landing-on-an-aircraft-carrier-the-fresnel-lens-optical-landing-system-flols-used-to-ensure-safe-lan

Question

Landing on an Aircraft Carrier The Fresnel Lens Optical Landing
System (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 pilot
sees on approach determines whether the plane is above, below, or
on the proper landing path. The basic idea behind a Fresnel lens,
which has the same optical properties as an ordinary lens, is shown
in 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 the
lens. 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 desired
to have the image of the light bulb at infinity. What object distance
will give this result for this particular lens?

Nathan Silvano · Accepted Answer

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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;re looking to the object of infinity. So this is the answer to this problem. 9.24 centimeters

Zachary Warner · Answer

so to begin our question were asked for part A in terms of in and other sphere and the sphere radius are what is the distance between the image produced by the sphere and the right side? Um, of us fear. So to do this, we&#x27;re gonna have the relationship between the object distance in the image distance given us follows its indicate this is part, eh? We have a next of refraction. One in one, divided by the distance. The image I plus into over p is equal to into minus in one over our okay, But here, we&#x27;re gonna consider p going to infinity. So the images coming from infinity or excuse me? The, uh, the object distance is coming from infinity, so this is going to go to zero. So what we end up with is, um, I look is equal to in one times are divided by into minus in one. So here we&#x27;re gonna let in one equal one. We&#x27;re gonna let in, uh, in two equal in, and the value of in is between one into so into is between one and two. If this is true, then I has to be greater than two times are so the image does does form before the race strikes the opposite side of the sphere. So we can still consider this as a sort of virtual object for the second image ing event where this virtual objects That distance then is given by, um two R minus one is equal to in minus two times are divided by in minus one. Um, where we have simplified the notation again by writing into his equal to end here on putting this in for p an equation. Att, The very beginning. So we go back to our beginning equation where in one over I plus into over p is equal in two minutes and one over arc. But now we&#x27;re not gonna let p equal infinity. We&#x27;re gonna let p equal, um, to our minus one. So for this part, P is equal to our minus one. Therefore, I will call this a p prime, right? So we have in one which is just one, so we can just right. It is as such, we have won over the new image will call I prime plus, um, up in overpay prime, which is to our minus one. But two are minus one was equal to into in minus two times are divided by in minus one so we can just plug that in. So we have end up within divided by and minus two times are divided by in minus one, and this is equal to into minus in one which is in minus one, divided by our So we can rearrange here we&#x27;re gonna try to solve for I prime. So we have won over. I prime is equal to and minus one over R minus in times in minus one, divided by in minus two times are. So I just really think that above equation and I simplified the term there. Here, I&#x27;ll circle it. I just simplified this term here by making by taking the value that&#x27;s at the very bottom and bring it up to the numerator. Okay, so we can continue to simplify this and what we have is and minus one because we need to get a common denominator and that common denominator is going to be in minus two times are so this is gonna be in minus one times and minus two divided by, uh, and minus two times are And since they have a common denominator now we can go ahead and use that common denominator. And we subtract in times in minus one. So we&#x27;re gonna, um, go ahead and carry out those operations in the, uh, in the numerator there. So we have won over I Crime is equal to and squared minus two in minus in plus two. So we carried out in minus one times in minus two, and then this is gonna be minus. Now we&#x27;re gonna carry out the operation in times in minus ones. This is gonna be minus and squared. Plus in. And this is all still divided by in minus two times are so we could do some simplification here. Let&#x27;s cancel out some terms. So this end squared cancels with this minus and squared, right? This minus in cancels with this plus in. So what are we left with? We&#x27;re left with this equaling to minus two in divided by in minus two. Times are which is equal to negative too. Multiplied by in minus one, divided by in minus two times are. So we need to simply invert these equations to solve for I prime. So now we find that I crime is equal to, um to minus in times are. So I&#x27;m just distributing that negative sign in front of the negative, too, through that value to get rid of that negative sign there. So this is two times in minus one when we can go ahead and box that in, Is your solution for part a part B, um, says is the image to the left or to the right side? Uh, so the images to the right side of the sphere and we just have to look at the value for, uh, for I prime. And since I prime is going to be, um, positive for any value of in between one and two, right? No matter what in is as long as end is between one and two. So it&#x27;s greater than one less than two I prime. It&#x27;s positive. So we can say that I prime is greater than one or is, um, greater than zero integrated in zero. That&#x27;s a part B. So because it&#x27;s greater than zero thus say thus, it is to the the right side of the spear making box set in as your solution to be in the final part of the question.

