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Derivation for magnification equation for convex mirrors

  1. Jun 8, 2012 #1
    Hello everyone, I really run into a problem here.

    The magnification equation for mirrors describes such a relation: M=-distance of image/distance of object = height of image/height of object. (M=(-i/o)=h'/h).

    I understand how this formula can be proved using a ray diagram for concave mirrors simply by proving similar triangles between the image distance and the object distance. However, it didn't quite work out for the convex mirrors as I worked for a few days trying to identify a similar triangle using the two. All I can get from the diagram is that h'/h= distance from object to the centre of mirror/distance from image to the centre of mirror. Interestingly this gives the same result as i compute h'/h, indicating that this is the correct magnification equation.

    I would like to ask the physics experts here to kindly give me some suggestions on how to mathematically solve this derivation here; greatly appreciated, thanks!
     
  2. jcsd
  3. Jun 9, 2012 #2

    Yes, but where are the similar triangles, and how do you proof them?
     
  4. Jun 9, 2012 #3

    Simon Bridge

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    The important part of this sort fo thing is finding the triangles an important skill.

    Lets label the diagram above:
    O M I F and C are the points on the horizontal axis for the object, mirror, image, focus and center or curvature. So |MF| = f right?

    A and B are the top of the object and the top of the image.
    D E and G are the three places the rays touch the vertical (mirror) axis.

    Using these - triangle AOF is similar to GMF ... now do you see them? I can see three similar triangles involving point C.

    Triangle AOF has height h and base o+f.

    To use them, write the formulas you get from considering that the ratios of corresponding sides are equal ... this will get you a set of simultanious equations: eliminate the unwanted variables and rearrange.
     
  5. Jun 9, 2012 #4
    Yes, I drew a diagram just like you described, and I did find a similar triangle of AOF and GMF, but pardon my slowness in math; I failed to see how the rest of the derivation goes. So can you please elaborate on the rest of the similar triangles and how the final result is derived?
     
  6. Jun 9, 2012 #5

    Simon Bridge

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    Well those two give you:[tex]\frac{h'}{h} = \frac{f}{f+o}[/tex]... or something like that right?

    That looks close - but you need to get rid of that pesky f, and you are missing an i ... can you see any other similar triangle sets you can use? Maybe one that lets you express f in terms of i?
     
  7. Jun 9, 2012 #6
    I appreciate your patience, but sorry, I can't see the other similar triangle that can help.
    On the other hand, the relation ΔAOF≈ΔGMF only gives you AO/GM=MF/OF which is h/GM=(f+o)/f in this case i think we cannot assume GM=h' because if you draw a bigger diagram you can see that GM is slightly bigger than h', the tangenline drawn at M and the incident ray that goes to F, and its reflected parallel ray do not all intercept at point G.
     
  8. Jun 9, 2012 #7
    sorry for the unclearness of the picture, my cellphone camera is very crappy.
     

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  9. Jun 10, 2012 #8

    Simon Bridge

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    That's not an assumption - it has to be the case by geometry. You just made a mistake in your drawing: you have drawn all your rays to the curved line representing the mirror surface - then bending the rays at that surface. This is not how you draw ray diagrams. (Also your horizontal lines are not very parallel... but that's not so important.)

    In all the optics you are learning you are using the par-axial approximation ... the mirror surface in that approximation is so small the entire curve would fit inside the thickness of the vertical line.

    You should always draw your rays so they bend at the vertical axis - that's why you draw it. Look again at the one I showed you - the rays seem to bounce off the vertical and not the "mirror". This is why. The big curve people like to draw on these diagrams is indicative only and not meant to be part of the analysis. It is actually better to leave the curve off the diagram completely... they'll make more sense.

    |GM| has to be h' since it is (by definition) the intersection with the vertical axis with a horizontal line drawn through B and |BI|=h'.
     
  10. Jun 11, 2012 #9
    I see, now I can derive the equation. Thank you so much for your help!
     
  11. Jun 12, 2012 #10

    Simon Bridge

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    No worries - you should find these diagrams easier to understand and use now.
     
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