Video: Intensity Ratio of Diffraction Pattern Maxima

For a three-slit interference pattern, find the ratio of the peak intensities of a secondary maximum to a principal maximum.


Video Transcript

For a three-slit interference pattern, find the ratio of the peak intensities of a secondary maximum to a principal maximum.

If we call the intensity of a principal maximum in the interference pattern created ๐ผ sub zero and we called the intensity of a secondary maximum ๐ผ sub two, then what weโ€™re looking for in this problem is the ratio ๐ผ sub zero to ๐ผ sub two.

Letโ€™s begin by drawing out a view of a three-slit interference pattern. If we have a set-up with three slits, that means that light from those three slits combines, interferes, and then forms a pattern on the screen to the far right. We can call the distance between the three slits ๐‘‘, and we can call the angle with the horizontal that a wave from the central slit makes ๐œƒ.

Now in this drawing, weโ€™ve just shown the waves from the three apertures or three slits meeting at one place on the screen. But of course waves from the three slits combine on all locations on the screen, forming a pattern of maxima, or intensity highs, and minima, or intensity lows.

If we call the wavelength of the waves passing through the three slits ๐œ†, then there is a mathematical relationship we can look up for the resulting intensity created by a three-slit interference set-up. In this equation, ๐ผ is the intensity that results from the combination of the waves from the three slits. ๐ผ sub zero is the intensity of one of the principal maxima. And that value divided by nine is multiplied by the expression we see inside the brackets.

Letโ€™s take a minute and look at that expression inside the brackets, in particular weโ€™ll look at the cosine function. Letโ€™s recall what the cosine function looks like graphically. When we look at a plot of the cosine function, we see that it has maxima where its argument is even multiples of ๐œ‹ and it has minima at odd multiples of ๐œ‹.

Now weโ€™re ready to apply this equation for intensity in a three-slit interference pattern to our scenario. Weโ€™ll do that by graphing this function; in particular, weโ€™ll be finding the maxima points of the function. There are two types of maxima as weโ€™ll see: there are principal maxima, which occur at even multiples of ๐œ‹ as we saw in the cosine function, and there are secondary maxima that occur at odd multiples of ๐œ‹.

Letโ€™s plot a few of the principal and secondary maxima on a graph. For the graph we make, on the horizontal axis weโ€™ll vary a value called ๐‘š, where ๐‘š represents the argument of the cosine function in our intensity formula. And on the vertical axis, weโ€™ll plot values according to the ratio, the intensity ๐ผ divided by the intensity of a principal maximum ๐ผ sub zero.

Letโ€™s plot our first point on the graph when ๐‘š is equal to zero. Looking back at our intensity function, the cosine of zero is one. And when we multiply that by two, we get two plus one gives us three. Then when we square that term, three squared is nine. And when we multiply nine by ๐ผ sub zero divided by nine, weโ€™re left simply with ๐ผ sub zero. When we plot that on our graph, because weโ€™re taking a ratio, that value is one.

If we jump ahead next and let ๐‘š equal two ๐œ‹, once again the cosine of that argument is plus one times two is two. Add one to it gives us three. And just like before, we end up with a value on our vertical axis of one. Next we look at four ๐œ‹, and we find the same result. These three points on our graph are principal maxima. They are the highest maxima created by our interference pattern.

Now letโ€™s look at the secondary maxima. These occur at odd multiples of ๐œ‹: one ๐œ‹, three ๐œ‹, five ๐œ‹, and so on. If ๐‘š equals ๐œ‹, the cosine of ๐œ‹ is negative one times two is negative two, plus one is negative one, squared is positive one. And in our intensity equation, we wind up with ๐ผ sub zero divided by nine times one or ๐ผ sub zero divided by nine. When we plot this secondary maximum on our graph, weโ€™ve seen that at a value of one-ninth of the principal maximum.

And when ๐‘š is equal to three ๐œ‹, we find the same result. So weโ€™ve now figured out the ratio of the principal maxima to the secondary maxima in this interference pattern. The ratio of the intensity of the principal maximum to the intensity of the secondary maximum is equal to nine to one, meaning the principal maxima have nine times more intensity than the secondary maxima.

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