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Phenomenon of Light Refraction

  1. Nov 28, 2012 #1
    I have read about effect of refraction due to slowing of light rays reaching the denser medium first and so light bends etc...but why can't light go as shown in the figure (attachment)?
    In the diagram (Sry, not a great artist ;-) ) the wavefront does change but not the path of light.....or is it because of apparent angle of light that reaches our eyes that cause refraction phenomenon? (Like if u trace back the changed wavefront...it will give different perception of the actual object ie the path doesnt change but the wavefront changes)

    Attached Files:

  2. jcsd
  3. Nov 28, 2012 #2
    Crude effective theories cannot explain this, light is a wave caused by acceleration of electric charges.
  4. Nov 28, 2012 #3
    The diagram is supposed to show the rays of light and the wavefronts.

    In the top part of the diagram the wavefronts are perpendicular to the rays as they must be.

    But in the bottom part of the diagram the same principle must be adhered to.
  5. Nov 28, 2012 #4
    The artist either made a mistake or his drawing is incomplete. I think that he should have added a few labels.

    As far as I can see, he did not draw the rays of light. The rays of light in a dielectric (i.e., electrical insulator) are always perpendicular to the wave fronts.

    I am not sure what he meant by those "bracket" lines at the ends of the wavefront which pass through the interface between the medium. I call them the bracket lines since they seem to bracket the wave fronts. I don't have anyway to draw the figures so let me use words.

    If the bracket lines were intended to represent the rays of light, then they were drawn incorrectly on below the interface. The rays indicate the direction of the light beam. The rays in an electrical insulator (e.g., vacuum, air, water, glass) are always perpendicular to the wave fronts. The bracket lines below the interface are not perpendicular to the wave fronts. Therefore, the bracket lines can't indicate the direction of the ray below the interface.

    It is hard to read a diagram without labels, especially if there is an unknown mistake in it. Apparently, you interpreted the bracket lines as rays. The bracket lines in this diagram do not show the direction of the light beam. I think that they should have been left out of the diagram altogether.

    I suggest that the rays of light should have been drawn in the middle of the wave fronts. The rays should be drawn perpendicular to the wave fronts.

    The important thing in a diagram is what the lines and curves in it mean. Therefore, I suggest that when one asks a question using a diagram, one should always label the parts of the diagram with words. It is much easier to discuss the physics of the diagram, because then one can use the words in the labels. Sometimes, a word can be worth a thousand pictures.
  6. Nov 28, 2012 #5
    Sorry about the diagram...
    Let me put my question in a different way using an analogy most commonly used.
    Consider a car going from a good road to sandy terrain where the speed decreases (say due to slipping) at an oblique angle. (like light from rarer to denser material). Now only one wheel of car enter the sandy region first and slows down while the other wheel, which is still on road is travelling at higher speed. So the car bends inwards due to speed difference until both wheels enter the sandy region thus having same speed.
    This analogy makes sense only when we consider the wheels to be joined together by say an axle which forces the car to bend.(If no connection was present then the wheels will continue their motion along respective straight lines with some lag in relative position changing distance between them)
    {use the previous figure considering the two rays as the lines along the wheels of car and the situation shown in figure is without the "connection" between the wheels}
    But what about light?? Do the photons have some connection between them?

  7. Nov 28, 2012 #6
    The diagram shown is considering light from the point of view of waves. The rays show the direction of these waves. So the two parallel rays incident ARE connected together. The connection is that rays and wavefronts are always perpendicular to each other.

