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B Thin lens approximation and Apparent depth

  1. Mar 3, 2017 #1
    So we are studying optics in school this semster, Very interseting topic I say but I just have a couple of question I want to ask.

    In concave and convex mirror, we study spherical ones where F = R/2. I was able to prove this and that it is only an approximation when ## R >> h_o ## or ## h_0## is small enough and as concave mirror and convex mirror are the same this applies to both. But in lenses I couldn't do that, I read in my book that they assume that the lens is thin and it only refract once in the center of the lens. I don't really understand that.
    What should happen is that first I have to measure the derivative of the spherical mirror at the point where the light hits the surface of the mirror then take the perpendicular and use snell's law, The light travels a bit inside the lens so it hits a different point on the other surface of the mirror so I need to do what I did once again.

    So What I am asking is how does the thin lens approximation make it refract at a single point? and a Text book that I have said that the focal point of lens is R/2 which is wrong I guess and Is there a derivation which consider lens purely mathematical to compute the focal length? without assumptions(Only thin lens)?


    The other thing is about apparent depth, Lets first assume an object which it's depth is ##y_o## and you make a perpendicular ray coming out of it and lets assume that the eye of the observer is near the perpendicular line.
    Since our eye is not that big that eye will only see two rays which the distance between them on the surface is small (Or you can say the angle between them is small whatever you like)
    So the 2nd ray will intersect with the perpendicular one at Point ## Y_i ## and it can be calculated through the equation I derived:
    ##(\frac{dx^2+ y_i^2}{dx^2 + y_o^2})^{0.5} = 1/n##
    Where this object is under for example water and the ray is going to air.
    Now as dx gets really small the approximation of ## n = \frac{y_o}{y_i} ## gets better. I guess even if there is difference in Yi the brain enhances it to appear coming of 1 point. All of this to answer one question what happens when you are exactly at the perpendicular? I read a text book that says that you see the actual depth since light doesn't refract. But the problem is the eye isn't a point, it can never know where the object is if you only have one ray.. However, If you take dx as infinitesimal you get the same equation above to approximate the situation or you can say that the center of the eye is on the object but since it has width and that it is small you get the same answer again.

    So If I have an exam, What answer should I consider?
     
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  3. Mar 3, 2017 #2

    Simon Bridge

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    Thin lens approximation:
    The thin lens approximation can be derived from working out the equations for a real lens, then using the par-axial approximation (that is the one where the distance of the rays from the optic axis is very small compared with the radius of curvature of all the surfaces you are considering ... like what you did for the spherical mirror) ... then you take the limit that the thickness of the lens is small compared to the radius of curvature.

    Apparent depth:
    ... for vision to work you need diverging light to arrive at your eyes. Everything you see is the image of some object - but that is usually OK because the image and object positions are usually the same.
    In all your optics things, you are dealing with the situation where the image and object may not be in the same place ... in this case, the distance to the image from the surface of the water is called the "apparent depth".


    Your text has given you an approximate analysis of this - which results in a model where the image is always directly above the object, and there is only one possible position. Then it confuses you by saying "except when you are directly above..." without giving a good explanation of how that is different. After all, you can use exactly the same derivation as they did for arbitrarily small angles to the normal and get the same result.

    You can test this - take a meter rule to a pool, and put half of it under the water ... try this at different angles. How does the spacing of the marks vary with the angle you hold the ruler? You can also see if you spot an apparent depth looking directly downwards into the pool ... and does the far end of the pool, under the water appear closer that the bit above the water?

    You can explore the math by picking a refractive index and carefully constructing many rays from an object and out through the surface of the pool ... do they all appear to come from the same place?

    Also see: https://www.researchgate.net/publication/234721887_Apparent_Depth [Broken]
    ... though I encourage you to test this out for yourself.

    As to exams ... I'd use the text book for internal exams, and use whatever the standard is for your country at your education level for external exams.
     
    Last edited by a moderator: May 8, 2017
  4. Mar 3, 2017 #3

    jtbell

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    First you derive an equation for refraction by a single spherical surface which is somewhat similar to the one for reflection from a spherical mirror. A Google search for "spherical surface refraction" turns up lots of hits such as this one:

    https://www.math.ubc.ca/~cass/courses/m309-01a/chu/MirrorsLenses/refraction-curved.htm

    A lens is two spherical surfaces, one after the other, separated by a distance d. The image formed by the first surface serves as the object for the second surface. To get the thin-lens equation, use the spherical refaction equation twice, combine them, and let d approach zero.
     
