Assume that you fixed a speaker to an inclined pipe as well the torch. You can hear sound from the other end of the pipe, but can't see the light from other end of the pipe, why?

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    $\begingroup$ In the above example the sound is most likely reflected, not bend $\endgroup$ – Lord_Gestalter May 15 '14 at 6:31

It has to do with wavelength. A tube is of the correct size for sound waves, not for light waves that have very much smaller wavelengths. An optical fiber does bend light.

  • $\begingroup$ I think even if the tube was of the correct size for light waves, there would have been no reflection as expected. Light waves would get absorbed (assuming the pipe to be of such sort by the OP's tone). I feel the bending or reflecting property depends on the material we use. In case of sound, which is the vibrations of air molecules has no way in getting absorbed, so they come out "bending". $\endgroup$ – Immortal Player May 15 '14 at 10:46
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    $\begingroup$ @Godparticle How about waveguides? en.wikipedia.org/wiki/Waveguide My answer is on those lines. $\endgroup$ – anna v May 15 '14 at 11:00
  • $\begingroup$ I saw the page of waveguide. It reads "As a rule of thumb, the width of a waveguide needs to be of the same order of magnitude as the wavelength of the guided wave." Then the page continues "The propagation inside the wave-guide, hence, can be described approximately as a "zigzag" between the walls. This description is exact for electromagnetic waves in a hollow metal tube with a rectangular or circular cross-section." So, I think even if the waveguide satisfied the first condition of wavelength, it is important to have zig-zag motion. If wave moves linearly, I think it will not bend. $\endgroup$ – Immortal Player May 15 '14 at 11:43
  • $\begingroup$ @Godparticle Sure, there should be reflections from the walls. If they are absorptive it will not work. $\endgroup$ – anna v May 15 '14 at 11:51
  • $\begingroup$ Even if the zigzag reflection may be a good low level exemplification, I think it's misleading in this context. The Maxwell equations result in wave guidance because the electric field all along the conductive surface is leveled due to the conductivity. Don't try to understand waveguides on a level below this (smattering and such) $\endgroup$ – Lord_Gestalter May 15 '14 at 12:34

As @anna v pointed out, it's a question of wavelength. Too add to this, if you move to radio waves they luckily bend, otherwise you'd need line of sight to the sender antenna

  • $\begingroup$ If you consider periscope which looks like the same pipe (with complicated two bending areas), it bends the light. Actually there will be mirrors at the bending areas of periscope which reflects light. So, it depends whether you use mirror or other things. One can also bend light using any material by the principle of total internal reflection. Optical fiber as noticed by anna v is one of the best example. $\endgroup$ – Immortal Player May 15 '14 at 10:34
  • $\begingroup$ @Godparticle Reflection, refraction (known as bending), and diffraction are distinct processes. The example above is reflection, not bending. Optical fiber actually use reflection (standard) or refraction (gradient fibers), depending on the refraction indexing along the core $\endgroup$ – Lord_Gestalter May 15 '14 at 11:25
  • $\begingroup$ In OP's tone I meant bending to be reflection. Sorry, for non-standard usage. I will take care. $\endgroup$ – Immortal Player May 15 '14 at 11:47

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