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 Tags math puzzles , physics puzzles

 1st December 2006, 08:00 AM #1 Dazed Muse   Join Date: Aug 2006 Posts: 586 Is this possible? (math) A converging lens has a focal length of 10 cm. A screen is placed 30 cm from an object. Where should the lens be placed, in relation to the object, to produce a focused image? You're supposed to be able to get the answer using the thin lens equation, F = 10cm di + do = 30cm 1/di + 1/do = 1/F But nothing I've tried clicks. Everything seems to indicate that when the focal length is 10cm, the image always appears at least 40cm away. Sorry to post such a crappy question but I'm stumped.
 1st December 2006, 08:34 AM #2 ChristineR Illuminator   Join Date: Jan 2006 Posts: 3,189 Behind the object. Where are you getting di+do=30cm? Why not di - do = 30 cm? __________________ Avatar (c) Neopets.com
 1st December 2006, 08:42 AM #3 Dazed Muse   Join Date: Aug 2006 Posts: 586 do is the distance from the object to the lens, di is the distance from the projected image to the lens. Since its given that the image and the object are 30 cms apart, and the lens is somewhere in between, then the sum of di and do must equal 30. Object--------------------Lens----------------Image <--------------do--------->< ---------di----------> <------------------------30cm--------------------> The problem is, the position of the lens is the variable.
 1st December 2006, 08:48 AM #4 Hellbound Abiogenic Spongiform     Join Date: Sep 2002 Location: In a handbasket Posts: 8,922 IF the lens must be between them, I keep getting an imaginary number. Specifically, I came up with 15+/-(7+iSQRT(3))/2 for di I used your second equation (di+do=30) to substitute do in terms of di (do=30-di), then plugged it into the third, reduced it to a quadratic (di2-30di+300=0), and applied the quadratic formula. Are you sure you have all the units and their values correct? (i.e.-centimeters vs. meters, 10 vs 1 or 1.0, etc)? DISCLAIMER: It's bene a while, so I may well have amistake somewhere. Last edited by Hellbound; 1st December 2006 at 08:56 AM.
 1st December 2006, 08:57 AM #5 ChristineR Illuminator   Join Date: Jan 2006 Posts: 3,189 Why does the lens have to be between the screen and the object? Put the object between the screen and the lens and the math works out. __________________ Avatar (c) Neopets.com
 1st December 2006, 09:04 AM #6 Dazed Muse   Join Date: Aug 2006 Posts: 586 "The system of equations 1/di+1/do=1/f=1/10 and di+do=30 will help you find two numbers that would satisfy the geometry of the question (distance between the image and the object is 30 cm) and the optical device used to produce the clear image of the object (the lens has focal distance of 10cm). To visualize the story, you may try to do the following experiment. Get the magnifying glass (it is a converging lens), lit the candle, and set up a screen (piece of paper). By moving the converging lens between the screen and the candle, you will find that there would be one specific distance between the candle and the lens that produces the larger image of the candle light. As you can see, the distance between the candle and the screen remains constant. "
 1st December 2006, 09:05 AM #7 Yllanes Muse   Join Date: Sep 2005 Location: Madrid Posts: 826 Originally Posted by Dazed But nothing I've tried clicks. Everything seems to indicate that when the focal length is 10cm, the image always appears at least 40cm away. Sorry to post such a crappy question but I'm stumped. You are right, the minimum distance object-screen with a real object and a real image is 4f'. If that distance is greater than that, there are two possible positions. The usual equation is $ \frac{1}{f'} = \frac{1}{s'}-\frac{1}{s} $ where distances to the left (right) of the lens are negative (positive). A real object has s < 0. Let's assume s' >0 (real image) and s= - |s| < 0. We want the minimum value of |s| + s'. I'm going to use your notation: |s| = do, s'=di. $ D=d_o + d_i = d_o + \frac{d_o f'}{d_o-f'} $ With dD/d(d_o) = 0 we get d_0 = 2f'. If we go back to our original relation, this implies d_i = 2f', so D= 4f' is the minum distance real object-real image. As an interesting additional exercise, consider Bessel's method for the determination of the focal distance. I said that if D > 4f', we have two positions for the lens where an image of the object is formed on the screen. If we call a the separation between those two positions, prove that $ f' = \frac{D^2-a^2}{4D} $ Last edited by Yllanes; 1st December 2006 at 09:12 AM.
