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Points reflected by a parabolic mirror create images that appear to be at specific positions on the other side of the mirror. I am attempting to use geometry to find the map from each point in space to the apparent position of their image.

Given the parabolic mirror $y(x) = x^2/(4f)$, my calculations [*] show that a point $\langle x, y\rangle$ on one side of the mirror will be reflected to an image point apparently at

$$\hat x = 2f \left[\frac{y-f}{x} + \sqrt{\left(\frac{y-f}{x}\right)^2 + 1}\right]$$

$$\hat y = f + \left(\frac{x^2/4f - f}{x}\right) \hat{x}$$

For example, the reflection of a grid takes the following apparent shape: enter image description here

I am attempting to figure out whether the bowed curves created by the horizontal lines form a particular named shape. To do so, I am trying to reparametrize the curve in terms of some $\hat{x}(t)$ and $\hat{y}(t)$, but I'm unsure of how to proceed. For example, I previously suspected that they might be confocal parabolas (though probably not) and would like to rearrange terms so as to show what they are.


[*] My calculations:

I found this formula using the property that a ray of light pointed at the focus will be reflected parallel to the axis of the parabola, and vice versa. Given the point $\langle x,y\rangle$, we drop one vertical (paraxial) line from $\langle x,y\rangle$, which is reflected at the point of intersection with the parabola to a line $\ell_1$ pointing toward the focus. We drop a second line from $\langle x,y\rangle$ itself toward the focus, which is reflected at the point of intersection with the parabola into a vertical (paraxial) line $\ell_2$. The intersection of $\ell_1$ and $\ell_2$ yields the apparent position of the image point $\langle \hat{x}, \hat{y}\rangle$. This method breaks down when $\langle x, y\rangle$ is situated on the axis of the parabola, because then both $\ell_1$ and $\ell_2$ are coincident vertical lines — but I expect this to be a manageable point discontinuity.

The line $\ell_1$ lies between $\langle 0, f\rangle$ and $\langle x, x^2/4f\rangle$. The line $\ell_2$ is a vertical line positioned at wherever the ray from $\langle 0, f\rangle$ toward $\langle x, y\rangle$ intersects the parabola (solvable by quadratic equation).


Note that as a sanity check, the map fixes points on the parabola: if $y = x^2/(4f)$, then $x^2 = 4fy$ and first equation becomes:

\begin{align*} \hat{x} &= \frac{2f}{x}\left[(y-f) + \sqrt{(y-f)^2 + x^2}\right]\\ &= \frac{2f}{x}\left[(y-f) + \sqrt{(y-f)^2 + 4fy}\right]\\ &= \frac{2f}{x}\left[(y-f) + \sqrt{(y+f)^2}\right]\\ &= \frac{2f}{x}\left[2y\right]\\ &= \frac{4fy}{x}\\ &= \frac{x^2}{x}\\ &= x. \end{align*}

Similarly, \begin{align*} \hat y &= f + \left(\frac{x^2/4f - f}{x}\right) \hat{x}\\ &= f + \left(\frac{x^2/4f - f}{x}\right) x\\ &= f + x^2/4f - f\\ &= x^2/4f\\ &= y \end{align*}


New insight — there's a convenient change of coordinates we can make. Instead of rectilinear coordinates $\langle x,y\rangle$, we can use $\langle x, \alpha\rangle$, where $\alpha$ is the angle formed between the point, the focus, and the x-axis. Using one rectilinear coordinate and one focal angle makes sense, because parabolas turn vertical lines into lines angled through the focus and vice-versa, with slope dependent on horizontal position.

We can specify any point in 2D by giving its coordinates $\langle x, \alpha\rangle$. Then the coordinates of the reflected map are simply:

$$\begin{align*}\widehat{x} &= 2f\left[\frac{1+\sin{\alpha}}{\cos{\alpha}}\right]\\\widehat{\alpha}& = \arctan \frac{\frac{1}{4}x^2-f}{x}\end{align*}$$

Wonderfully, $\widehat{x}$ depends only on $\alpha$, and $\widehat{\alpha}$ depends only on $x$.

