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Landing on an Aircraft Carrier The Fresnel Lens Optical LandingSystem (FLOLS) used to ensure safe landings on aircraft carriers consists of a series of Fresnel lenses of different colors. Each lens focuses light in a different, specific direction, and hence which light a pilotsees on approach determines whether the plane is above, below, oron the proper landing path. The basic idea behind a Fresnel lens,which has the same optical properties as an ordinary lens, is shownin FIGURE $27-30 .$ Suppose an object is 17.1 $\mathrm{cm}$ behind a Fresnel lens, and that the corresponding image is a distance $d_{1}=d$ in front of thelens. If the object is moved to a distance of 12.0 $\mathrm{cm}$ behind the lens,the image distance doubles to $d_{1}=2 d .$ In the FLOLS, it is desiredto have the image of the light bulb at infinity. What object distancewill give this result for this particular lens?

9.24 $\mathrm{cm}$

Physics 103

Chapter 27

Optical lnstruments

Wave Optics

Simon Fraser University

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Okay, so in this problem, we have two different situations. The first situation we have. I like to move place at a distance off 17.1 ST Emitters. And in this situation, the images produced by a distance. Okay, The second situation. We plays the same light bulb at a distance off 12. Same team enters off the Fresnel lens, and this time, the distance off the image is too. Do you? Okay? And the objective off this problem is to discover the focal length off this fractional ends. So let's think about this. Your first situation. We have, uh, one divided by D, which is the position of the image, but its Sequels too. One divided by half miners, one divided by the zero. This is just if England's equation. So if we put the values that we have, we know that first situation one divided by D is equal one divided by F minus one, divided by 17.1. Okay, so let's think about the second situation. The second situation. We have one divided by two d, which is the distance of the image equals. Want if I buy f same f now divided one divided by 12 which is the new distance off the object. But if we look here, this is just one divided by two times one divided by D, and we know that one divided by D is just this equation. And here, So it's put desecration in this one. So this is going to be one divided by two that multiplies one divided by F minus one, divided by 17.1 that is equal to one divided by half miners, one divided by 12. So now we have a question for Defoe Collyns. This equation can be rearranged so we have f equals half multiplies one divided by 12 miners, one divided by journey 4.2 the power of minus one, which gives us a photo off nine point 24. Same team enters. And this is the answer to this problem because the problem once the object distance equals to the focal lens. We're looking to the object of infinity. So this is the answer to this problem. 9.24 centimeters

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