RELATIONS BETWEEN MEASUREMENTS 33 



In this equation, the electric signal power delivered the medium and is sufficiently large so that diffrac- 

 by the hydrophone in a matched circuit is lion can be neglected. 



Introducing these values in the above equation for 



p _ e s /'',,, we obtain 



47" 



/•',, = R n - 10 log A - 10 log r + 10 log 9 —^— 

 For the acoustic input power it is possible to take the 



product of the free field intensity, p-/pc, and A, the (30) 



effective diaphragm area, and obtain, Comparing this expression with equation (29), 



P, = P A X 10" watts, E p = R ]C - 10 log A - 10 log r + 10 log P ~^, 



where p is the free field pressure. P, is actually the the important result is obtained, that E n = E p . Thus, 

 available acoustic power in the water, which equals a projector has the same efficiency on transmitting 

 the actual acoustic input power if the device matches and on receiving. 