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    $\begingroup$ Is your reflection formula correct? Presumably, points on the parabola reflect to themselves. In particular, $(0,0)$ should remain fixed; however, substituting $x=0$ into your formulas is ... problematic. It might help if you showed how you derived your formulas. (Also, to be clear: when you write "$x^2/4f$", you mean "$\frac{x^2}{4f}$". Correct?) $\endgroup$ – Blue Oct 13 '18 at 9:26
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    $\begingroup$ Have a look at this sc.ehu.es/sbweb/fisica3/ondas/parabolico/parabolico.html $\endgroup$ – Cesareo Oct 13 '18 at 11:01
  • $\begingroup$ @Blue. I've added a note about my derivation in case the formula itself is wrong. Yes, $x^2/4f$ means $x^2/(4f)$. $\endgroup$ – user326210 Oct 13 '18 at 15:09
  • $\begingroup$ @Cesareo. Thanks for the reference. My formulas use the optical property of parabolic mirrors as mentioned on that site, but I'm unsure how to take the next step to find the transformation map as mentioned in my question. $\endgroup$ – user326210 Oct 13 '18 at 15:11
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Here's a derivation based using a left-opening parabola with the focus at the origin, with $f$ the distance from vertex to focus. That is, we take the parabola to have Cartesian equation

$$4 f ( x - f ) + y^2 = 0 \tag{1}$$

In polar form: $$r = \frac{2f}{1+\cos\theta} \tag{2}$$


enter image description here

Now, let $P = (p,q) = r (\cos\theta,\sin\theta)$ be our reflecting point. The horizontal line through $P$ meets the parabola at $A$; line $\overleftrightarrow{PF}$ meets the parabola at $B$. Then, the horizontal line through $B$ meets the $\overleftrightarrow{AF}$ at the reflected point, $P^\prime$. The reader can verify these calculations:

$$A =\left(\frac{4f^2-q^2}{4f},q\right)=\left(\frac{4 f^2 - r^2 \sin^2\theta}{4 f}, r \sin\theta\right) \qquad B = \frac{2f}{1+\cos\theta}\left(\cos\theta,\sin\theta\right) \tag{3}$$ so that $$P^\prime = \left(\frac{4 f^2 - r^2 \sin^2\theta}{2r (1 + \cos\theta)}, \frac{2 f \sin\theta}{1 + \cos\theta}\right) = B - \frac{\left(\;r(1+ \cos\theta)-2f\;\right) \left(\;r(1-\cos\theta)+2f\;\right)}{2 r(1 + \cos\theta)}\;(1, 0) \tag{4}$$

Since points on the parabola satisfy $(2)$, we see immediately that, for such points, $P^\prime$ reduces to $B$, which coincides with $P$. $\square$

Note that the Cartesian form of $P^\prime$ is $$P^\prime = \frac{1}{2\left(p+\sqrt{p^2+q^2}\right)}\;\left(\;4 f^2 - q^2, 4fq\;\right) = -\frac{p-\sqrt{p^2+q^2}}{2q^2}\;\left(\;4 f^2 - q^2, 4fq\;\right) \tag{5}$$ This can be put into agreement with OP's formulas by appropriate coordinate transformations. (Move the origin to the vertex via $p\to p+f$; then rotate $90^\circ$ via $p \to -q$ and $q\to p$.)


As for the reflections of lines perpendicular to the parabola's axis ... Such a line has polar equation $r = k\sec\theta$ for some $k$, so that the corresponding reflected curve comes from eliminating $\theta$ from the system $$(x,y) \;=\; \left(\;\frac{4 f^2 \cos^2\theta - k^2\sin^2\theta}{2 \cos\theta (1 + \cos\theta)}, \frac{2 f \sin\theta}{1 + \cos\theta}\;\right) \tag{6}$$ The identity $\sin^2\theta+\cos^2\theta = 1$ lets us write our system in terms of $\cos\theta$. Eliminating that (with the help of Mathematica's Resultant[] command) yields $$\left(2kx + y^2-4f^2\right)^2 = 4 k^2 \left(x^2+y^2\right) \tag{7}$$

Here are the associated curves for $k=f$, $2f$, and $3f$.

enter image description here

Whether such a curve has a name, I do not know. It is definitely not a parabola.