  5. Mar 4, 2017 #4
    No no no, you got me wrong. I derived that it is above the object for small angles near the perpendicular. I made an equation which tells me where the diverging light intersect with the perpendicular then took the limit as the angle gets smaller and smaller. (dx in this situation). The textbook didnt even talk about the topic of deriving it and I thought that it is just impossible to have one solution and the text book even said that if you look directly from above. you will see the actual depth because no refraction happens which is wrong. You either take the limit for small angles near the perpendicular or just consider the center of eye is on the perpendicular and you the eye has a width which is very small so you will still get apparent depth just the same as small angles.
    That is why I asked you whether I should follow my reasoning.

    And I took your advice and now I will try to derive an equation to give me the location of the image if it is not above the actual object. I checked the paper and saw that he was going to do that so I stopped reading and thought I should compare the results later when I finish mine. Thank you so much for the link


    Yes I actually did that, However my approach was a bit long, I used a circle equation and got the derivatives.. But it became a mess. I will try to do it once more and see what I get at.
    i searched on google for like a week before asking here, Never found those links you gave me. Thank you, appreciate it.
     
  6. Mar 4, 2017 #5
    This website discusses the Lensmaker's Equation: https://www.boundless.com/physics/t...lenses-170/the-lensmaker-s-equation-615-4333/

    NaJv-B4CrcVhQ-e8TwIiMXllFwCzfFH2iBBldc2TAdOxP9nGc8r0Y2-zSMQr1Ei3Yt88Y2FDoLszf4jhNfplacA=w632-h61.jpg
    where d is the thickness of the lens as measured on the optical axis. Under the assumption that the lens is thin, the last term can be ignored. Now knowing the focal length and the location of the lens, rays can be traced parallel to the lens and through the center in order to find the image. Under the assumption that the thickness of the lens is zero, the ray is considered in that approximation to enter and exit the lens at the same place.
     
    Last edited: Mar 4, 2017
  7. Mar 4, 2017 #6
    I didn't follow all the detail in your question, but you are correct that the eye cannot be exactly at the perpendicular as it has a finite aperture. The non-perpendicular rays from the object refract at the surface of the water. When followed back they intercept the perpendicular ray at a point below the surface. That is where the image appears to be located and it is closer to the surface than the object.
     
  8. Mar 4, 2017 #7

    jtbell

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    Most people have two eyes, right? That's where your depth perception comes from. Only one of them (at most) can be at the perpendicular.
     
    Last edited: Mar 4, 2017
  9. Mar 4, 2017 #8

    Drakkith

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    While true, I don't think pixel is referring to depth perception.

    Traced backwards, both the normal and non-normal rays intersect at the object. The image of the object is where the rays appear to come from, which will not be located at the object. Maybe you meant this, but I wanted to clarify.


    Let's be clear here. There are two things to account for: The non-zero size of your eye's pupil, and the non-zero size of the object. If your pupil were of zero diameter (or near enough so that only a single ray could get through, which I think is a decent approximation for this discussion) the image of a non-zero size object would still have nearly the exact same dimensions on your retina as it currently does and you would see apparent depth. The rays from the outer areas of the object not on your optical axis would refract at the surface of the water, altering the apparent depth of the object.

    A zero-size object (or again, near enough to zero-size) would appear as a point-source and your eye would not be able to see its depth at all since the image would be essentially the same regardless of the depth of the object. Granted, this is just talking about your eye being placed direction above the object. You could see apparent depth because of binocular vision or if you just move your head around and look from different angle.

    Well, I think so at least. I haven't done any calculations so take it with a grain of salt.

    If in doubt, ask your instructor!
     
  10. Mar 4, 2017 #9

    jtbell

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    On further reflection, I agree. It's been several years since I covered this in class, so it took a while to remember it.
    This is actually not as simple as it appears! What happens is best seen by drawing a careful ray diagram. Place a point object below the flat horizontal surface of the water. Draw a fan of rays from the object to the surface, at various angles. Where each ray hits the surface, measure the angle and use Snell's law to calculate the angle of the refracted ray. Draw the refracted rays above the surface. Extrapolate them back into the water as dashed lines. You'll find that no two of them intersect at the same point! That is, there is no single "point image" like we get in idealized lens and mirror ray diagrams.

