Question

Two wires having different densities are joined at $x=0.$ An incident wave $y=A_{0}sin(\omega t-k_{1}x)$ travelling to the right in the wire $x\le 0.$ If $K_2$ is the wave vector of the transmitted wave, then the ratio of the amplitude of the reflected wave to that of the incident wave is:

• A. $\frac{A_r}{A_0}=\frac{k_1-k_2}{k_1+k_2}$
• B. $\frac{A_r}{A_0}=\frac{k_1+k_2}{k_1-k_2}$
• C. $\frac{A_r}{A_0}=\frac{2k_1{k}_2}{k_1+k_2}$
• D. $\frac{A_r}{A_0}=\frac{2k_1}{k_1+k_2}$

Answer 1

The correct option is A $\frac{A_r}{A_0}=\frac{k_1-k_2}{k_1+k_2}$

The given incident wave equation is $y=A_{o}sin(wt-k_{1}x)$

where $A_o$ is the amplitude of the incident wave.

Let $A_r$ and $A_t$ are the amplitudes of the reflected and transmitted waves respectively.

Thus the reflected wave is $y_r=A_{r}sin(wt+k_{1}x)$ for $x\le 0$

And the transmitted wave is $y_t=A_{t}sin(wt-k_{2}x)$ for $x\ge 0$

Now applying the boundary conditions:

At $x=0$, $y+y_r=y_t$ ............(1) for all values of $t$

$⟹A_o+A_r=A_t$ ..............(a) (for $t=0$)

Also differentiating (1) w.r.t $x$, gives $\frac{dy}{dx}+\frac{d{y}_r}{dx}=\frac{d{y}_t}{dx}$ for all values of $t$

$⟹-k_1{A}_o+k_1{A}_r=-k_2{A}_t$ (at $x=0$ and $t=0$)

Thus $A_o-A_r=\frac{k_2}{k_1}{A}_t$ .............(b)

From (a) and (b) eliminating $A_t$ $⟹\frac{A_r}{A_o}=\frac{k_1-k_2}{k_1+k_2}$