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Silvered Lens and Optics

  1. Feb 12, 2005 #1
    Hello

    I've been reading optics (both the cartesian and real/virtual sign convention) and I have some queries:

    1. I was trying to work out problems involving silvered lenses (spherical lens with one side silvered) and first I tried to reason to myself how the imaging would take place. This what I believe:

    for a real object in front of a silvered lens encountering first its nonsilvered side, a ray of light leaving the object would first undergo refraction from the nonsilvered side, reflection from the silvered side and then refraction from the first refracting surface again. This amounts to two refractions and one reflection.

    However, the solved problems book I use reasons this way: a silvered lens can be considered as a normal spherical lens (nonsilvered that is) plus a spherical mirror (the curvature of which is identical to the silvered side of the given lens) at zero distance from each other.

    I personally feel that my reasoning is correct because taking first a lens instead of one refracting surface means that the ray will encounter refraction at the second surface, followed by reflection from the mirror which should not happen as the distance between the virtual boundary of the refracting medium and the spherical surface which has been silvered is zero so the ray immediately suffers reflection on reaching the geometric boundary of the surface. Who is correct?

    To find out, I compared my equations (3 in number) with the three equations given as part of the solution and the equations are certainly different. So who really is correct?

    2. As I am frequently required to change my point of view :wink: to accomodate rays incident from the left and from the right and also several lenses, prisms and mirrors combined and am required to find image parameters, I opted to use the sign convention outlined in Resnick, Halliday and Crane (virtual side/real side sign convention) as it is much more convenient than the cartesian convention which becomes very tedious and unwieldy. Yet, I would be grateful to receive some general opinions about the choice of sign convention for complex problems.

    Thanks and cheers
    Vivek
     
  2. jcsd
  3. Feb 13, 2005 #2

    Andrew Mason

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    As far as the first problem is concerned, it seems to me that the text is correct. Why would there be any refraction at the second surface? Refraction results from the light passing from one medium to another having different indices of refraction. Since the second surface is silvered, the light never passes from the lens to another medium (ie air).

    What is the basis for your contention that there is a second refraction before the reflection?

    AM
     
  4. Feb 13, 2005 #3
    First off, thanks for your reply Andrew. It is the book (not me) which says that there is a second refraction before the reflection (pray read my post again). That is bothering me because in case of a degenerate lens--a slab--silvered at one end, there is no such problem as we treat the ray refracted at the first face to reflect at the second face which is in direct contact with a mirror. The "controversy" if you will, arises in the case of a spherical silvered lens.

    I would be grateful if you could suggest an authentic site/resource/book which I could quote to my instructors to convince them.

    Also any ideas for the second question? While surfing the internet, I came across several websites discussing sign convention but everyone seems to have assumed that rays will be incident from the left so there is no general sign convention (minus cartesian which I already know) which is sophisticated enough to handle all cases cleanly (except the Crane convention--R side and V side).

    Thanks and cheers,

    Vivek
     
  5. Feb 13, 2005 #4
    I think both of you (you and the book) are correct. If the equations are different, that doesn't mean that the solution will be different. The system of equations
    x+y=2, x-y=4 and the system x=3, y=-1 have the same solutions but they look totally different. So see if your answer is matching with the book's answer.
     
  6. Feb 13, 2005 #5

    Andrew Mason

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    Well in that case, I find your question really confusing. Perhaps you can clarify what the text is actually saying and what you think it is saying.

    I don't see why you think that a lens followed by a spherical mirror at zero distance will have refraction at the second surface.

    AM
     
  7. Feb 13, 2005 #6
    Mate, I'm sorry to say this, but that is a very poor analogy and I think you should understand the distinction between the two ideas. Surely the final answer is not as important as the way you approach it. Two general linear systems may indeed have the same solution but that does not mean their component equations represent the same planes in three-dimensional space!! :eek:
     
  8. Feb 13, 2005 #7
    Okay here's the clarification (apologies for any confusion caused):

    First, there's an ideal spherical lens. It forms a real or a virtual image of a real object, the location of which can be computed using the lens equation. Now, if the surface opposite to the side of incidence (of light) is silvered completely to make it a reflecting surface, the problem asks to find the characteristics of the final image formed and given the refractive index of the medium inside the lens, we have to find the focal length of this lens.