Ryan Kutayiah · Answer

in problem 78. We have a screen, uh, which is given over here. This is the screen. And then we have an object given green, and the object and the screener are basically in the same are the same distance away from a converging lens and 60 centimetres away from this object and screen. Isa plainer is a flat tire, right? And then in between them, somewhere there is a converging lens that&#x27;s placed, and we&#x27;re told that at 22 centimeters from the object, one replaced the converging lens there than the image that&#x27;s reflected of the director goes through the land and that is reflected back from the murder is sharply focused on the screen, so it&#x27;s started sharply focused on the same spot where the object is and a party. We were asked to draw a diagram indicating basically what&#x27;s going on so right away from knowing that thie, the object in the mirror, are 60 centimetres apart and that the lenses 22 centimeters from the object we can find out the distance between the lens and the mirror, which is nothing more than the difference between the distance from the object to the mirror minus the distance from the object to the let&#x27;s. So that gives us the distance from the lens to America and we&#x27;ll need us. We&#x27;ll need this number. So I just kind of threw it in. And it&#x27;s 38 centimetres when you take the difference. And now for this con verging lands here we have an object. And what will happen when it goes through When the light passes through the lens from the object, Uh, which the distance is going by ass one, which is 20 centimeters. There&#x27;s an image that has form, mind is converging lens and I have been noted the image that&#x27;s form by by the lens of of this object by us, one primed now. So that&#x27;s the first thing that happens. Not the next thing that happens is that this image now becomes the object for the mirror. Right, so So now we have a so we have over here as one prime. The distance between Let&#x27;s say, the marriage is placed at some distance sm over here. So now this object is sm away from from the mirror, and it&#x27;ll form a reflection that&#x27;s pretty much equidistant the same distance, sm, which a class on prime so right away. Because it&#x27;s a plain Mary we know S m is equal to s m prime, right? Ok And now what will happen is that this lens because the light is being reflected back through we&#x27;ll see effectively on object now, right? That&#x27;s at s too. So then Sdu But we have s two is s one primed plus s m plus s m prime But we know for a plain married we have sm is equal Teo sm prime Tio Then we have s two is equal to s one prime plus to S m and let&#x27;s let&#x27;s just same this result Tio basically won. The light is coming back now for being reflected from the mirror Thie object distances s too which is given by this equation here and now The image is going to be formed as to prime and as to prime mister distance from the lands to the screen. But the screen and the objects are located in the same place. So s two Prime is the same as this one prime which is from the object to the lens, right? So weaken right here that there s one is the same as as two prime. Okay, So these air too important? No. Well, we have three important pieces of information that we&#x27;re going to use, uh, in part B, which is to find the focal length of this particular lens. So let&#x27;s remember, we have this equation one ever s plus one over X. Brian is equal to one over f. So we will apply this equation two times, right? One set for for S and s one prime. And again for S and s to prime. Right, Tio, we have won over Asplund plus one over X one. Prime is equal to one over F. The focal length is that face is not changing because we have the same lens that we&#x27;re using And then again, we can apply it another time. One of us two plus and won over asked to Prime is equal to one over f. Now, Uh, what we could do is, um, substitute for, let&#x27;s say, uh, use this equation. Right? So one over ass is one over ass phone post, one of ass to prime and substitute for this one over half hair. So I&#x27;ll just have this one of her has to plus one over s two prime. Does he go to and I&#x27;ll substitute this on the right hand side Won over s one plus one of her as one primed. Okay. And now we&#x27;re going to use the fact that s one, uh, her estate as two prime is equal to us one. So will this keep me But over here. So one ever asked to, uh plus one over As two Prime does not ask one as one. They want to one over as one plus in one of our s one prime. Right? So I just the right hand side, eh? Put it over here and the left hand side I replaced has two prime my ass one. And if I move s one from this side over to the right side. But if I subtract that&#x27;s one from both sides. That tram goes away and then I have won over s too is equal to one over ass One primed Which tells me that SDU is the same as as one prime. Okay, so as to is the same as us one prime. So let&#x27;s look at our picture. What does that mean for us? So here is your s to distance where they images as being reflected. The image is formed by the reflection in the mirror and we&#x27;re told that Okay, that&#x27;s actually the same distance as this has one prime. So that means the marriage Really wait right here. And so the marriage and be right here. And if the mirror&#x27;s right here, that means s m and therefore sm prime zero. Right? But another way to do that to see that, like, if you wanted to just use the equations is if you substitute, we just found That s to assess one prime if we substitute that in here, right? So if I substitute for us one primed. So I used these to acquaintance and that tells me that OK asked to is equal. Do That&#x27;s one prime, which is as to plus to S m. And if I move this too, the left side, then I have to S m to S M is equal to zero, which means, as Emma&#x27;s equals zero, right? So what does that mean? What is sm as I&#x27;m Mrs Descends from the object of mirror and if that distance zero, then the mirror and the object. Coincide. They&#x27;re the same location. So we have sort of ah, physical idea of what&#x27;s going on here. And now that we have the spread of information, we can we can pretty much solver our problem. Okay, so how will we solve this particular problem? Well, we&#x27;ll use this fact that we have the distance from the lens in a mirror is 38 centimeters in the fact that the distance Ah, uh, basically es Juan and s two are the same. That means the mirror is over here at s one primed and Andy lm really so that we said the mirrors Over here, the lime is the distance between the lands in the mirror. So this distance is 38 centimetres. So that means, as one primed is the same as the LM. So, yes, one prime does equal do D l m, which is 38 centimeters. Okay, so now we can use this equation. So we know ass one, right? This one is 22 centimeters. We know s one primed, which is to yell m and hopefully was Claire. How I figured out that as one prime is the same as the distance between the lens and America. Right? Because, remember, we found that Thank you. The image in the mirror location coincide. So let&#x27;s let&#x27;s solve on the next page for the focal length one over at Swan plus one over S one prime is equal to one over. And then we can, you know, do some some stuff here so I could multiply top and bottom by asked one prime this one of us one plus one of us one primed top and bottom. Buy us one, and that&#x27;s what over f. I&#x27;m basically just doing a procedure here. Teo, get common denominators. So then I have s one primed plus s one over. As one prime times asked. One is equal to one of the F, which means F is equal to s one primed. It&#x27;s, uh it&#x27;s one crime in this one s one over s one prime plus us one. And I have the values for these. Let&#x27;s go back to page one. So has one. Prime does 38 and Esteban is 22 both in units of centimeters. So that&#x27;s 38 centimeters times 22 centimeters. Divided by 38 centimeters plus 22 centimeters and I&#x27;ll leave it to you as an exercise toe. Basically, plug this number and a calculator, and then you&#x27;ll get your final answer for focal length.

