Optics – Proofs for Off-Axis Focal Lengths of a Concave Mirror

Question

I am reading "Fundamentals of Optics" the 4th edition, by Francis Jenkins. The book mentions a conclusion for off-axis concave mirrors:

That is, depending on:

• mirror radius of curvature $r$,
• off-axis incident angle $\phi$,

the equivalent focal lengths of an off-axis mirror are:
$$
\begin{align}
f_T & = \frac{r}{2}\cos\phi, \tag{focal length at the tangential plane} \\
f_S & = \frac{r}{2}\frac{1}{\cos\phi}. \tag{focal length at the sagittal plane}
\end{align}
$$

I would like to verify this by hand. The book attaches the below figure. Deriving $f_T$ is easy, but I am struggling with $f_S$.

Can anyone provide complete proof for deriving the two focal lengths?

Answer 1

As mentioned in the comment of @Farcher, the proof by Monk is on the Page 424-425 of link. Monk's proof requires a set of law of sines and multiple steps for calculations.

Fortunately, as shown below, my proof is much concise compared to the original one by Monk.

Assumptions and conventions

1. We consider parallel rays hitting the concave mirror.
2. We consider the mirror's radius of curvature $r$ is much larger than the mirror's diameter, and use the small angle approximation: $\tan\theta \approx \theta \approx \sin\theta$ for $\theta \ll 1$.
3. We denote the 3D space in $(x,y,z)$, with the $x$-$z$ plane and the $y$-$z$ plane being the sagittal and the tangential plane, respectively.
4. We abuse the sign convention in optics and $r$ is always positive.

Given these, we denote the following vectors:

• Incident ray direction $\mathbf{d} = (0, \sin\phi, -\cos\phi)$.
• Central ray reflection direction $\hat{\mathbf{d}} = (0, \sin\phi, \cos\phi)$.

Proof for the tangential focal length $f_T = \frac{r}{2} \cos\phi$

In this case the incident ray and surface normal are on the tengential plane and can be considered in 2D. Consider a ray intersects the mirror with $(0, y, z)$:

• Surface normal $\mathbf{n} = (0, -y/L, r/L)$, where $|y| \ll r$ and $L = \sqrt{y^2 + r^2} \approx r$.

The reflectance direction $\mathbf{r}$ is given by the law of reflection: \begin{align} \mathbf{r} & = \mathbf{d} - 2(\mathbf{d} \cdot \mathbf{n}) \mathbf{n} \\ & \approx \left( 0, \sin\phi - 2\frac{y}{r} \cos\phi, \cos\phi + 2\frac{y}{r} \sin\phi\right), \end{align} where in the approximation we employ $L \approx r$. Let $\theta = 2(y/r)$ and $\mathbf{r} = \left( 0, \sin(\phi - \theta), \cos(\phi - \theta) \right)$. Compared to the central ray output direction $\hat{\mathbf{d}}$, the angular difference is: \begin{align} \Delta\theta & \approx \tan(\phi - \theta) - \tan\phi \\ & \approx -\frac{\mathrm{d} \tan\phi}{\mathrm{d} \phi} \cdot\theta \\ & = \frac{-1}{\cos^2\phi} \theta. \end{align}

This angular difference projection to $y$ is: $$ \Delta\theta_y = \Delta\theta \cos\phi = -\frac{2y}{r\cos\phi} = -\frac{y}{f_T}, $$ thus $f_T = (r / 2)\cos\phi$.

Proof for the sagittal focal length $f_S = \frac{r}{2} \frac{1}{\cos\phi}$

In this case the incident ray and surface normal are not on a plane and needs consideration in 3D. Consider a ray intersects the mirror with $(x, 0, z)$:

• Surface normal $\mathbf{n} = (-x/L, 0, r/L)$, where $|x| \ll r$ and $L = \sqrt{x^2 + r^2} \approx r$.

The reflectance direction $\mathbf{r}$ is given by the law of reflection: \begin{align} \mathbf{r} & = \mathbf{d} - 2(\mathbf{d} \cdot \mathbf{n}) \mathbf{n} \\ & = \left( -2\cos\phi \frac{x r}{L^2}, \sin\phi, -\cos\phi + 2\cos\phi \frac{r^2}{L^2} \right) \\ & \approx \left( -2\cos\phi \frac{x}{r}, \sin\phi, \cos\phi\right), \end{align} where in the approximation we employ $L \approx r$. Compared to the central ray output direction $\hat{\mathbf{d}}$, the angular difference is: \begin{align} \Delta\theta & \approx -2\cos\phi\frac{x}{r} = -\frac{x}{f_S}, \end{align} which is also the $x$ projection. Thus$f_S = r / (2\cos\phi)$.