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A diffraction pattern is formed on a screen when light of wavelength 410 nm is passed through a single slit of width 1 μm. If the source light is replaced by another light of wavelength 700 nm, what should be the width of the slit so that the new light produces a pattern with the same spacing?
  1. $0.6 \textrm{ }\mu\textrm{m}$
  2. $1 \textrm{ }\mu\textrm{m}$
  3. $1.4 \textrm{ }\mu\textrm{m}$
  4. $1.7 \textrm{ }\mu\textrm{m}$
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OpenStax College Physics for AP® Courses Solution, Chapter 27, Problem 11 (Test Prep for AP® Courses) (2:26)

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This is College Physics Answers with Shaun Dychko. We have a diffraction pattern formed when 410 nanometer light is passed through a slit of with one micrometer. Now if this light is then replaced by a different one with wavelength 700 nanometres what should the width of this lid be adjusted to in order have the same spacing in the diffraction pattern? So the diffraction pattern is given by this formula. This tells us what the angle is to each of the dark fringes and it has the slit with capital <i>D</i> times sine of that angle equals <i>m lambda </i>where <i>m</i> is some integer one, negative one and so on and Lambda is the wavelength of light. So in the first case with the first slit with we’ll have <i>D1</i> times <i>sine theta</i> equals <i>m lambda 1 </i> and in the second case we'll have <i>D2</i> <i>sine Theta</i> and I intentionally did not rate a subscript 2 there because this angle is meant to be the same in the second case because we want to have the same pattern. So we have the same <i>Sine Theta</i> equals <i>m lambda 2</i>. Now from this first formula we can solve for <i >sine theta</i> and say that it <i>m lambda 1</i> over <i>D1</i> by dividing both sides by <i>D1</i> here and then we'll use that to replace <i>Sine Theta</i> in the second expression. So I've written <i>m lambda 1</i> over <i>D1</i> in place of <i>Sine Theta</i> in this expression in the second scenario and so we have <i>D2</i> over <i>D1</i> equals <i>lambda 2</i> over <i>lambda 1</i> because we can multiply both sides by one over <i>m lambda 1</i> and the <i>m</i> cancel and so <i>D2</i> over <i>D1</i> is <i>lambda 2</i> over <i>lambda 1</i> and then we'll multiply both sides by <i>D 1</i> to solve for <i>D2</i> so the diameter 2 such that we have the same pattern as we did with diameter or slit with one. I used the word diameter because of the <i>D</i> but it's not a really diameter it's the slit width so <i>D2</i> is <i>D1</i> times <i>lambda 2</i> over <i>lambda 1</i> so that's one micrometer initially times a new wavelength of 700 nanometres divided by the initial wavelength of 410 nanometres giving a new slit with of 1.71 micrometers to give the same diffraction pattern and so that's which one is that, that's answer <i>D</i>.