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The well known simple explanation of the Herschel-Quincke interference tube is inadequate. It is herein shown that the ratio of the transmitted to incident acoustic energy is $[4({sin({{α}}₃+{{α}}₂)}{2})({cos({{α}}₃{-}{{α}}₂)}{2})]{{{[1{-}2cos({{α}}₃+{{α}}₂)+cos({{α}}₃{-}{{α}}₂)]}²+4{sin}²({{α}}₃+{{α}}₂)}}^{{-}1/2}$ wherein ${{α}}₂$ and ${{α}}₃$ are the phase changes over the two branches of the tube. This ratio is zero not only when ${{α}}₃{-}{{α}}₂=(2n+1){π}$, as formerly explained, but also when ${{α}}₃+{{α}}₂=2n{π}$, provided ${{α}}₃{-}{{α}}₂{≠}2{n}₁{π}$, where $n$ and ${n}₁$ are independent integers.Comparison of the new theory with experiment.---The transmission in an interference tube having branches 10.0 and 14.3 cm in length was studied and showed satisfactory agreement with the new theory. Practically zero transmission was found at frequencies of 1100 and 3300 d.v. and this is in accord with the simple theory of difference in path, as well as the revision here presented. But there was also serious interference at 1000, 2000, 3000 and 4000 d.v. and this fact is in accord only with the revised complete theory.
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G. W. Stewart (1928) studied this question.
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