

Subject:
Sound deflection in Air
Category: Science > Physics Asked by: rbairosga List Price: $10.00 
Posted:
06 Feb 2004 01:07 PST
Expires: 07 Mar 2004 01:07 PST Question ID: 304043 
Is it possible to momentarily deflect a beam of light with a single cycle of a sound wave? The cycle will compress/decompress the air as it travels, changing the index of refraction. I am not interested in information regarding AO devices, which create standing RF waves in a crystal. The idea I am curious about is more akin to a mirage on a hiway on a hot summer day. Except, instead of being created by a layer of heat, it would be created by a controllable repeating sonic pulses (sound) Specifically I would like to know if such a setup is feasible and exactly what this formula/relationship is between sound intensity / frequency and maximum angle of deflection in regular atmospheric conditions. Thank you. 

There is no answer at this time. 

Subject:
Re: Sound deflection in Air
From: racecarga on 06 Feb 2004 14:48 PST 
Atmospheric pressure is about 10^5 Pa (Pascals). Sounds, even loud sounds, cause pressure fluctuations only on the order of mPa (millipascals). For example, a 100 dB sound (referenced, as is custumary, to 20 micropascals), corresponds to a pressure fluctuation of about 3 mPa. Thus for a 100 dB sound, the pressure fluctuates by only a few parts in 10^8, and I would be surprised if any optical effects could be observed. On the other hand, larger pressure changes would have an optical effectfor example, the shock wave formed by a supersonic object can be seen in photos. 
Subject:
Re: Sound deflection in Air
From: rbairosga on 06 Feb 2004 15:46 PST 
Thats what Im learning from other sources, the effects are miniscule. If though, an extremely high frequency sound source could be used, and the light beam could travel in parallel with the source, it then would experience several cycles of refractions to perhaps cause a noticeable displacement. Then again.. perhaps it would need to be a very long path indeed. A 100khz sound produces 3mm wavelengths so how many would be needed to translate a beam 1mm I wonder. 
Subject:
Re: Sound deflection in Air
From: hfshawga on 06 Feb 2004 18:08 PST 
If you propagate the light in exactly the same direction as plane sound wave, there will be no deviation of the light at all, because the light ray encounters regions of changing refractive index at a zero angle of incidence. Snell's Law tells us that sin(theta_refracted) = [n(incident)/n(refracted)]*sin(theta_incident). If theta_incident is zero, sin(theta_incident) is also zero, and so is theta_refracted. If you try to get around this by propagating the light at an angle to the direction of propagation of a plane acoustic wave, there will be (very small, see below) deviations in the direction of propagation due to the small changes in refractive the refractive index of air as a function of pressure, but over any distance corresponding to one full wavelength of the acoustic wave, there will be no net displacement of the light (assuming that the acoustic wave has a symmetrical waveform, e.g., a sine wave). Remember that a sound wave contains regions of both compression and rarefaction. In the compressed regions, the refractive index is higher, and in the rarefacted regions, the refractive index is lower. Over one full cycle, the light beam will be refracted a distance x in the region of high pressure, but a distance x in the region of low pressure, so although the direction of propagation "wiggles" back and forth, there is no accumulated displacement. To see how large an effect one might be able to see, let's do a back of the envelope calculation. A 0dB sound wave corresponds to fluctuations in pressure of +/ 20x10^6 Pa, or 2 x 10^10 atm. A 160 dB sound (an intensity that will literally rupture your eardrums) corresponds to fluctuations of +/2x10^10 * 10^(160/20) = 2 x 10^2 atm, which is a +/2% change in ambient pressure. If one were able to make a 160dB square wave, the light beam would alternately encounter regions of pressure at 0.98 atm and 1.02atm To second order, the refractive index of 25 C air varies as: n(P) = 1+(26334.56*P + 9.697*P^2)/10^8, where P is in atmospheres (Owens, 1967, Optical refractive Index of Air: Dependence on Pressure, Temperature, and Composition. Applied Optics v6, #1, pp5159) In this, I have neglected the temperature changes accompanying the adiabatic compression/rarefaction of the air because for the effect on the refractive index is negligible. at P = 0.98 atm, n = 1.0002583 at P = 1.00 atm, n = 1.0002634 at p = 1.04 atm, n = 1.0002686 Using Snell's Law and a little trig, one can derive the formula for the horizontal deviation of a light ray that leaves a region of refractive index n_i and enters a region of refractive index n_r and thickness L at an angle t_i to the normal of the interface: deviation = L*{tan(t_i) tan[arcsin(n_i/n_r*sin(t_i))]} where t_i is in radians For a light ray propagating at an angle of 45 degrees to the direction of propagation of sound, and a 160dB square wave at a frequency of 1000Hz (= wavelength of 0.33 m), the light beam will be deflected back and forth by about 3.4 microns as it traverses the alternating 0.165 m wide regions of high and low pressure. Higher angles of incidence will obviously result in larger deflections on each half cycle. (someone should check my arithmetic!) 
Subject:
Re: Sound deflection in Air
From: rbairosga on 06 Feb 2004 19:12 PST 
