52 



RAY ACOUSTICS 



We restrict ourselves to the case where the sound 

 velocity is a function only of the depth coordinate y. 

 The ray pattern in the xy plane can be computed 

 according to the methods of Section 3.3. We can get 

 the entire ray pattern in space by rotating the ray 

 pattern of the xy plane about the vertical {y) axis; 

 because the velocity depends only on y, the ray pat- 

 terns in every plane through the y axis will be identi- 

 cal in size and shape. 



We assume that the projector is a point source 

 located on the y axis at the depth 2/0, which radiates 

 energy at the rate of F energy units per unit solid 

 angle per second. Then, energy will be projected into 



Figure 14. Specification of solid angle. 



a very small solid angle dQ. at the rate of Fd^ energy 

 units per second. The rays bounding this solid angle 

 will curve in some fashion depending on n{y) and 

 the angle of emission; suppose at the point P some- 

 where out along the ray bundle, the cross-sectional 

 area of the bundle is dS. Then the intensity at P will 

 be the energy crossing this area dS per second, which 

 equals FdQ,, divided by the cross-sectional area dS. 



do. 

 Intensity at P = F—- (49) 



dS 



Because of the cylindrical symmetry of the rays 

 with respect to the y axis, we shall find it convenient 

 to define our small solid angle as indicated in Figure 

 14. It is the solid angle swept out in space by rotat- 

 ing the portion of the xy plane between the angles Oq 

 and ^0 + dd about the y axis. On a unit sphere the 



solid angle so defined intercepts a spherical zone of 

 radius cos Ba and width dda which is therefore of 

 area 2ir cos dodda. Thus our solid angle dQ, is given by 

 dfi = 27r cos (9orfflo. (50) 



We wish to calculate the intensity for the ray of 

 initial direction d^, in the xy plane, at the horizontal 

 range R. By equations (49) and (50), this intensity is 

 given by 



F ■ {2t cos d^ddn) 



i{R,eo) = 



dS 



(51) 



where dS, the cross-sectional area, is clearly the area 

 swept out when the segment PP' in Figure 15 is 

 rotated about the y axis. We proceed to calculate dS. 



Figure 15. Diagram used in deriving intensity 

 formulas. 



The horizontal range R is clearly a function of the 

 depth h and Bq: 



R = Rih,Bo). (52) 



Therefore, the horizontal separation dR at a fixed 

 depth is given by 



dR = dBo. 



dBo 



(53) 



The minus sign is inserted because R at fixed depth 

 decreases as Bo increases. 



Let Bh be the direction of the ray at the point 

 (R,h) as in Figure 15. Then PP', the shortest (normal) 

 distance between these two rays near P is dR sin dh = 

 — (dR/ dBo) dBo sin Bk. By rotating PP' about the y 

 axis we get dS, the desired cross section of our bundle. 

 This area is clearly that of a spherical zone with 

 radius R and thickness— (3i2/3&o) sin BhdBo; its area 

 dS is therefore 



dR 



dS = -2irR — sinBkdBo. 

 dBo 



(54) 



