Intensity of parabolic reflector as a function of the radius.

Question

I'm trying to make sense of the intensity of light leaving a parabolic reflector, given that the light is a point source at the focus. (In the diagram below, this might be the intensity of light at $Q_1$ vs $Q_3$.)

For ease of discussion, say that the reflector is described by $z = \frac 14(x^2 + y^2) - 1$ with the focus at the origin.

My thinking is to instead try the two-dimensional version $y = \frac 14 x^2 - 1$. Under the assumption that the light source is equally intense in all directions, I suspect I'd start by determining where the ray intersects the parabola as a function of angle.

How do I determine the intensity of the light as a function of $r$ (in the 3D case) or $x$ (in the 2D case)? (For what it's worth, I'm only interested in the intensity of the part of the light that has been reflected.)

Answer 1

Let $y=ax^2$ be the equation of the parabola. Angle $\alpha=\angle VFP$ is twice the angle $\theta$ formed by the tangent at $P=(x,y)$ with $x$-axis (see figure below): $$ \tan\theta={dy\over dx}=2ax. $$ Hence: $$ {d\alpha\over dx}=2{d\theta\over dx}={4a\over1+4a^2x^2}. $$ Let's now switch to 3D, with paraboloid $z=ar^2$. The fraction of total power $P$ emitted by the source in the solid angle $d\Omega$ comprised between $\alpha$ and $\alpha+d\alpha$ is $$ dP=P{d\Omega\over4\pi}={P\over4\pi}2\pi\sin\alpha\, d\alpha. $$ This power, after reflection, is spread on the surface $dA=2\pi r\,dr$ of the annulus comprised between $r$ and $r+dr$. Hence the flux of power (a.k.a. intensity of light) is: $$ {dP\over dA}={1/2 P\sin\alpha\,d\alpha\over2\pi r\,dr}= {P\over4\pi}{\sin\alpha\over r}{d\alpha\over dr}. $$ But $d\alpha/dr$ has been computed above as $4a/(1+4a^2r^2)$ and $$ \sin\alpha=\sin2\theta={2\tan\theta\over1+\tan^2\theta}={4ar\over1+4a^2r^2}. $$ Inserting these into the preceding equation finally leads to $$ {dP\over dA}={P\over4\pi}\left({4a\over1+4a^2r^2}\right)^2. $$

Answer 2

I am indebted to @Intelligenci Pauca for valuable remarks.

Due to the rotational symmetry of this issue, it is convenient to consider a 2D treatment in an axial section plane.

Let us take the focus of the parabola as the origin, allowing to represent it with the following polar equation (see Remark 1 below):

$$r=\dfrac{1}{1-\sin \theta}\tag{1}$$

[equivalent cartesian equation $y=\tfrac12(x^2-1)$].

Polar coordinates will prove effective in the depiction of the correspondence between an angular sector with aperture $\Delta \theta$ and the length $\Delta x$ covered by the reflected ray on an horizontal line.

Let us work at the infinitesimal scale, i.e., look for the connection between $d \theta$ and $dx$, knowing that the intermediate arc length $ds$ has to be considered:

Fig. 1: Angle $\alpha$ made by $ds$ with the horizontal axis is in fact equal (by orthogonality) to the angle of incidence (and therefore to the angle of reflection), both equal to $\tfrac12\theta+\tfrac{\pi}{4}$ by angle chasing. The horizontal "screen" where we can see the "spot" $dx$ reflected by the ray with angular width d$\theta$ is positionned arbitrarily ; it should be thought as being placed at infinity.

Differentiating (1), one gets:

$$dr=\dfrac{\cos \theta}{(1-\sin(\theta))^2}d\theta\tag{2}$$

Besides, the infinitesimal arc length $ds$ hit by the ray has (classical) square length given by (formula (6) here):

$$ds^2=dr^2+r^2 (d \theta)^2=\left(\dfrac{(\cos \theta)^2}{(1-\sin\theta)^4}+\dfrac{(1-\sin\theta)^2}{(1-\sin \theta)^4}\right)(d\theta)^2$$

Therefore:

$$ds=\dfrac{\sqrt{2-2\sin \theta}}{(1-\sin \theta)^2}d\theta\tag{2}$$

Besides, $dx$ being the projection of $ds$ onto the horizontal "screen", we have

$$ds=\dfrac{1}{\cos(\tfrac12\theta+\tfrac{\pi}{4})} dx \tag{3}$$

Quotienting (2) and (3), we obtain:

$$\dfrac{d\theta}{dx}=\dfrac{(1-\sin \theta)^2}{\cos \theta \sqrt{2-2\sin \theta}} \tag{4}$$

Now, what are we looking for ? For a density $y=f(x)$, i.e., we want this function such that

$$\int_0^x f(\xi)d \xi=\theta \tag{5}$$

Differentiating (5) wrt $x$ gives $f(x)=\dfrac{d\theta}{dx}$ i.e., expression (4).

Therefore a set of parametric equations for the looked for density is:

$$\left(x= r \cos \theta=\dfrac{\cos \theta}{1-\sin \theta}, \ \ \ \ y=\dfrac{(1-\sin \theta)^2}{\cos(\tfrac12\theta+\tfrac{\pi}{4})\sqrt{2-2\sin \theta}}\right)\tag{6}$$

yielding the very simple cartesian equation (see Remark 2).

$$y=\dfrac{2}{1+x^2}\tag{7}$$

Surprizingly, it is the same relationship as the Lorentzian/Lambertian equation (classical direct illumination curve of a plane by a light above it).

Fig. 2: The illumination curve with equation (7) . The area under the curve is $4 \pi$.

Remarks:

1. (1) is a particular case of the general polar equation of a conic curve with the origin at its focus (or one of its foci for an ellipse of hyperbola):

$$r=\dfrac{p}{1+e \cos(\theta-\theta_0)}.$$

Here $p=1$, eccentricity $e=1$ and $\theta_0=-\pi/2$.

2. The rather tedious computations from (6) to (7) are eased by the fact that one can express $\sin \theta=\dfrac{x^2-1}{x^2+1}$ giving $\cos \theta=\dfrac{2x}{x^2-1}$. Remark: we recognize in these formulas that setting $\beta=\tfrac{\pi}{2} - \theta$, $x=\tan(\tfrac{\beta}{2}$...