Question

An ultrasound plane wave of amplitude \(P_0\) hits the semi-infinite boundary of two media having acoustic impedances \(Z_1\) and \(Z_2\). The sum of the amplitudes of the reflected and the incident waves at the boundary is equal to __________________.

• A. \(\displaystyle \frac{2P_{0}Z_{2}}{(Z_{1}+Z_{2})}\)
• B. \(\displaystyle \frac{P_{0}(Z_{2}-Z_{1})}{(Z_{1}+Z_{2})}\)
• C. \(\displaystyle \frac{P_{0}Z_{2}}{Z_{1}}\)
• D. \(\displaystyle \frac{P_{0}Z_{1}}{(Z_{1}+Z_{2})}\)

Hint:

- For normal incidence, \(R_p=\dfrac{Z_2-Z_1}{Z_2+Z_1}\) and \(T_p=\dfrac{2Z_2}{Z_2+Z_1}\) for pressure amplitudes. - Use continuity of pressure and velocity (\(u=p/Z\)) at the interface.

Answer 1

The Correct Option is A

Step 1: Let the incident, reflected and transmitted pressure amplitudes be \(P_i (=P_0)\), \(P_r\) and \(P_t\) respectively. At the boundary, continuity of pressure and particle velocity gives \[ P_i+P_r=P_t,\qquad \frac{P_i-P_r}{Z_1}=\frac{P_t}{Z_2}. \] Step 2: Eliminating \(P_t\): \(Z_2(P_i-P_r)=Z_1(P_i+P_r)\Rightarrow (Z_2-Z_1)P_i=(Z_2+Z_1)P_r\). Hence \[ \frac{P_r}{P_i}=\frac{Z_2-Z_1}{Z_2+Z_1}. \] Step 3: Therefore, the sum of the incident and reflected amplitudes at the boundary is \[ P_i+P_r=P_i\!\left(1+\frac{Z_2-Z_1}{Z_2+Z_1}\right) =P_i\,\frac{2Z_2}{Z_2+Z_1} =\frac{2P_0Z_2}{Z_1+Z_2}. \]