Katie Mcalpine · Answer

Okay, so this problem is about the lens of the eye, and so it&#x27;s a double convex lens, so I&#x27;ll go ahead and drawl it like this. And it has an index of refraction of 1.44 So and is equal toe one point for four. And, um, the focal length is eight millimeters, so eat groups and the radius of curvature has the same magnitude, and then we want to find the radius of curvature of the lens. So to do that, we can use our creation than lens of creation, which is won over half is equal. Thio n minus one times one over r one. I&#x27;m not plus minus one over r two, okay. And so these are the same. So that&#x27;s gonna be on minus one, all multiplied by, um, two over are. And so if we want to solve for are we got that? That&#x27;s equal to, um, half times and minus one. It&#x27;s minus times two. So I&#x27;m gonna go ahead and plug that into a calculator and see 16 times 160.44 7.4 millimeters. Okay. And then what else does the question I ask if the object is placed, so we know that&#x27;s object to 16 centimeters tall. Gonna do this on another page? Um, why is 16 centimeters and we know that it&#x27;s position is 30 centimeters, so us is equal toe 30 centimeters, and and we won&#x27;t basically just want to know where it&#x27;s focused and how tall the image would be. And so the F is equal to eight millimeters. I&#x27;ll go ahead and transpose out here, so af is equal. Thio point to those 0.8 millimeters now eight millimeters. So that&#x27;s point eight centimeters, and Okay, great. So now I can use that then lens equation to get us prime the position of the image. So we know one of us plus one over s prime is equal to one over half. So we get hit one over s prime, bring this to the other side at the fractions, and then cross multipliers some other version of that algebra. Um, we&#x27;ll give you half times us divided by, uh, minus us. Did I do that in my head? Right. Oh, Whoops. To Owen as prime equals, bring that over us. It would be one of her half. I&#x27;m just gonna read, write it out, having trouble visualizing it. So one over us prime equals one over F minus one over s. And then yeah, then a B s minus f and then flip it. Okay, so I was wrong down here, so that&#x27;s gonna be s minus f great and plugging in f and s. Is this something I could do in my head? I can&#x27;t. Okay, 30 times plain age divided by, um 29.2. I got 1.25 centimeters. So that&#x27;s where the image gets focused and makes sense. And it&#x27;s such a short distance because it needs to reach your cornea, um, toe to your brain on. So now we want to get the height of the image so we can do why? Over why? Prime equals minus Um Oh, oops. That should be. That&#x27;s prime over s. Excuse me. A y prime over. Why you do this? I&#x27;ll go ahead and to block this off. Now we&#x27;re trying to find the height. So s prime negative s prime over asked equals y prime over. Why and why? Prime is therefore why times us crime groups don&#x27;t forget the negative divided by us. So I&#x27;m gonna plug that into a calculator. Okay. 1.25 provided by Bernie and everyone I multiplied by the height of the original. Just 16 on dhe bra means that the average height is 0.667 Thank you. Centimeters. Okay, let&#x27;s see him getting different answers. I&#x27;m gonna pas this and, um reinvestigate. Okay? I need a calculator error when I calculated as prime. So let me go ahead and fix that. Um, so I mean, yeah, it was just a calculator error. So if I re calculated, I got 0.8 to 2 centimeters. And then if I go ahead and redo this calculation, see what I got? So 0.8 to 2 two divided by Bernie. Well, supplied by 16. I got point for 38 centimeters. So when we were by Assad. Great. And what else is asked? Um, that&#x27;s the height. And then if s prime is positive. The images Riel thinks you really e think all like you would have to be really like the photons would really have to focus on your cornea for, um, that&#x27;s right. Part of the eye. You&#x27;re the photons would have to focus to create an image that your mind could read. And, um so is it Inverted. So magnification is negative s over s prime and us and as primer both positive, um, and therefore magnification negative. And therefore it&#x27;s inverted. So I&#x27;ll go ahead and write that down. So it&#x27;s inverted and huh? No pause this role, okay? Yeah, I stand by. My answer is that was this double checking everything.

