From energy conservation principle prove the relations,

$A_r=\left(\frac{v_2-v_1}{v_1+v_2}\right)A_i \text{ and } A_t=\left(\frac{2v_2}{v_1+v_2}\right)A_i$

Here, symbols have their usual meanings.

 By conservation of energy, the average incident power equals the average reflected power plus the average transmitted power

or

$P_i=P_r+P_t$

$\therefore$ $\frac{1}{2}{\mu }_1{\omega }^2{A}_i^2{v}_1=\frac{1}{2}{\mu }_1{\omega }^2{A}_r^2{v}_1+\frac{1}{2}{\mu }_2{\omega }^2{A}_i^2{v}_2$ 

or $\left(\frac{T}{v_1^2}\right)$  ${}^2{A}_i^2{v}_1=\left(\frac{T}{v_1^2}\right)A_r^2{v}_1+\left(\frac{T}{v_2^2}\right)A_t^2{v}_2$ …….(i)

Further

$A_t-A_r=A_i$…….(ii)

Solving these two equations for $A$, and $A_t$, we get

$A_r=\left(\frac{v_2-v_1}{v_1+v_2}\right)A_{,}$

and

$A_1=\left(\frac{2v_2}{v_1+v_2}\right)A_i$

Hence proved.