roughing-it-with-science-a-professor-shipwrecked-on-hooligans-island-decides-to-build-a-telescope-fr

Question

Roughing It with Science A professor shipwrecked on Hooligan's
Island decides to build a telescope from his eyeglasses and some
coconut shells. Fortunately, the professor's eyes require different
prescriptions, with the left lens having a power of $+5.0$ diopters
and the right lens having a power of $+2.0$ diopters. (a) Which lens
should he use as the objective? (b) What is the angular magnification of the professor's telescope?

Nathan Silvano · Accepted Answer

So in this problem, we were beauty a telescope using two lances. Uh, the first left the left lens. Let&#x27;s go. The left lens has ah, refractive our off to Diop tres And they&#x27;re right. The right lens has a reflective power off five Diop tres. So the first question of the problem we need to explain which off this lances should should be used as the objective one. Okay, well, we know that reflective. Our is just the inverse of the focal lens. So if a reflective our off the left, let&#x27;s call on l here is too the focal lens. The focal land needs to be 0.5 and the right one if their focal interview effective power is five. So the focal lens let&#x27;s go after it&#x27;s going to be zero point one. So, as we know with the theory, we know that the objective lens should be, by definition, the lens with the smaller focal lens. So we need to use the focal and 0.1. So this is they&#x27;re right blends. And that&#x27;s the answer to the first item. The second item. We need to calculate the magnification of this telescope. So we know that the magnification is just f objective, divided by the F off the eyepiece. And we know here that the F objective is 0.1. So just, you know, 0.1 divided by 0.5. This has to be close to, Let&#x27;s see, two point five. So this is the magnification off this Dallas group, that&#x27;s all. Thanks for watching.

Netra Sharma · Answer

So in this problem, the person has two lenses of different powers. One is positive, 1.2 doctors and the other one is positive. Nine doctors. So from here, we can actually find the focal length of both of the lenses, which is basically the reciprocal of the power of each. And I have already shown the focal length F one enough to hear. So if one is going to be a zero point 833 0.833 meter and after is going to be 0.111 meter. So if the sailor wants to be like telescope, that actually f one has to be the focal length of the objective. Okay. Hey, 33 and F two has to be the focal length of the eyepiece. Now, from here, we can easily find the magnification for party. So far, for part a, the magnification of the angular magnification of the telescope is for collective objective. Over focal length off eyepiece. This is going to be a 0.833 over 0.111 And this will give me a 7.5. All right now for part B the length of the telescope basically would be the focal length of the objective and focal length of the I Piss, which is 0.833 plus 0.111 meters because 0.944 meters.

Carlos Henrique De Lima · Answer

has two lenses that has powers off Judy and 4.5. And we want to know where is the mats move magnification telescope that to compute and which lands were used for the eyepiece. So the magnification for part A is the ratio off the powers off the ID based divided by P zero. So the best magnification is when the eyepiece has the greater power. So this will be minus 4.5, divided by two. This is managed to 0.3. And for part B, we are ready. Discover which one is greater to get a positive magnification greater than one magnification. We need that. The eyepiece has a great apartment. The objective Lance. So we need to use Ah, the lance with power equal for 25.