"If you propagate the light in exactly the same direction as plane sound wave, there will be no deviation of the light at all" Thats a good point to remember. I was envisioning both waves travelling in the same direction, but the acoustic wave front to be spherical in shape. In reality though, it will be most planar for such short distances. "Over one full cycle, the light beam will be refracted a distance x in the region of high pressure, but a distance x in the region of low pressure, so although the direction of propagation "wiggles" back and forth, there is no accumulated displacement." This is something Im not quite sure I understand. Though the light wave would experience a net angular displacement of zero, I thought it would emerge each compression slightly displaced though still heading in the same direction. The pictures below illustrate my (potentially erroneous) thinking: http://www.physics.fsu.edu/courses/spring99/ast1002h/telescopes/refraction.gif http://www.carsoncity.k12.mi.us/~chuck_peirce/Parfrt03.gif Do you mean to say, because of the sinusoidal shape of the wavefront, the above images do not apply? If that is so, Ill give it some more thought. Does your argument still hold true for your square wave example further on in your post? I much appreciate your initial calculations, as thats giving me a great system to plug numbers into. If you could confirm your nethorizontaldisplacement assertion to me, Id be happy to consider this question answered Mr. hfshaw. 
Subject:
Re: Sound deflection in Air
From: hfshawga on 09 Feb 2004 17:25 PST 
>>Over one full cycle, the light beam will be >>refracted a distance x in the region of high pressure, >>but a distance x in the region of low pressure, so >>although the direction of propagation "wiggles" back and >>forth, there is no accumulated displacement." >This is something Im not quite sure I understand. >Though the light wave would experience a net >angular displacement of zero, I thought it would >emerge each compression slightly displaced though >still heading in the same direction. What I said is only true for the very small changes in refractive index and correspondingly small angles of refraction. Remember that sound waves are composed of alternating regions of higher AND lower pressure (or alternatively, regions of higher and lower density). By mass balance, the mass or pressure excess in the regions of compressed air are balanced by the mass or pressure deficit in the regions of low pressure. In the regions of high pressure, the refractive index is higher (but only by a smidge) than it would be in the absence of the sound wave, and in the regions of low pressure, the refractive index is lower (but only by a smidge) than it would be. As a light beam traversed the sound field, it will alternately encounter regions with refractive index higher and lower than that of soundfree (constant pressure) air. Thus, the physical picture is somewhat different than the images you posted; a better representation of the beam traversing a full cycle of the sound wave would show a ?slab? of lowrefractive index material beneath the slab of highindex material, and the light ray would be refracted back toward the original direction of propagation as it crossed the interface between the high and lowindex regions. The refraction resulting from crossing into the region of high index is (nearly) cancelled out by the refraction that occurs when the beam crossed into the region of low index. The dependence of refractive index on pressure, however, is not exactly linear. In fact, the equation I posted in my earlier message for the refractive index of dry air as a function of pressure (n(P) = 1+(26334.56*P + 9.697*P^2)/10^8, where P is in atmospheres) includes a secondorder term, so the absolute value of the change in refractive index in the high pressure regions (relative to ambient pressure) is not exactly the same as the absolute value of the change in the low pressure regions. This effect is quite small, though ? on the order of 9.697/26334 = 0.036%, so for the small changes we?re dealing with here, we can assume that the change in refractive index is proportional to the pressure to a very good approximation. Similarly, the angle of refraction is not linearly related to the change in refractive index. Again, however, for the small angles we?re talking about here, (and assuming that the initial angle of incidence of the light ray with respect to the direction of propagation of the sound wave is small) a linear approximation is quite good. For a light beam crossing M full cycles of a plane sound wave with wavelength 2*L, and initially propagating at an angle T_1 to the direction of propagation of the sound wave, the net displacement is equal to (draw a picture of the situation to see where this comes from): Delta = M*[L*tan(T_2) + L*tan(T_3) 2*L*tan(T_1)], Where the displacement is measured relative to a light ray transiting a region with unperturbed pressure. T_2 is the angle the refracted ray makes with respect to the direction of sound propagation in the highindex regions, and T_3 is the angle the refracted ray makes with respect to the direction of sound propagation in the lowindex regions. T_2 and T_3 are related to T_1 by Snell?s Law: n_1*sin(T_1) = n_2*sin(T_2), n_2*sin(T_2) = n_3*sin(T_3), (which can also be written as n_1*sin(T_1) = n_3*sin(T_3)) where n_1, n_2, n_3 are the refractive indices of the unperturbed, highpressure, and lowpressure regions, respectively. If we assume, per the paragraph above, that the refractive index is directly proportional to the pressure, at least for small changes in pressure, then we can say that n_2 = n_1 + dn, and n_3 = n_1 ? dn. Using these approximations and solving the Snell?s Law conditions for T_2 and T_3 we get: T_2 = arcsin[(n/(n+dn)*sin(T_1)] T_3 = arcsin[(n/(ndn)*sin(T_1)] You can set up a spreadsheet to play around with different values of n, dn, T_1, L, and M to see how large a displacement you might expect. For the sorts of changes in refractive index (dn) we?re talking about (parts in 10^6), wavelengths on the order of a meter, and incident angles (T_1) up to 45 degrees, I get that the net deviation per full cycle (M=1) of the sound wave is on the order of Angstroms. Note that if the direction of propagation of the light is nearly *perpendicular* to the direction of sound propagation, then it's possible to effect more substantial deviations. In the extreme case, where (sin(T_1) ~ (n_1n)/n_1), the sound field (if intense enough) could act as a sort of waveguide. 
Subject:
Re: Sound deflection in Air
From: rbairosga on 17 Feb 2004 00:48 PST 
Thank you, if you would like to change your comment to an answer, Id be happy to mark it as answered. 
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