains what is diffraction. $\endgroup$
    – Sofia
    Commented Feb 16, 2015 at 1:13

2 Answers 2


You should read this page :http://en.wikipedia.org/wiki/Huygens%E2%80%93Fresnel_principle

The Huygens principle explain intuitively why the wave will spread after being "cutoff" by an obstacle, as the spherical sources at the edge will not interfere anymore with the adjacent ones (those being stopped by the obstacle) to form a plane wave. As shown in the figure in the link, this principle also explain refraction.

The diffraction by a slit can also be explained as an effect of the uncertainty principle (velocity-position): when the photon passes through the slit, its position in along the slit direction (vertical if you look at the figure in Rnr1410 answer) is known more precisely (the uncertainty on its position is reduced). So, in virtue of the uncertainty principle the velocity along the same direction will be increased. Before the slit, the velocity along the slit dimension was zero (assuming perpendicular incidence), after the slit it could be 0 plus or minus some vertical component : the photon will thus appear on the screen at a position higher or lower thatn the slit's "shadow". If this is done with many individual photons (one by one) the overall distribution of the photon on the screen will reproduced the interference pattern, see http://en.wikipedia.org/wiki/Double-slit_experiment#Variations_of_the_experiment


diffraction only occurs when the width of the slit is less then wavelength λ of the source. we can prove this in lay-man terms also.

each photon ( light-particle ) will be assumed as a wave too.
so imagine you cut the slits into infinitely many slits, which were placed side-by-side. then each slit acts as a souce of light. each slit makes amplitude waves on the screen. if the width of all the slits is more than λ , then the sumation of all the waves on the screen is made analogous to what is guessed by classical physics. but if the slit width is small, then the quantum mechanics comes into dominance.
Note: if the width is big, light is approximated as falling only on center, but not exactly.

see superposition of waves - wikipedia enter image description here

  • 4
    $\begingroup$ This is incorrect: "diffraction only occurs when the width of the slit is less then wavelength λ of the source". A wave can and does diffract at the edge of anything, no need for a slit. See Sommmereld's classic solution of the edge diffraction problem puhep1.princeton.edu/~kirkmcd/examples/sommerfeld.pdf . $\endgroup$
    – hyportnex
    Commented Jan 20, 2015 at 12:27

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