Kristela Garcia · Answer

in this problem, you are asked to determine the distance between these two lenses if you&#x27;re given the focal lengths for both of these lenses and you know the original object distance from the first Linz and the image distance for the second leads. So we&#x27;re gonna do that in two parts. So we want to know what this distances we&#x27;re going to first a solve for the object distance for the second. Let&#x27;s So we&#x27;re gonna sell for D. Oh, shoot now, because all of my numbers are most of my numbers aren&#x27;t centimeters. I do want to just leave it in centimeters and actually convert this d 012 centimeters when I need it. So I&#x27;m gonna use be thin lens equation. So that&#x27;s one over d o plus one over d. I equals one of her f and we want to solve before D Oh, So that&#x27;s one over d o equals one of her f minus one over d i and then solving for D O by inverting the left side and then inverting everything off the right side again because this is talking about the second lens. I&#x27;m gonna go back and add little twos to all of this. And then this is one over went over. Is there a 0.6 minus one over the image justice for the second object for the sick blends, which is negative 100. It does tell you that it is to the left of the second lens. So if it tells agents to the left of the second lens, and that&#x27;s only true for images, but to the left of the lens, it is negative. If it&#x27;s to the right of the lens, it&#x27;s positive again. That is only for the images again plugging all of that into your calculator. You should get a distance of the object of 0.59640 point six of you round up. What that tells you is that might image of the first Molins is located a distance of 0.6 centimeters from the second lands. So here this is this point is de I to and D Oh, it&#x27;s you. Well, correction, that&#x27;s d I want d I one equals d o to. So now we know what the image distance is. That&#x27;s just diodes here. But now that we know that it&#x27;s 0.6 centimeters from the second lens. We want to know what this distances and this distance is actually going to be d I won and we can calculate the distance using again the same equation. So instead of solving for D Oh, I&#x27;m gonna sell for D I. So one over D I equals one of her F minus one over d o the I equals 1/1 over minus one over dio again, All of these are ones. And then I get a distance of three centimeters here that tells you that if d I one is three centimeters and d o to this 00.6 centimeters than the total distance between these two lenses must be 3.6 centimeters again, that&#x27;s distance between these two lenses. Okay, so for the next part, for angular magnification, we actually need to know what my leaning your magnification is before you consult this. So we are going to write down some of our values. We said again, this is d I. One and then this year was d 02 We&#x27;re going to make some room, but clearing some of this out so again for this next part we&#x27;re gonna sell firmly near magnification first. That&#x27;s gonna help us. Sulfur are angular magnification A little bit later. So two sulphur, My linear magnification This is de I one times d i to all over d 01 times d 02 So d I one was three centimeters d I, too, was negative 100 d 01 waas 20 times 10 to the to and D O to result to be 0.6 sense leaders. And this gets you a magnification of 0.25 negative. So my height of my image is negative 0.25 times my height of my object. Now, I don&#x27;t know what my heights are, but I&#x27;m gonna use this relationship in my angular magnification. A question which is How did the image times d o one divided by D. I to times the height of the object again. So for h I, instead of using h, I am just going to put negative 0.25 h o. And the reason that&#x27;s helping us so this is 20 times 10 to 2 is because these h oes are eventually going to cancel out and I get an angular magnification advice

Abhishek Jana · Answer

Okay, So to get the magnification of the telescope, we can use the question twenty five flying three on. We see that magnification is given as, ah negative f over f e. And if we use the question twenty three point seven, we can write the focal lengths in terms of power. So that&#x27;s gonna be B E over f. So being over p o. So power is inverse a focal length. Now, to get the maximum magnification, we need the greater number on top and the less number at the bottom. And we don&#x27;t care about the negative sign because we always take the absolute value for magnificient. It&#x27;s going to be so the bigger than it is five point five, doctor and the smaller valley is to laughter. So the man in creation is negative. Two point eight x now for part B. Since we know that if we need magnification greater than one, we need the eyepiece with smaller focal length are greater power. So we see that five point five doctor is greater than two after, so that should be used as an eyepiece. Thank you

Sean Dougherty · Answer

So you have these two lenses of our classes. The left side has a power of 4.5 Diop tres in the right side of three point owned by actors. And we need to find out which one is the would be the better objective lands in our telescope. Well, we know that lens power is equal to one over the focal length. So for our are left lens, our focus is one over pl were won over 4.5, which is equal to about 0.22 meters are left are Excuse me, right lens because he won over p oh are which is equal to one over a 3.0 or 0.33 meters. Now we know that, um London. The bigger focus is a larger diameter. It has a larger diameter and a larger focus. It can focus in a result, objects farther away than a lens that has a smaller focus and sugar diameter. So we conclusively say that f R is the objective. Clint&#x27;s it&#x27;s also makes sense. You think about a telescope you generally imagine at the bigger lends being the one where the light comes in and the smaller one where you view. That&#x27;s not necessarily physically sound all the time, but that is a helpful way to look at it. Next part asks us to find the telephone. The telescope magnification em is equal to negative f objective over. If I, which for us is negative zero point 33 meters for objective lens over F high, which is our focal length of our left lens, was your point to two meters. So our magnification is going to be negative 1.5 times now The last part asks us How far away should the lenses be? Well, we&#x27;re focusing on distant objects. We know that our object distances infinity and when we have a converging lens and something infinitely far away, we know it focuses on the focus of that lens. So that would be the focus for objective. And we also know that the eyepieces oh, almost always positioned in a place near the image of the first lens. So the different distance between the image the first lines in the second eyepiece is the focal length of that eyepiece Hey which in our case, his 0.33 meters plus syrah point to two meters. So the distance train the two lenses is equal to 0.56 meters

Problem

Galileo's Telescope Galileo's first telescope use…

Problem 75 Hard Difficulty

Answer

a) right lens
b) 2.5

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James S. Walker

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Wave Optics - Intro

Multiple Aperture Interference - Overview

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a) right lensb) 2.5

Physics

Elyse G.

Liev B.

Zachary M.

Aspen F.

Wave Optics - Intro

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James S. WalkerPhysics

Elyse G.

Liev B.

Zachary M.

Aspen F.

a) right lens
b) 2.5

James S. Walker

Physics