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The intensity of light with wavelength $\lambda$ traveling through a diffraction grating with $N$ slits at an angle $\theta$ is given by $I(\theta)=N^{2} \sin ^{2} k / k^{2}$, where $k=(\pi N d \sin \theta) / \lambda$ and $d$ is the distance between adjacent slits. A helium-neon laser with wavelength $\lambda=632.8 \times 10^{-9} \mathrm{~m}$ is emitting a narrow band of light, given by $-10^{-6}<\theta<10^{-6}$, through a grating with 10,000 slits spaced $10^{-4} \mathrm{~m}$ apart. Use the Midpoint Rule with $n=10$ to estimate the total light intensity $\int_{-10^{-6}}^{10^{-6}} I(\theta) d \theta$ emerging from the grating.

$M_{10}=2 \times 10^{-7}[I(-0.0000009)+I(-0.0000007)+\cdots+I(0.0000009)] \approx 59.4$

Integration Techniques

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Okay, so this question wants us to consider multi slipped a fraction where light is going through a lot of slits. And we want to look at how the intensity pattern changes based on the angle. Fada. So there are a lot of variables here, So to make things easier, let's just keep track of all of them. So and is the number of slits and how many do we have? Well, it's history of 10,000 D is the spacing of the slits, which is 10 to the minus forth. Lambda is the wavelength, which is 6 32.8 nanometers, and that is all we need for now. So now we could sub this in decay because you know that Kay vehicles and D hi signed data all over Lambda and we can plug in on the old unknown values here and we're just left for the function that depends on data. So we get that the total light intensity coming from the grating is the integral from negative turn to the negative six to positive 10 to the negative. Sixth of I have faded if data and that's equal to the integral from negative tending the minus six to tend to this. My sixth of end squared times. Sign of cave data over K of data all squared D theta. And we know how to calculate K of data because we have known values. So let's call this whole function if data now that we know what we're subbing into, So he wants us to use midpoint with an equals 10. So are Delta acts. It's just Well, we're ending at 10 to the minus six, and we're starting at 10 to the minus sixth, but a negative version. So we get two times 10 to the negative six divided by 10. What should give us two times 10 to the negative seventh. So are integral should equal delta x Times f of Delta X over too, plus f of Delta X over two plus Delta X plus dot all the way up. So we should get that our eye of Fada should be two times 10 to the negative. Seven times I of well, Delta access to times 10 to the negative seventh. This formula is assuming that we started zero, but we don't So it's negative. 10 to the negative sixth plus 10 to the negative seventh, which is negative nine times 10 to the negative seven plus I of negative seven times 10 to the native seventh. Plus, I got that all the way up to the final I value, which is 10 to the minus sixth minus two times 10 to the negative set up. And if we plug all these values into our calculator, I'd suggest using a table or calculator program. We'll see that in terms of NNK, our answer turns out to be about 1.2 times 10 to the 15th and squared over K squared. And you can evaluate this out if you like, But the book leaves it in this form, so that's what we're gonna write it as.

University of Michigan - Ann Arbor

Integration Techniques