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Explain why the amount of bending that occurs during diffraction depends on the width of the opening through which light passes.
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OpenStax College Physics for AP® Courses Solution, Chapter 27, Problem 2 (Test Prep for AP® Courses) (2:48)

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This is College Physics Answers with Shaun Dychko. Here's the formula that describes the diffraction pattern that you see when a single wavelength of light goes through a single slit with some width D and the question is asking us explain why the amount of bending that occurs due to the diffraction depends on the width of the opening? So it depends on this D. So let's figure out the angle to the minimum and we can say that the greater the angle to the minimum, the greater the amount of bending around the corner. So we have this D times this sin of the angle is where m is 1, negative 1, 2 and negative 2 and so on and this is a formula telling us about the destructive interference positions here. So if we divide both sides by D, we get a formula for sin Θ and then take the inverse sin of both sides and then we get an expression for the amount of bending Θ. So the further Θ is away from this central maximum, the more the light has been bent so Θ is the inverse sin of over D. Now here's a graph of y equals sin Θ so to take the inverse sin of something is to say given a certain y value, what value of the x-axis or Θ-axis in this case does that y value correspond to? So we have m this number times the wavelength divided by the slit width D and if D is really large that makes this fraction really small in which case the y value on this graph would be really small and the angle corresponding to that small y value would also be really small so not much bending in other words but if D is small so we have a small slit width compared to the wavelength then this fraction becomes bigger and so we move up this graph and the fraction has to be less than 1 by the way but the bigger the fraction, the bigger the angle that corresponds to that certain y value so this is over D is the y value and Θ is the angle that corresponds to that y value... it corresponds to that minimum, in other words. So we can see that having small slit widths result in a large angle to the diffraction minimum and so there we go!