Simple Nature - Light and Matter

valid solutions. In the following subsection, we prove that a validsolution can always be found in which a reflection occurs. Now inphysics, we normally assume (but seldom prove formally) that theequations of motion have a unique solution, since otherwise a givenset of initial conditions could lead to different behavior later on,but the Newtonian universe is supposed to be deterministic. Sincethe solution must be unique, and we derive below a valid solutioninvolving a reflected pulse, we will have ended up with what amountsto a proof of reflection.h / A pulse being partially reflectedand partially transmittedat the boundary between twostrings in which the wave speedis different. The top drawingshows the pulse heading to theright, toward the heaver string.For clarity, all but the first andlast drawings are schematic.Once the reflected pulse beginsto emerge from the boundary,it adds together with the trailingparts of the incident pulse. Theirsum, shown as a wider line, iswhat is actually observed.Intensity of reflectionI will now show, in the case of waves on a string, that it is possibleto satisfy the physical requirements given above by constructing areflected wave, and as a bonus this will produce an equation forthe proportions of reflection and transmission and a prediction asto which conditions will lead to inverted and which to uninvertedreflection. We assume only that the principle of superposition holds,which is a good approximations for waves on a string of sufficientlysmall amplitude.Let the unknown amplitudes of the reflected and transmittedwaves be R and T , respectively. An inverted reflection would berepresented by a negative value of R. We can without loss of generalitytake the incident (original) wave to have unit amplitude.Superposition tells us that if, for instance, the incident wave haddouble this amplitude, we could immediately find a correspondingsolution simply by doubling R and T .Just to the left of the boundary, the height of the wave is givenby the height 1 of the incident wave, plus the height R of the partof the reflected wave that has just been created and begun headingback, for a total height of 1+R. On the right side immediately nextto the boundary, the transmitted wave has a height T . To avoid adiscontinuity, we must have1 + R = T .Next we turn to the requirement of equal slopes on both sides ofthe boundary. Let the slope of the incoming wave be s immediatelyto the left of the junction. If the wave was 100% reflected, andwithout inversion, then the slope of the reflected wave would be −s,since the wave has been reversed in direction. In general, the slopeof the reflected wave equals −sR, and the slopes of the superposedwaves on the left side add up to s − sR. On the right, the slopedepends on the amplitude, T , but is also changed by the stretchingor compression of the wave due to the change in speed. If, forexample, the wave speed is twice as great on the right side, thenthe slope is cut in half by this effect. The slope on the right istherefore s(v 1 /v 2 )T , where v 1 is the velocity in the original mediumand v 2 the velocity in the new medium. Equality of slopes gives366 Chapter 6 Waves