Suppose a wave with amplitude $A_i $ in incident on an intersection of two media where we get a reflected wave with amplitude $A_r$ and a transmitted wave Amplitude $A_t$.

The velocity of the wave in medium 1 iv $v_1$ and in medium 2 it is $v_2$.

Then $$A_r = \frac{v_2 - v_1}{v_2 + v_1}A_i$$ and $$A_t = \frac{2v_2}{v_2 + v_1}A_i \, .$$

Proof: enter image description here

I understand the math but I am not able to understand the line where it is written:

Now as the wave is continuous so at the boundary(x = 0). Continuity of displacement requires $ y_i $ +$ y_r $ = $ y_t $ at x = 0.

What do they mean by this? Are they trying to say the reflected and the incident ray superimpose to form transmitted wave?Or Are they trying to say that the displacement of particle due to reflected and incident wave (on the left) must be equal to the displacement of particle due to transmitted wave (on the right)?

I am not able to get an intuition. There is also written:

Also at the boundary the slope of wave will be continuous, i.e. slope of reflected wave + slope of incident wave = slope of transmitted wave.

Why is this true? I am not able to understand this.

  • $\begingroup$ The point at the intersection can't have 2 displacements at one time so it must satisfy the waves that roll on the left(reflected+incident) and the ones which roll on the right(transmitted). $\endgroup$ Feb 23, 2018 at 15:03
  • $\begingroup$ This derivation is done for sinusuidal wave. Will it be true for any wave?? For eg., i got a wave like y= 3-(x-5t)^2 then what will be the amplitude it it goes in a medium with c= 5/2 ? If we use direct result, ans obtained it correct but if I try to do by deriving, I am unable to obtain any result. How do we prove for all waves? $\endgroup$
    – Ashpect
    Jul 7, 2020 at 15:13

2 Answers 2


The displacement of each particle on one side of the boundary with be equal to the sum of the displacement due to the incident wave $\vec y_{\rm i}$ and the reflected wave $\vec y_{\rm r}$ and particles on the other side of the boundary will have a displacement due the transmitted wave $\vec y_{\rm t}$.

If at the boundary $\vec y_{\rm i}+\vec y_{\rm r}\ne \vec y_{\rm t}$ there would be a gap between the particle immediately one side of the boundary and the particle on the other side of the boundary which is not possible.

The slope condition is a much more subtle condition and it might help to see how the wave equation is derived for transverse waves on a string.
For a more detailed explanation please have a look at one of the many derivations that you can find on the Internet.

The diagram below shows a section of a string $AB$ of mass per unit length $\mu$ with one end at position $x$ displaced from the equilibrium position by $y_1$ and the other end at position $x +\Delta x$ displaced from the equilibrium position by $y_2$.

$T$ as the tension in the string.

The diagram is greatly exaggerated in that the angles are small and their difference is small.

enter image description here

The net force in the y-direction on the element $AB$ is $T \sin \theta_2 - T \sin \theta_1$.

$T \sin \theta_2$ is the gradient of the string at $x+\delta x$ which I shall write as $\left(\dfrac{dy}{dx}\right)_{x+\Delta x}$ and similarly the gradient of the string at the other end is $\left(\dfrac{dy}{dx}\right)_x$

So the net force on the element is
$$T\left(\dfrac{dy}{dx}\right)_{x+\Delta x} - T\left(\dfrac{dy}{dx}\right)_x = T\Delta x \dfrac {\left(\dfrac{dy}{dx}\right)_{x+\Delta x} - \left(\dfrac{dy}{dx}\right)_x}{\Delta x}$$

Now using Newton's second law the acceleration $a$ in the y-direction can be found.

$$\mu \Delta x \,a = T\Delta x \dfrac {\left(\dfrac{dy}{dx}\right)_{x+\Delta x} - \left(\dfrac{dy}{dx}\right)_x}{\Delta x}$$

$$\Rightarrow a = \dfrac{T}{\mu} \dfrac {\left(\dfrac{dy}{dx}\right)_{x+\Delta x} - \left(\dfrac{dy}{dx}\right)_x}{\Delta x}$$

So the acceleration of the string depends on the rate of change of the gradient of the string.

For successive elements of the string the gradient and hence the acceleration of the string will change.

However imagine that the element was at a boundary and it had a kink in it which is saying that the gradient changes in a discontinuous manner as one crosses the kink.
This would mean that the gradient of the string on one side of the kink would be very different from the gradient of the string on the other side of the kink.
Now the element of the string with the kink in it would be subjected to a very much larger acceleration than elements on either side of it.
This is not possible and so there can be no kink in the string and at a boundary the gradient of the string must be the same on both sides.


You’re asking about intuition, so let me try to answer it that way:

There’s really only on wave, existing everywhere. Because of the surface, we can call the part on one side “incident” and another part “transmitted”, but that’s just us: Nature is dealing with the whole thing all the time.

So that overall wave has to be consistent everywhere. It has to retain its properties everywhere, including right at the surface.

Draw an (imaginary) surface 1cm before the real surface. You imagined it, so it can’t affect the wave, right? So what boundary conditions are needed there? The two waves (left & right of the imaginary boundary) have to match up in frequency, intensity, etc.

Now you can repeat that at the real boundary.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.