Results 1 to 2 of 2

Thread: Mirrors

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    81

    Mirrors

    Hey guys,

    I got a question here:

    Facing a slightly steamed-over mirror, hold one eye shut and trace the outline of your face in the mirror. Explain why the outline is exactly 1/2 the width and 1/2 the height of your face.

    I have no idea how to approach this question and anyyy help would be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello GreenDay14
    Quote Originally Posted by GreenDay14 View Post
    Hey guys,

    I got a question here:

    Facing a slightly steamed-over mirror, hold one eye shut and trace the outline of your face in the mirror. Explain why the outline is exactly 1/2 the width and 1/2 the height of your face.

    I have no idea how to approach this question and anyyy help would be greatly appreciated.
    Study the attached diagram.

    The line $\displaystyle AB$ represents the width of your face, with $\displaystyle E$ as your right eye. $\displaystyle A'B'$ is the reflection of your face in the mirror, which lies along the line $\displaystyle MPQ$.

    Then it is well known that the distances of the object and its reflection from the mirror are equal. So
    $\displaystyle AM = MA'$

    $\displaystyle \Rightarrow AM = \tfrac12AA'$
    A ray of light from $\displaystyle A$ hits the mirror at $\displaystyle P$ and is reflected into your eye, $\displaystyle E$, producing an image of $\displaystyle A$ at $\displaystyle A'$. So $\displaystyle EPA'$ is a straight line, with $\displaystyle P$ the apparent point on the mirror of the position of $\displaystyle A'$.

    Similarly for $\displaystyle Q$: this is the apparent point on the mirror of the position of $\displaystyle B'$.

    By similar triangles, since $\displaystyle AM = \tfrac12AA'$, we can easily prove that $\displaystyle PQ = \tfrac12A'B' = \tfrac12AB$.

    Grandad
    Attached Thumbnails Attached Thumbnails Mirrors-untitled.jpg  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mirrors
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: May 12th 2009, 09:48 PM
  2. mirrors
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: May 7th 2009, 04:08 AM
  3. Mirrors question
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Apr 30th 2008, 11:15 AM
  4. Circular mirrors are expensive
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: May 27th 2007, 09:32 AM

Search Tags


/mathhelpforum @mathhelpforum