Waves Reflection Transmission

Each medium which a wave travels offers a characteristic impedance to wave motion. When a wave travelling in a particular medium of impedance z₁ meets a normal boundary beyond which the impedance changes to z₂, the incident wave may get partially reflected and partially transmitted. In other words, part of the incident flux is transmitted while the remaining is reflected back into first medium. Let us find out the reflected and transmitted flux.

For example, consider a long string which consists of two pieces joined smoothly at point x = 0. The entire string is stretched having uniform tension T. The part of the string left to x = 0 has linear mass density ρ1, hence impedance ; the right part has density ρ2 and impedance .

Suppose the incident transverse harmonic wave travelling along the string (from left) is given by

y_i ( x, t ) = A_i cos ω (t – (x/v₁ ) ,             x < 0

where subscript i stands for incident wave; is the wave velocity in the left string. [Note that frequency ω = 2 π v of the wave is decided by the frequency of oscillations of the source while wavelength λ = v/v depends upon the medium.]

The incident wave meets the discontinuity in impedance suddenly at x = 0. Let us assume that total incident energy is partially reflected back and partially transmitted in the form of reflected and transmitted harmonic waves given by
         
y_r ( x, t ) = A_r cos ω (t + (x/v₁ ) ,             x < 0

and    y_t (x, t) = At cos ω (t – x/v² ) ,                x > 0

In the above expressions, note the following:
    
(i) y_r is the wave moving towards –ve X-axis; y_t is moving towards +ve X-axis.
    
(ii) Amplitudes have to be in general different; remember that energy flux is proportional to square of amplitudes.
    
(iii) The frequency ω is same for all the three waves. The particles at x = 0 are made to oscillate with frequency ω by incident wave; in turn, these particles become the source of reflected and transmitted waves.

Now, there are two geometrical boundary conditions that must be satisfied (for all values of t) at the point x = 0.
    
1. First, the net displacement of string must be equal immediately to the left and right of position x = 0. In other words, at any instant t,

y_i + y_r = y_t               at        x = 0.

[Note that net displacement on left of x = 0 is y_L = y_i + y_r, and that on right y_R = y_t.]

This implies    A_i + A_r = A_t
    
2. Secondly, the slope of string y/x must be same immediately to the left and right of point x = 0. That is,

/x (y_i + y_r) = yt/x at x = 0, for any t

This implies, – 1/v₁ ( A_i – A_r ) = – 1/v₂ A_t

By solving these eq. we find
         


where we used the relation T/v = z in order to get the last terms.

We define the reflection and transmission coefficients R₁₂ and T₁₂ for the displacement y (or amplitude A) as

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