Khoobchandra Agrawal · Answer

Hello, friends. This question is based on barking off pinhole camera. It is given in it off. Pinhole camera has a small circular purchase off the army. 30 light from distant object passes through the aperture and to where otherwise dark box falling on the screen located at a distance. And if he is too large, that display understand will be for the because off right point in the field off of you will send life onto a circle off time eater slightly larger than the on the other end. Yet if these two small diffraction bill blood that display on the state, the screen sources are reasonable Serve made the diameter off Central discover diffraction pattern is equal to the actor script. So in the first part we have toe justify that for more diplomatic life. The plane different and must be very large compared Toby the condition for serve. You fulfilled a piece Where is going toe this So in the part A justified that you showed that for monochromatic light on monochromatic light big plane away front and help Toby much greater than the the condition for a shot of you, it&#x27;s fulfilled. If the square Toby equals 2.44 Lambda Times in tow and deeper. Find optimum pin hole diameter off a month. Been hold diameter. What light off prevalent? 500 nanometers. Rejected all ice cream at 15 centimeter of eight. Let us start solving what part? The minimum. And it&#x27;s one point. Toto Lambda Bitey radios off diffraction desk. Why? So we will get the square. Toby. 2544 Lambda into it. All right now be. But we have to find the deep for black. Devlin&#x27;s is given 500 nanometers on L Toby 15 m substituting developed. Of these cultural 2.44 babe length off light is given 500 nanometers and screen is after the space off 15 centimeters. But this will be for 80 four countries, 4.28 and tend to the power minus 4 m. That&#x27;s one for this problem. Thanks for watching it

Netra Sharma · Answer

So in this problem for a pinhole camera in part a. If the distance between the pressure of the pinhole camera and the screen is too large compared to the porter diameter of the camera, which is L very, very greater than D the pressure diameter that, first of all, from Raleigh criterion they don&#x27;t mean is 1.22 Lambda over D. We know this part. It&#x27;s called this equation one also. I&#x27;m trying to show in the figure here they&#x27;re mean for a small angle. He&#x27;s also are over l right, because in this case that I mean basically going to be cool to r or l here, where r is the the radius of the diffraction circle. Okay, the fresh diffraction disk. You can call it that the radius of the diffraction disk and L is the distance between the pressure of the camera to the diffraction spot or screen. Now it is stated in the statement of the problem that we will obtain the sharp image when the pressure diameter is closely quoted the diameter of the diffraction disk. That means two are the diameter of the diffraction disk. It cause pressure diameter in this case we get a sharp image. Okay, so let&#x27;s just substitute this in Equation one and equation too. And whatever gonna get here, we&#x27;re going to get They don&#x27;t mean cause 1.22 lambda over Capital D equals de by two over L. Right. So from here, we&#x27;re going to see that the square is going to be called a 2.44 Landale from this two guys over here. Right? So that proofs the part? A no, in part B of the problem to find the optimum pinhole diameter for Lambda it calls 500 at a meter that is projected on the screen. That is at L. Because 15 centimeter away from the away from the do you preacher from part A, we can see that d is going to be cool, too square root of 2.44 Lambda l. So that means square root of 2.44 times 500 times 10 to the negative, nine times, 15 times 10 to the negative too. And skirt of this whole term over here. And this is going to give us 4.28 to a times 10 to the negative four meter. Or you can also say zero point for 28 millimeter. So that&#x27;s ah, pressure diameter.

Problem

A Cassegrain astronomical telescope uses two mirr…

Problem 93 Hard Difficulty

Answer

9.24 $\mathrm{cm}$

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