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Homework Help: Confusing ray optics problem!

  1. Jun 1, 2010 #1
    1. The problem statement, all variables and given/known data
    A young lady wants to see herself in a small mirror, the ear to ear width of her face is 6 in and eye to eye is 4 in. What should be the minimum width of a mirror so that she can see her face completely


    2. Relevant equationsThere a few equations which when directly applied give the required answer but i want this to be solved using pure ray optics....



    3. The attempt at a solutionI've applied simple ray optics such as angle of incidence being equal to angle of reflection,but what i don't understand is that don't we require the distance of face from the mirror .....please help
     
    Last edited: Jun 2, 2010
  2. jcsd
  3. Jun 1, 2010 #2
    One thing it does depend on is whether or not the young lady wants to be able to see herself with both eyes or just one. If both, it would need a wider mirror than if just one.
    Which is it? (She can see her face completely in both cases)
     
  4. Jun 2, 2010 #3
    I think it has to be with both eyes...although i'm not sure exactly!
     
  5. Jun 2, 2010 #4

    Doc Al

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    Staff: Mentor

    No. If she is a distance D from the mirror, how far away is her image? That's all you need. (Draw yourself a diagram.)
     
  6. Jun 2, 2010 #5
    BUT the main requisite of the question is that she should she her entire face in the mirror so the mirror has to be kept away from a certain distance from her face!! Please Explain!!
     
  7. Jun 2, 2010 #6

    Doc Al

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    That's your assumption. Instead, draw a diagram showing the location of her image in the mirror.
     
  8. Jun 2, 2010 #7
    Yes I got it but not through ddiagram but by seeing my face in the mirror!!
     
  9. Jun 3, 2010 #8
    I've tried doing it many times but still can't find the correct solution. Although I know the answer but I have wet not been able to solve this....please help!
     
  10. Jun 3, 2010 #9
    I have done a ray diagram for the case where only one eye needs to see the whole face.
    X and Y are the images of the ears.
    I have indicated how it works if both eyes need to see the whole face.
    It's not quite to scale to show the rays more clearly.

    mirror.png
     
  11. Jun 3, 2010 #10

    Doc Al

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    Unfortunately, Stonebridge just did it for you. But the way to do it for yourself is to first figure out where the image of someone would be in the mirror. (If you don't know, then your first step is to figure that out using a ray diagram.) Then you'd be able to see that the answer doesn't depend on the distance.
     
  12. Jun 3, 2010 #11
    I'm not sure this is so "unfortunate" as the poster was clearly unable to do this with the hints given so far. The poster wanted the case for both eyes, I think. I've done the case for just the one eye. I hope he will be able to adapt the diagram for the other scenario, and thus learn something.
     
  13. Jun 3, 2010 #12

    Doc Al

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    We have no idea what the poster was or was not capable of, since no attempt was shown. The only way to know what someone is thinking and where they are getting confused is to have them show what they've tried.
     
  14. Jun 3, 2010 #13
    The answer for one eye has been given 5in is actually incorrect for both the eyes.I feel so stupid now, a little bit of more effort on my part would certainly have solved the problem(the answer is 1in for both the eyes)any way thanks for the ray diagram it showed me where I went wrong. I think you all will agree that sometimes we brand a problem as beyond our scope and don't give in full effort even if it could be solved with a little bit of more of effort. Thank you I'll try from now not to label any question and have an open mind in solving it.
     
  15. Jun 4, 2010 #14

    ehild

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    Homework Helper

    Schumi,

    You can solve such problems if you know the basic laws. Those for reflectance from a plane mirror are:

    The reflected ray is in the plane defined by the normal of the mirror and the incident ray.
    The reflected ray makes the same angle with the nomal as the incident ray, but travels on the other side of the normal. (See the left figure)

    The image of a point is there where all the reflected rays cross.

    You see an image in the direction of the light rays reaching your eye.

    See the right figure that shows the image of a point-like light source. Light rays emerge in all direction and those which reach the mirror, will be reflected. The extension of the reflected rays cross each other at the other side of the mirror: You see the image at the point from where the reflected rays apparently emerge. The yellow triangle is an isoceles triangle: therefore the image is at the same distance from the mirror as the object.

    ehild
     

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    Last edited: Jun 4, 2010
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