 1st December 2006, 09:10 AM #8 I less than three logic Graduate Poster   Join Date: Dec 2005 Location: St Cloud, MN, USA Posts: 1,464 Originally Posted by Dazed A converging lens has a focal length of 10 cm. A screen is placed 30 cm from an object. Where should the lens be placed, in relation to the object, to produce a focused image? Just shooting from the hip off of intuition here, but wouldn't the distance of the lens, from either the object or the screen, to produce a focused image be dependent on the size of the object and the intended size of the projected image of the object on the screen? __________________ “There is perhaps no better demonstration of the folly of human conceits than this distant image of our tiny world.” - Carl Sagan “The fact that we live at the bottom of a deep gravity well, on the surface of a gas covered planet going around a nuclear fireball ninety million miles away and think this to be normal is obviously some indication of how skewed our perspective tends to be.” – Douglas Adams
 1st December 2006, 09:13 AM #9 Dazed Muse   Join Date: Aug 2006 Posts: 586 Ah.. So then the image must be virtual? But then how would it project onto the screen?
 1st December 2006, 09:14 AM #10 Merko Graduate Poster     Join Date: Nov 2006 Posts: 1,901 Ilessthanthree: No. That would be determined by the distance between the object and the screen, not the lens position.
 1st December 2006, 09:14 AM #11 Dazed Muse   Join Date: Aug 2006 Posts: 586 Originally Posted by I less than three logic Just shooting from the hip off of intuition here, but wouldn't the distance of the lens, from either the object or the screen, to produce a focused image be dependent on the size of the object and the intended size of the projected image of the object on the screen? Those variables aren't given.
 1st December 2006, 09:15 AM #12 Merko Graduate Poster     Join Date: Nov 2006 Posts: 1,901 Dazed: You could always re-arrange the object into a virtual one. Or get another lens.
1st December 2006, 09:23 AM   #13
Dazed
Muse

Join Date: Aug 2006
Posts: 586
The problem with the virtual image is it appears on the same side of the lens as the object, rather than being focused onto the screen on the other side, which is the goal.
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 1st December 2006, 09:25 AM #14 ChristineR Illuminator   Join Date: Jan 2006 Posts: 3,189 Case 6 dude. You can see the image. I'm trying to find a pic of one of those "magic boxes" where the image floats in the air. __________________ Avatar (c) Neopets.com
 1st December 2006, 09:28 AM #15 Hellbound Abiogenic Spongiform     Join Date: Sep 2002 Location: In a handbasket Posts: 8,922 Try looking around here. If you have do equal about 7.91288cm, you can produce a di of -37.91291cm, which gives 30cm between image and object. I didn't see where the question said it must be a real image, as opposed to a virtual image.
 1st December 2006, 09:34 AM #16 Yllanes Muse   Join Date: Sep 2005 Location: Madrid Posts: 826 Originally Posted by Huntsman I didn't see where the question said it must be a real image, as opposed to a virtual image. The described setup will not work with this (and it implies real object and real image, because the light comes from the very candle we are trying to image and the lens is between the object and the screen). Also, virtual image does not form a visible projection on a screen (you can image it and record it in many ways, but this implies an additional optical system). The suggested setup is impossible with these values. Is it from a book? Maybe an erratum. Last edited by Yllanes; 1st December 2006 at 09:39 AM.
 1st December 2006, 09:38 AM #17 Orangutan Graduate Poster Tagger     Join Date: Sep 2004 Location: Florida. Posts: 1,175 Originally Posted by Huntsman I didn't see where the question said it must be a real image, as opposed to a virtual image. You're right but it is implied by the mentioning of the screen. If I were to answer the question I would start by explaining why a real image cannot be produced on the screen, before you assume it has to be a virtual image. That way you demonstrate your knowledge of the equation and the lens theory. I would also expect that a lazy tester just "changed a few numbers" in an old question and assumed it would still produce a valid answer. Did you post the >exact< text of the question?
 1st December 2006, 09:43 AM #18 ChristineR Illuminator   Join Date: Jan 2006 Posts: 3,189 The virtual image doesn't form on the screen, but you can see it at the appropriate place. I'm not finding any photos, alas. I think this would depend on the exact wording of the problem and how you interpret "focus on screen." I think it would be possible to give the appearance of an object plastered onto a screen. __________________ Avatar (c) Neopets.com
 1st December 2006, 09:57 AM #19 Hellbound Abiogenic Spongiform     Join Date: Sep 2002 Location: In a handbasket Posts: 8,922 I dunno, this is my first time with this equation so I'm figuring it out as I go Okay, so you set up a small wormhole generator, so the distance from the lens to the screen is longer than the distance from the object to the screen...