Note that each curve is doubly-asymptotic with horizontal lines through the ends of the parabola's latus rectum (since the $y$ coordinate of $(6)$ approaches $2f$ as $\theta$ approaches $\pi/2$). Also, when $y=0$, equation $(7)$ reduces to $x = f^2/k$; as $k$ grows without bound, the "vertex" of the curve approaches the origin, i.e., the parabola's focus.

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  • $\begingroup$ Thanks for your thorough solution --- I look forward to going through it in detail. Does the final figure include physically impossible solutions? It looks like vertical lines are on the same side of the mirror as some branches of their reflections. Maybe when you take the line through the focus, you include both intersection points with the parabola instead of just one? Or maybe I'm misreading the figure, or you're including all solutions for completeness. $\endgroup$ – user326210 Oct 13 '18 at 20:46
  • $\begingroup$ @user326210: Yes, I plotted the entirety of the reflections curves $(7)$ for completeness. The full curves include branches that don't correspond to actual reflections. The branches you care about are the ones "inside" the parabola. $\endgroup$ – Blue Oct 14 '18 at 4:07
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Proceeding according to your calculations [*] I got the transformation formulas a bit different with yours.

I got for the reflected point $(x_r,y_r)$ and the real point $(x,y)$

$$ x_r(x,y,f) = -\frac{2 f \left(\sqrt{(f-y)^2+x^2}+f-y\right)}{x}\\ y_r(x,y,f) = \frac{\left(\sqrt{(f-y)^2+x^2}+f-y\right) \left(f \left(\sqrt{(f-y)^2+x^2}+f+y\right)-\frac{x^2}{2}\right)}{x^2} $$

Follows a plot for $f = 0.5$. The parabola is in red. The parabola focus is a black dot, In green is the real grid and in light blue is the reflected grid. It is shown also a green point (real) and a light blue point (reflected)

I hope this helps.

enter image description here

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Actually, according to the sketch you made, thus with a negative $f$, the equation of the curves are: $$ \left\{ \matrix{ x = 2f\left( {{{v - f} \over u} \mp \sqrt {\left( {{{v - f} \over u}} \right)^{\,2} + 1} } \right) \hfill \cr y = f + {1 \over u}\left( {{{u^{\,2} } \over {4f}} - f} \right)x \hfill \cr} \right. $$ and for positive $u$ we shall take the $-$ sign and v.v..
It is then understood that for $u=0$ we shall take the limit of the above expressions.

Let's rewrite them putting $f =- h$, so not to get confusion with the signs $$ \left\{ \matrix{ h = - f \hfill \cr x = 2h\left( {\sqrt {\left( {{{v + h} \over u}} \right)^{\,2} + 1} - {{v + h} \over u}} \right) \hfill \cr y = - h + {1 \over u}\left( {h - {{u^{\,2} } \over {4h}}} \right)x \hfill \cr} \right.\quad \left| {\;0 < x,u,h} \right. $$

Then you already have the parametric equations:

  • at constant $u$, you have the line $$ y = - h + {1 \over u}\left( {h - {{u^{\,2} } \over {4h}}} \right)x $$ passing through the focus;

  • at constant $v$, you get $$ \eqalign{ & x = 2h\left( {\sqrt {\left( {{{v + h} \over u}} \right)^{\,2} + 1} - {{v + h} \over u}} \right) = \cr & = {h \over {h + v}}u - {h \over {4\left( {h + v} \right)^{\,3} }}u^{\,3} + O\left( {u^{\,5} } \right) \cr & y = - h + {1 \over u}\left( {h - {{u^{\,2} } \over {4h}}} \right)x = \cr & = - {h \over {h + v}} - {{h^{\,2} + h\left( {h + v} \right)^{\,2} } \over {4\left( {h + v} \right)^{\,3} }}u^{\,2} + O\left( {u^{\,4} } \right) \cr} $$ and thus the curves are "simil-parabolas".

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