1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Paraxial approximation on concave mirror

  1. Jan 18, 2010 #1
    1. Find the length of the curvature of a concave mirror of 20cm that comply with paraxial approximation for all incident rays


    2.conventional geometry formula , sinθ≈θ or tanθ≈θ for paraxial rays


    3. I had try drafting out the diagram , labeling all the unknown angle with symbol and trying to work out on them but I just can't find the angle CAB of the triangle because I don't know height of the triangle.

    This is a sketch:
    question.jpg

    Did I miss anything?
    Any suggestion on how to solve this problem?
     
  2. jcsd
  3. Jan 18, 2010 #2

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    I'm pretty familiar with basic optics, but have trouble understanding this question.
    What is meant by "length of the curvature"?
    Also, what is the condition for complying with the paraxial approximation?
     
  4. Jan 18, 2010 #3
    The incoming rays is parallel to the ray axis.
    According to the definition from the book : Rays that make small angle (such that sinθ≈θ ) with the mirror's axis are called paraxial rays.In the paraxial approximation only the paraxial rays are considered.
    The length of the curvature is the curve where reflected ray is comply with the paraxial approximation (the curve next to the height of the triangle I draw).The main problem is I don't know how to find the angle of CAB to ascertain the length of the curvature.
     
  5. Jan 19, 2010 #4

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Okay, but you (or somebody) has to decide: what is the maximum angle for which sinθ≈θ is true? Is it when sinθ and θ differ by 1%? Or 10%? Or something else? Does your professor or textbook provide the criterion? If not, the problem cannot be solved.

    Thanks for clarifying.
    The angle of the reflected ray is CAO, where O is at the center of the lens.
    As an approximation, you could assume the ray passes through the focal point. You'll have to use some trigonometry to figure out the angle.
     
  6. Jan 19, 2010 #5
    Well,there are only 2 things I know about this triangle which is AB = 10 cm and AC = 20 cm , all the rest are remain unknown.I try to solve it using trigonometry method such as A/sinA=B/sinB=C/sinC and theorem pythagoras but still I am stuck in this question.

    l is the length of the whole triangle, h is the height of the triangle and [tex]\theta[/tex] is the angle of CAO.

    [tex]\frac{20}{sin90}[/tex]=[tex]\frac{l}{sin(90-\theta)}[/tex]=[tex]\frac{h}{sin\theta}[/tex]
    since sin(90-[tex]\theta[/tex])= cos[tex]\theta[/tex]
    after rearranging the formula
    l=20cos[tex]\theta[/tex] and h=20sin[tex]\theta[/tex]

    CB2=h2+(l-10)2

    I am stuck here and don't know how to relate them in this step,any clue?
     
    Last edited: Jan 19, 2010
  7. Jan 20, 2010 #6

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    My apologies. The reflected ray is CB, not CA. And angle CBO is the reflected angle θ (not CAO as I said earlier). So let angle CBO be θmax, the largest angle for which sinθ≈θ is an acceptable approximation.

    You can proceed as you were doing, solving triangle ABC to get the length of side BC. You know the lengths of sides AB and AC, and also that angle ABC=180-θmax

    Once you have the length of BC, you can use sinθmax to find h.
     
  8. Jan 20, 2010 #7
    CB2=h2+(l-10)2

    =h2+l2-20l+100
    =(h2+l2)-20l+100
    since sin2θ+cos2θ=1
    =202-20l+100
    =500-20l

    For triangle ABC , there is another unknown angle(ACB) and I will label as [tex]\theta[/tex]ACB

    [tex]\frac{20}{sin(180-\theta)}[/tex]=[tex]\frac{CB}{sin\theta}[/tex]=[tex]\frac{10}{sin\theta_{ACB}}[/tex]

    so by substituting CB = 500-20l

    [tex]\frac{20}{500-20l}[/tex]=[tex]\frac{sin(180-\theta)}{sin\theta}[/tex]

    so am I doing it right?
     
  9. Jan 20, 2010 #8

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    While this is correct, I think sinθ=h/CB is probably more useful.
    Problem: on the left-hand term, you are using θ for <CBO. But in the middle term, you are saying θ is <CAB.
    I was using θ = <CBO, as that it is the angle-from-horizontal for the reflected ray CB. I.e., it is the angle for which the condition sinθ≈θ is to be applied.

    No, CB2 = 500 - 20l.
    Instead, I recommend using the law of cosines here, using the angle <ABC which is 180-θ.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Paraxial approximation on concave mirror
  1. Concave mirror (Replies: 1)

  2. Concave Mirrors (Replies: 4)

  3. Concave Mirrors (Replies: 9)

  4. Concave Mirror (Replies: 3)

Loading...