Superposition and Coherence

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Two light waves, each of intensity I nought, reach the same spot on a screen. So together they give two I nought, right? Well, not always, and that is the surprise. At some spots you get four I nought, and at others, complete darkness. One plus one gives four, and also zero! And two ordinary lamps never show this at all. Today we find out why. Questions on this come up often in JEE, so let's begin.

0:34Two waves at one point: the cross term

So, let us start. Two waves, of intensities I one and I two, meet at a point; the small one and two written below the I are subscripts. Their phase difference, phi, the Greek letter on screen, stays constant. Then the resultant intensity, I, equals I one plus I two, plus two times the square root of I one times I two, all times cos phi; the tick-shaped sign is the square root. Okay? Look at the last term, the cross term. It is the only part that depends on phi. So tell me, which value of cos phi makes I the largest? It is cos phi equal to one, yes, exactly! Then the three terms make a perfect square: I max, with the small word max below, equals root I one plus root I two, whole squared. Right? And I is the smallest when cos phi equals minus one. So I min, with the small word min below, equals root I one minus root I two, whole squared. Clear? Now, you might worry that energy vanishes at the dark spots. It does not. Over the whole pattern, cos phi averages to zero, so the average intensity is just I one plus I two. What goes missing at the minima appears at the maxima, so energy is only redistributed. Got it?

1:50From path difference to phase difference

Now, phi comes from the extra distance one wave travels, the path difference, written capital delta, the little triangle on screen. Phi equals two pi over lambda, times delta, where lambda, the Greek letter, is the wavelength. With me? So let me ask you this. If one wave travels exactly one extra wavelength, what is phi? It is two pi, that is correct! And cos of two pi is one, so that point is bright. So we get maxima when delta equals n times lambda, where n is an integer. Okay? And for half a wavelength extra, phi is pi, and cos of pi is minus one, so that point is dark. So we get minima when delta equals, n plus a half, in brackets, times lambda. Right?

2:38Equal intensities, and the amplitude ratio

Now a special case that comes up often. Let both waves have the same intensity, I nought, that is I with a small zero below. Put this in our formula, and I equals four I nought, times cos squared of phi over two. Okay? Now tell me, for two equal waves, what is the brightest value of I, and what is the darkest? The brightest is four I nought, when cos squared is one, and the darkest is zero, completely dark! That is our puzzle solved. One plus one gives four at the bright spots, and zero at the dark ones. Clear? Next, intensity is proportional to amplitude squared; the curly sign on screen means is proportional to. So with amplitudes small a one and small a two, I max over I min equals, a one plus a two, over a one minus a two, whole squared. With me? And for slits, the intensity from each slit is proportional to its width. So a wider slit sends out more light. Got it?

3:37Coherent sources

Now, the second half of our puzzle, the lamps. Two sources are coherent if the phase difference between them stays constant in time. Okay? So take two separate lamps, exactly alike, side by side. Are they coherent? They are not coherent, and many students get this wrong. The lamps are independent, so their phase difference changes randomly, about ten to the power eight times a second; the wavy sign means about. Right? So cos phi keeps changing at random, and the cross term averages to zero. Then I equals I one plus I two everywhere. The intensities simply add, and there are no fringes. Clear? To get a coherent pair, we take both beams from one single source, as in Young's experiment. Then the two beams change together, so their phase difference stays fixed. Got it? And for a sustained, clear pattern, the sources should also have the same frequency and nearly equal amplitudes, and they should be narrow and close together. Nearly equal amplitudes make the dark fringes really dark. With me?

4:43Example 1: resultant intensity at 60 degree phase difference

Now our first example. Two coherent waves, each of intensity I nought, meet with a phase difference of sixty degrees; the small circle means degrees. Find the resultant intensity. Pause the video and try it first. Are you done? Okay, let's check. Many students write four I nought, times cos squared of sixty degrees. Is that right? No, that is exactly the trap! The formula has phi over two, and sixty over two gives thirty degrees. So I equals four I nought, times cos squared of thirty degrees. Okay? Now, cos thirty degrees is root three over two. Square it, and you get three over four. Right? So I equals four I nought, times three over four, which gives three I nought. That is the answer. Clear?

5:37Example 2: source intensity ratio when Imax/Imin = 25

Now a harder one, where we go backwards. In an interference pattern, I max over I min is twenty five. Find the ratio of the intensities of the two sources. Pause the video and try it first. Are you done? Okay, let's check. Start from our amplitude formula: a one plus a two, over a one minus a two, whole squared, equals twenty five. Take the square root; the double arrow on screen means this gives, so the fraction equals five. Okay? Cross multiply, and a one plus a two equals five times a one, minus five times a two. So four times a one equals six times a two. Right? Divide, and a one over a two equals six over four, which is three over two. With me? Now, many students stop here and write three to two. Is that the answer? Oh no, careful, three over two is only the ratio of the amplitudes! Intensity goes as amplitude squared, so I one over I two is three over two, whole squared, which gives nine over four. So the ratio of the intensities is nine to four, written with a colon. Got it?

6:49Quick revision: resultant intensity, 4 I0 bright, coherence

So, quick revision. First, I equals I one plus I two, plus two times root of I one times I two, times cos phi, with phi equal to two pi over lambda, times delta. Second, two equal waves give four I nought when bright and zero when dark, and intensity goes as amplitude squared. Third, coherent means a constant phase difference, and two independent lamps simply add their intensities. Now try the practice problems on this page. All the best!