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Find the largest wavelength of light falling on double slits separated by $1.20\textrm{ }\mu\textrm{m}$ for which there is a first-order maximum. Is this in the visible part of the spectrum?
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$1.20\textrm{ }\mu\textrm{m}$
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OpenStax College Physics Solution, Chapter 27, Problem 14 (Problems & Exercises) (2:05)

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This is College Physics Answers with Shaun Dychko. We need to find the wavelength of light that will be incident on a double slit that's separated by 1.20 micrometers and have a first order maximum at 90 degrees. Now the question doesn't say 90 degrees but when you rearrange this to solve for the wavelength— this is the double slit equation for constructive interference or for maxima— when we solve this for λ by dividing both sides by m—the order— we have λ is the separation between slits—d—times sin of Θ over m Now because it's a first order maximum and it's definitely 1, the separation between slits is given to us—1.20 micrometers— and so the only thing we can vary is this angle Θ and so we are finding a wavelength such that Θ to this maximum the angle between the central maximum and the first order maximum is going to be as large as possible and the largest possible angle you can imagine here is almost 90... I mean 90 precisely of course wouldn't work because it would never... light at a 90 degree angle to this central maximum would be parallel to the screen and never touch it but it's just almost 90 so we'll plug in 90 and so that's the way to maximize sin of Θ or another way to put it is the range of sin Θ is between negative 1 and 1 and so its maximum is 1 and so we can just plug in the number 1 here if we wanted but this is how we get the number 1 from this picture which is to plug in the number 90 for the angle so sin of 90 is 1. Alright! So the maximum wavelength then is 1.20 micrometers and this is infrared radiation that's not visible.