    Approach #1 (What the text follows): To simplify matters break the silvered lens into two optical elements:

    (a) a nonsilvered spherical lens
    (b) a spherical mirror

    Assume that light is incident from the left on the first curved surface which is nonsilvered.

    Write three equations:
    1. Refraction from the lens. The image formed by the lens is the object for the spherical mirror.
    2. Reflection from the spherical mirror. The image formed by the mirror is the object for the spherical lens.
    3. Refraction from the spherical lens once again. The image formed by this lens is the so called "final image".

    Please note the use of the term "lens" in this case. We have NOT treated each surface of the lens separately. In this case therefore, there are 4 refractions and 1 reflection.

    The book gives a formula to solve this problem:

    [tex]\frac{1}{f_{eff}} = \frac{2}{f_{lens}} - \frac{1}{f_{mirror}}[/tex]

    As I do not have access to authentic answers or solutions to the questions, I cannot comment on the correctness of this method except that this book is not a textbook and can go wrong in places (its a collection of problems after all).

    Approach #2: (what I was trying to justify)...write three equations again, but this time,

    Equation 1: Refraction from the first spherical refracting surface. The image formed by this surface is the object for the spherical mirror.

    Equation 2: The second surface with which light interacts is a spherical reflecting surface (the silvered surface). The image formed by this mirror is the object for a second refraction from the first spherical refracting surface.

    Equation 3: Second refraction from the first refracting surface, defining the position of the "final image".

    Note the use of "spherical refracting surface" instead of lens. Here there are 2 refractions and 1 reflection, clearly different (conceptually) from the above approach.

    I hope this is abundantly clear. As you can see I have been justifying that there should be no refraction at the second surface right from the start.

    What do you think?

    Cheers
    vivek
     
    Last edited: Feb 13, 2005
  9. Feb 14, 2005 #8

    Andrew Mason

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    I don't follow you there. Why would there be 4 refractions? Where does this suggestion come from? (that there is something equivalent to a refraction at the silvered surface). The path of the light at the second surface and reflecting is optically identical to light continuing through a thicker non-silvered lens.

    AM
     
  10. Feb 14, 2005 #9
    Well if you have a Lens isolated from a spherical mirror then if you use a single equation for it you are assuming (even if you do not say it explicitly) that there are two refractions (a lens means two refractions and a surface means one).
     
  11. Feb 14, 2005 #10

    Andrew Mason

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    But that is only if you have a surface at which the index of refraction changes. If you have a spherical convex lens surface and a spherical concave lens of complementary shape, with zero distance between, you have no refraction even though you have two lens surfaces. The light just keeps going straight. And that is optically the same as silvering the lens surface.

    AM
     
  12. Feb 14, 2005 #11
    Precisely Andrew thats what I am trying to tell you (and you continue to think I'm on the other side!! :smile:) but I have no book or resource which has addressed this problem mathematically, to make me convince the book authors and others who are using the <wrong> idea.
     
  13. Feb 14, 2005 #12

    Andrew Mason

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    It is not that I think you agree with that. I realize that you don't. What I don't understand is why you think that is what the 'other side' is saying (ie. that there is a refraction at the silvered surface).

    A lens with two spherical surfaces one of which is silvered lens is equivalent to a spherical lens and a spherical mirror with no separation, but that does not imply a refraction at the silvered surface. There obviously isn't a refraction at the silvered surface but I don't think your teachers are suggesting that there is.


    AM
     
  14. Feb 15, 2005 #13
    Well first of all, I am teaching myself (preparing for some exams actually) so there are no real instructors. But the books I am consulting are not really textbooks and they have actually mentioned that there is a refraction at the silvered surface and since I have come across three different books which say this, I have been perturbed. Thanks for your help. Unfortunately (as I recently discovered) some exams have a weird marking pattern which allows some wrong methods to be rewarded. God save physics.
     
  15. Feb 19, 2005 #14
    Hi again...just wanted to say that I've figured this out using the piecewise treatment I had been wanting to do all these days. It works but it is different from the "formulae" given in books. I'll post the treatment on my webpage and put up the link here sometime in the future (when I have more time at hand) so that it is useful to everyone.
     
  16. Jan 5, 2008 #15
    silvered lenses

    hey!
    I havent understood silvered lenses and am having exactly the same problem as you posted. My book says there are 4 refractions.
    I would like someone to explain this.

    Thanks
     
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