Image Formation by a Spherical Mirror: angle change from $-\theta_2$ to $\theta_2$? I am currently studying the textbook Fundamentals of Photonics, Third Edition, by Saleh and Teich. In a section titled Paraxial Rays Reflected from Spherical Mirrors, the authors derive the mirror equation (the authors refer to it as the "imaging equation (paraxial rays)":
$$\dfrac{1}{z_1} + \dfrac{1}{z_2} = \dfrac{1}{f}$$
The following images accompany the derivation:



A spherical mirror of radius $R$ therefore acts like a paraboloidal mirror of focal length $f = R/2$. This is, in fact, plausible since at points near the axis, a parabola can be approximated by a circle with radius equal to the parabola's radius of curvature (Fig. 1.2-5).
All paraxial rays originating from each point on the axis of a spherical mirror are reflected and focused onto a single corresponding point on the axis. This can be seen (Fig. 1.2-6) by examining a ray emitted at an angle $\theta_1$ from a point $P_1$ at a distance $z_1$ away from a concave mirror of radius $R$, and reflecting at angle ($-\theta_2$) to meet the axis at a point $P_2$ that is a distance $z_2$ away from the mirror. The angle $\theta_2$ is negative since the ray is traveling downward. Since the three angles of a triangle add to $180^\circ$, we have $\theta_1 = \theta_0 - \theta$ and $(-\theta_2) = \theta_0 + \theta$, so that $(-\theta_2) + \theta_1 = 2\theta_0$. If $\theta_0$ is sufficiently small, the approximation $\tan \theta_0 \approx \theta_0$ may be used, so that $\theta_0 \approx y/(-R)$, from which
$$(-\theta_2) + \theta_1 \approx \dfrac{2y}{(-R)}, \tag{1.2-1}$$
where $y$ is the height of the point at which the reflection occurs. Recall that $R$ is negative since the mirror is concave. Similarly, if $\theta_1$ and $\theta_2$ are small, $\theta_1 \approx y/z_1$ and $(-\theta_2) = y/z_2$, so that (1.2-1) yields $y/z_1 + y/z_2 \approx 2y/(-R)$, whereupon
$$\dfrac{1}{z_1} + \dfrac{1}{z_2} \approx \dfrac{2}{(-R)}. \tag{1.2-2}$$
This relation holds regardless of $y$ (i.e., regardless of $\theta_1$) as long as the approximate is valid. This means that all paraxial rays originating from point $P_1$ arrive at $P_2$. The distances $z_1$ and $z_2$ are measured in a coordinate system in which the $z$ axis point to the left. Points of negative $z$ therefore lie to the right of the mirror.
According to (1.2-2), rays that are emitted from a point very far out on the $z$ axis ($z_1 = \infty$) are focused to a point $F$ at a distance $z_2 = (-R)/2$. This means that within the paraxial approximation, all rays coming from infinity (parallel to the axis of the mirror) are focused to a point at distance $f$ from the mirror, which is known as its focal length:
$$f = \dfrac{(-R)}{2}, \tag{1.2-3}$$
Equation (1.2-2) is usually written in the form
$$\dfrac{1}{z_1} + \dfrac{1}{z_2} = \dfrac{1}{f}, \tag{1.2-4}$$
which is known as the imaging equation. Both the incident and the reflected rays must be paraxial for this equation to hold.

Note that the author said that the angle $\theta_2$ is negative since the ray is travelling downward.
Immediately following this derivation, the authors present the following exercise and accompanying image:

Now notice that the angle $-\theta_2$ in the first figure becomes $\theta_2$ in the second figure. This is where my confusion arises, because I don't understand why it was changed (the ray is still travelling downward, right?), and doing it one way or the other leads to a different solution (see below).
The solutions manual proceeds as follows:

In order to clarify the solution from the solutions manual, I will now provide my calculations.
The angle of incidence at the mirror is $\phi$, and we have that
$$\begin{align} &\theta_1 + \phi + (180^\circ - \psi) = 180^\circ \ \ \text{(Where $\phi + \theta_A = 180^\circ - \psi$, where $\theta_A$ is the remaining/unmarked angle.)} \\ &\Rightarrow \phi + \theta_1 - \psi = 0 \\ &\Rightarrow \phi = \psi - \theta_1 \approx \dfrac{y}{-R} - \theta_1 \ \ \text{(By the paraxial approximation.)} \end{align}$$
The reflected ray makes an angle $\theta_2$ with the $z$-axis. And we also have that
$$\begin{align} &180^\circ = 2\phi + \theta_1 + \theta_A \\ &\Rightarrow \theta_A = 180^\circ - 2\phi - \theta_1 \end{align}$$
Now, we get a different result for the following calculation depending on whether we have $-\theta_2$ or $\theta_2$ (I use $\theta_2$):
$$\begin{align} &\theta_A + (\theta_2) = 180^\circ \\ &\Rightarrow \theta_A = 180^\circ - \theta_2 \end{align}$$
Therefore, we have that
$$\theta_2 = 2\phi + \theta_1$$
I would greatly appreciate it if people would please take the time to explain why the authors change the angle from $-\theta_2$ to $\theta_2$. Is this an error?
 A: There is no error. This is a matter of notation. The $-\theta_2$ of the proof and the $\theta_2$ of the manual are in fact the same angle. You started solving the exercise for the angle $\theta_2$ showed in the Figure of the manual and if you continue your solution in this way you will end up with the same results. Just stick to either one notation, dont mix them.
Personally,I found it easier to maintain the notation of the book because I can straight forward use the formulas without having to 'translate' them to my new notation.
For example, you will have to use the formula $(-\theta_2)+\theta_1=2\theta_0$ proved in the book above eqn 1.2-1. In your new notation it would be $\theta_2+\theta_1=2\psi$ (to see this resemblance, look at the diagrams 1.2-6 and that of the manual), which is not something you have written down but you can easily get by playing around a bit: Use $\theta_2=2\phi+\theta_1$ and $\phi=\psi-\theta_1$ both of which you have proved.
Using this relation, the identity $y_1z_2-y_2z_1=0$ you are asked to prove and by following in the footsteps of the book ($\theta_1=\frac{y-y_1}{z_1},-\theta_2=\frac{y+y_2}{z_2}$ using the notation of the book ), you can prove 1.2-4.
