– 38 –
(iii) Geometrical attenuation.- If a rectangular crystal is crookedly placed in a plane parallel beam, the tilt being such that the one edge of the crystal is advanced in phase by an angle -/~ then the attenuation due to the tilt is -ein 3° ———--. With a square crystal whose side is 1 cm and a frequency of 15 Mc/s this would mean that we get the first zero in the response for a tilt of about 26', The setting is probably not really as critical as this, owing to curvature of the wave fronts. If the crystals are operating in a free medium without the tube this effect is easily estimable and we find that, for crystals sufficiently far apart the allowable angles of tilt are of the order of the angle subtended at one crystal by the other. It has been found experimentally with tubes operating at 15 Mc/s that tilts of the order of half a degree are admissible.
Now let us consider the loss due to boundary effects. We assume
a wave inside the tank of form p = Jol flortpe tint and assume a
boundary condition of form - — = { where we do not know * nor even
whether it is real or complex. The radius of the tank is a, so that
the boundary condition becomes ‘- as Ya. Let the solution Foe) 2 wae) apse? of Ra sy be u(y). Then we have 42 + ( | = -5 and ots) e a oo a, fy 1 z therefore Be Jas Ku Jasco. But since MES is small this ' rare &e y and the 208s in a length £ of the Be Ride we sea pry YY there are many means approximately N/, = ined. tank is abe ay solutions of Se but there is a bounded region of the u plane Fo in which there is always a solution whatever value 5a my have. This means to say that for any boundary condition there is always a mode in
which the attenuation does not exceed T ~s- where UT is some numerical constant.
The value of 7 is about 1.9. It is the largest value of xy such that (x + dy) Jy(x+ iy\/o(x + dy) is pure imaginary and y > 0, 0 < x < 2.4.
Taking Lo/awo = 0.31 (as p. 41) the maximum loss in this mode is 6 dB. We should however probably add a certain amount to allow for the fact that not all of the energy will be in this mode. A total loss of 20 dB would probably not be too small.
(iv) Attenuation in the medium.- The attenuation coefficient is given
by -r-oe where D is the dynamic coefficient of viscosity, i.e., the
ratio of viscosity to density. With water ( = .013 c = 1.44 Km/sec.) at a frequency of 10 megacycles and a delay of 1 ms we have a loss of
12 dB. With mercury under the sane circumstances the loss is only 1 dB.