    If the point of view of photons is to be considered then one has to talk about the probability amplitude that the photon will go this way and that way.
  8. Nov 28, 2012 #7


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    That analogy is not the best. If you want a better 'mechanical' analogy then consider ranks of soldiers marching across a parade ground. They are all marching in time with the music and taking equal length strides. Each rank represents one wave front. They come to the edge of the parade ground (at an oblique angle, with the field slightly to their right) and enter a ploughed field. They still keep in time with the music but they have to take shorter steps if they want to keep in time. (Frequency and speed come into this model, which makes it a good one imo)

    The rule is that when each soldier crosses over the boundary, he needs to keep the neighbour, who steps onto the ploughed region, at his shoulder. This will mean that, to avoid overtaking the guy, he will turn a bit to the right as he slows up. As a rank crosses the boundary, in order to obey the above rule, it will be taking shorter steps (slower wave speed) and must have changed direction slightly to the right to accommodate this. Observing this from above, the shape of the marching body will have a kink in it just like the usual diagram for refraction (not yours, though, I'm afraid because yours doesn't follow the right rules).
  9. Nov 28, 2012 #8
    No. Photons do not have some connection between them. The theory linking photons to light waves is a bit more complicated than that. However, the wave picture alone is sufficient to understand the basics of refraction.

    A photon is a particle of light. Therefore, it isn’t always useful in describing the behavior of light waves. Concepts clearly defined in terms of waves are generally ambiguous with respect to particles. And vica versa.

    The concepts of wave front and photon are mutually exclusive. There is an uncertainty relation between photon number and uncertainty in phase. It is a bit weaker that the other uncertainty relations, so there are strange exceptions. However, it is a good approximation under most conditions. Using this relation, the following can be shown. Under conditions where the light acts precisely like a wave, the concept of photon breaks down.

    That is quantum mechanics for you. Quantum mechanics assumes wave-particle duality. However, going from a wave picture to particle picture involves probability and statistics.

    If the wave front is defined precisely, then the phase of the wave is known precisely. If the phase is known precisely, than the photon number is completely uncertain. If the photon number is uncertain, then the concept of photon is not applicable.

    At , from Figure 5, simple geometry gives . From this we can see that there is a tradeoff between number uncertainty and phase uncertainty , which sometimes can be interpreted as the number-phase uncertainty relation. This is not a formal uncertainty relation: there is no uniquely defined phase operator in quantum mechanics [11] [12] [13] [14] [15] [16] [17] [18]

    Another version of the uncertainty relation between photon number and phase is presented in equation 19 of the following link.
    “We study a simple setup for non{estructive detection of photon number. We in particular
    compare the device's inaccuracy and its phase disturbance to the uncertainty
    principle. As the traditional uncertainty principle involves neither inaccuracy nor disturbance,
    a new type of uncertainty relation is employed”

    For historical completeness, the theory of Newton should be mentioned. Newton tried to explain many properties of light using the assumption that light was made of particles. These are now called Newtonian corpuscles. He explained refraction in terms of the pressure of the medium on the particles.

    Newton totally ignored the wave nature of light. His theory didn’t precisely explain all the properties of light known at the time. Therefore, his theory was replaced by the wave theory of light. Einstein developed the theory of photons as a particle. However, Einstein applied his relativity theory to the photons.

    Newtonian corpuscles are very different from photons. Photons travel at the speed of light. Therefore, relativity is needed to explain their properties. Photons do not satisfy Newton’s original, unmodified equations.

    In any case, Newtonian corpuscles are also inconsistent with the wave nature of light.
  10. Nov 29, 2012 #9
    Another analogy to refraction is shown below. I think I read this in a book by R.P. Feynman.

    Attached Files:

  11. Nov 29, 2012 #10


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    Refraction is easier understand using to the wave model of light. Bringing photons (particle model) into it just confuses the issue. Check out:


  12. Nov 29, 2012 #11
    tangential component of E and normal component of D should be continuous across the boundary,which imposes some restriction.Since B is very small,one can consider E in this case for getting ideas.
  13. Dec 4, 2012 #12
    Thank you guys for your guidance and help on the issue but I am not finding it crystal clear yet. So I think I would return to this issue when I have more knowledge on Light, EM waves etc, in the future.
    Anyways Thank you all. :approve:
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