 1st December 2006, 10:40 AM #20 Merko Graduate Poster     Join Date: Nov 2006 Posts: 1,901 Originally Posted by Orangutan I would also expect that a lazy tester just "changed a few numbers" in an old question and assumed it would still produce a valid answer. Or maybe a good tester phrased an impossible question to see if the takers really understood anything about the subject, or were just mechanically using the formulas!
 1st December 2006, 10:48 AM #21 Dazed Muse   Join Date: Aug 2006 Posts: 586 Brilliant answers, thanks. I couldn't figure it out I was starting to feel amazingly stupid.
 1st December 2006, 10:58 AM #22 skeptifem is not beauty 2K compliant     Join Date: Nov 2006 Posts: 3,259 Divide By Zero, Oh Shi-
 1st December 2006, 12:58 PM #23 Art Vandelay Illuminator     Join Date: May 2004 Posts: 4,790 Originally Posted by I less than three logic Just shooting from the hip off of intuition here, but wouldn't the distance of the lens, from either the object or the screen, to produce a focused image be dependent on the size of the object and the intended size of the projected image of the object on the screen? I think that you're thinking of a person looking through magnifying glass, in which case one doesn't need to produce a focused image, because one's eye will focus the image (as long as the image isn't too far from being focused). Or you might be imagining a projector, which can be moved a bit and still produce a readable image. However, that's only if the projector is moved only a small distance. If you focus a projector at one distance, then try to move it a significant distance away, it won't be focused anymore. Originally Posted by Dazed "To visualize the story, you may try to do the following experiment. Get the magnifying glass (it is a converging lens), [light] the candle, and set up a screen (piece of paper). By moving the converging lens between the screen and the candle, you will find that there [is] one specific distance between the candle and the lens that produces the larger image of the candle light." If they can't get the English right, that supports the supposition that they got the math wrong.
 1st December 2006, 01:35 PM #24 Schneibster Illuminator   Join Date: Oct 2005 Posts: 3,966 Originally Posted by Dazed A converging lens has a focal length of 10 cm. A screen is placed 30 cm from an object. Where should the lens be placed, in relation to the object, to produce a focused image? You're supposed to be able to get the answer using the thin lens equation, F = 10cm di + do = 30cm 1/di + 1/do = 1/F But nothing I've tried clicks. Everything seems to indicate that when the focal length is 10cm, the image always appears at least 40cm away. Sorry to post such a crappy question but I'm stumped. OK, the first thing you need to do is consider precisely how the lens works. Focal lengths for lenses are for an object at infinity. Convex lenses form real images for objects farther than the focal length, and virtual images for objects closer than the focal length. The equation to use is the thin lens formula: 1/F = 1/s1 + 1/s2 where, F is the focal length of the lens; s1 is the distance from the lens to the object; and s2 is the distance from the lens to the real image. Note that this formula is only valid if the distance to the object is greater than the focal length. In this case, since the focal length is 10cm, and there is 30cm between screen and object, there might be enough distance to form a real image on the screen if we place the lens between the object and the screen. To find out, we have to determine what the sum of the distance to the object and the distance to the screen must be, and it must be less than 30cm. Now, we also know that s1 + s2 = 30cm and F = 10cm If s1 is 10cm, then 1/10 = 1/10 + 1/s2 which would make s2 undefined. Let's try s1 = 11cm. 1/10 = 1/11 + 1/s2 so, 1/s2 = 1/10 - 1/11 = 1/110 Nope, that won't work. In fact, as s1 varies from 11cm to 10cm, s2 varies from 110cm to infinity. How about if we make s1 = 30cm? 1/10 = 1/30 + 1/s2 so, 1/s2 = 1/10 - 1/30 = 1/15 And that won't work either. A brief look at this shows that as s1 varies from 10cm to 30cm, s2 varies from infinity to 15cm. You cannot bring the object to focus with only 30cm total to work with. The minimum distance is 40cm. You get that with s1 = s2 = 20cm. I'll leave it to you to prove that. Last edited by Schneibster; 1st December 2006 at 02:06 PM. Reason: typo
 1st December 2006, 01:49 PM #25 Yllanes Muse   Join Date: Sep 2005 Location: Madrid Posts: 826 Originally Posted by Schneibster The minimum distance is 40cm. You get that with s1 = s1 = 20cm. I'll leave it to you to prove that. I already did... post #7.
 1st December 2006, 02:06 PM #26 Schneibster Illuminator   Join Date: Oct 2005 Posts: 3,966 Originally Posted by Yllanes I already did... post #7. So you did. I didn't look closely enough. Well, at least the querent is certain it's right.

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