    Nevertheless, if we take only rays that are very close to the vertical, they come close to intersecting at a single point. If we take another group of rays, close together, but clustered around (e.g.) 45° above the water, they also come close to intersecting at a single point, a different point at a different depth from the nearly-vertical rays.

    When you look into the water, the rays entering your eye pupil span a narrow range of angles, so they form a slightly fuzzy image of the point, and its apparent location and depth depend on the viewing angle.
     
  11. Mar 4, 2017 #10
    Yes, this is what I meant by saying followed back, not traced back. Good to point out the careful choice of wording needed in this context.
     
  12. Mar 4, 2017 #11

    Drakkith

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    That's interesting. I don't have the time right now to look into this, but I'll try to remember later.
     
  13. Mar 4, 2017 #12
    Yes, That is why I made an equation dx dependent where dx is the distance from the perpendicular line, and yi is where it intersects with the perpendicular.

    I Painfully did the first part using derivatives (:cry:). I am not sure how you want me to do the rest. I first thought of using the same way but however first I have to find where the refracted light from the first surface will hit the second surface which means I would need a lot of concentration and A4 pages.
     
  14. Mar 5, 2017 #13

    Drakkith

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    I played around in Matlab and managed to make this graph of the intersection of the virtual ray (the part of the ray traced backwards after it exits the surface of the water) with the vertical ray (Y-axis). I don't know if it's an issue with Matlab that I don't understand or something else, but there's a sharp discontinuity between the initial zero-degree ray and the very next ray. All rays initially started from a point located on the Y-axis -10 units below the surface of the water.

    Here's the attached graph.
    Ray Intersection vs Underwater Ray Angle.jpg
     
    Last edited: Mar 5, 2017
  15. Mar 5, 2017 #14
    I think it is an issue with matlab, I changed the equation above to angles and made a graph. Weird issue though.
     

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  16. Mar 5, 2017 #15

    Drakkith

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    Did you make this in Matlab? If so, would you mind posting your code so I can see where I might have gone wrong?
     
  17. Mar 5, 2017 #16
    Do you mean "traced" backward? :smile:
     
  18. Mar 5, 2017 #17

    Drakkith

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    Traced "backward"?
     
  19. Mar 5, 2017 #18
    No sorry, I didn't use Matlab. But I think the problem is when you asked the program to trace backward the perpendicular it self and somehow it chose the value -10. However the function there is discontinuous and I don't think matlab interprets discontinuous functions, so It just makes vertical lines. I have googled it and it seems this what it does
     
  20. Mar 5, 2017 #19

    jtbell

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    OK, so what you guys have are the vertical positions at which the refracted rays intersect the vertical axis (on which underwater object lies), when extended "backwards" into the water. I did the calculation in MS Excel and got a graph similar to Drakkith's, but with discrete points instead of a continuous curve. I maintain that these positions should not be considered to be apparent image locations. For that, you need to consider a narrow bundle of refracted rays, and see where they (approximately) intersect.

    Try calculating the refracted rays corresponding to e.g. the underwater rays at 10° and 11°, and find where they intersect each other. Repeat for various other pairs of 1°-separated underwater rays. If the angles are small, the intersection points should be close to the vertical axis, and close to the vertical positions you've already calculated. For larger angles, they will be away from the axis, and at significantly different vertical positions.
     
  21. Mar 6, 2017 #20
    Deja vue all over again, to quote Yogi Berra. I posted the following in this previous thread: https://www.physicsforums.com/threads/apparent-depth-conceptual-question.866441/#post-5439807

    Thinking about this some more - what I wrote was true if you just used the vertical ray and a non-vertical (oblique) ray from the point on the fish, using a giant eye. If we just consider the oblique ray, as you increase the angle from the vertical the refracted ray, when followed back into the pool, will be higher and higher above the fish. So the answer to your question is no - they rays would not trace back to a single point above the fish when a large spread of rays from the fish are considered, which is what BvU may have been alluding to.

    Realistically, as you look into the pool, rays with a single oblique angle and a small spread about that angle are entering your eye. You follow them back into the pool to locate the image of the point on the fish. As you move further away i.e larger ray angles to the vertical, the image moves up toward the surface of the pool. I've been playing with a spreadsheet that I created to calculate the location of the image and it does appear that it shifts to the right , as in your third diagram, as you move more to the right. Maybe someone else can confirm this.
     
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