Why didn't I reach $\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$?

In optics for convex and concave mirrors there is the following formula: $$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$$ where $$p$$ is the distance between the object and the mirror, $$q$$ is the distance between the picture and the mirror, and $$f$$ is the focal length.

I used the cartesian plane to reach the formula: I used the equation of the parabola (as a concave mirror) to be $$cy^2+dx=0$$. So its focus is located at the point $$(-\frac{c}{4d},0)$$. Then I put an object in a distance from the mirror with the point of its tip to be $$(a,b)$$ where $$a$$ is the distance between the object and the mirror, and $$b$$ is the height of the object.

The light comes from the tip of the object parallel to the $$x$$ axis and hits the mirror in the point $$(-\frac{cb^2}{d},b)$$. Then it reflects and goes through the focus. So the equation of the reflecting light is $$y=\frac{4bd}{c(1-4b^2)}x+\frac{b}{1-4b^2}$$.

Also, a light beam comes from the tip of the object and hits the mirror in the point $$(0,0)$$. The reflecting light has the equation $$y=-\frac{b}{a}x$$.

The intersection point of the two reflecting lights have the coordinates $$(-\frac{ac}{4ad+c(1-4b^2)},\frac{bc}{4ad+c(1-4b^2)})$$, where the $$x$$ axis coordinate is the distance between the picture and the mirror, and the $$y$$ axis coordinate is the height of the picture

Finally, I tried to reach the formula in the question using my findings: $$\frac{1}{a}-\frac{4ad+c(1-4b^2)}{ac}=-\frac{4d}{c}$$ (as you remember $$a$$ was the distance between the object and the mirror in my experiment; the others are also reciprocated).

After simplifying it more I reached $$cb^2=0$$ ! How is it possible? $$b$$ can't be zero. It's the height of the object. $$c$$ can't be zero as well. It's the coefficient of $$y^2$$ in the equation of the parabola.

Did I do a mistake here or the famous equation in the question is false at all?

• Your first equation applies (as an approximation) to the case of spherical mirrors. I don't believe you will be able to reproduce this equation with a parabolic mirror.
– d_b
Commented Jul 23, 2023 at 1:41
• @d_b A spherical mirror doesn't have a true focus. Also, the fact that a light beam parallel to the x-axis is reflected to go through the focus is only in parabolic mirrors. Commented Jul 23, 2023 at 1:47
• Yes, but the mirror equation uses the paraxial approximation. Can you provide a reference stating that this equation applies to parabolic mirrors as well as spherical mirrors? I have been unable to find one.
– d_b
Commented Jul 23, 2023 at 2:18
• I have not worked through the math, but I do observe that the starting point is (-a,b), provided you are taking "a" to be a distance. If you graph the reflected line, you will see that it has a positive slope. If "a" is the negative initial position, then it is not the object distance. Commented Jul 23, 2023 at 3:06
• If we assume the starting point to be $(-a,b)$ we are already assuming that the parabola (or our concave mirror) is opening to the left (or more technical, we have a positive $d$). But with just an $a$ there is no such assumption. If we give a positive number to it the parabola is opening to the right and if we give a negative number, the parabola is opening to the left. Commented Jul 23, 2023 at 3:26

The equation you linked to applies, as comments have said, only in the paraxial approximation. It is therefore no wonder that you have not managed to derive it.

In fact, what you derived is correct! (It seems to me.) After simplifying a little, we get

$$\frac 1 p + \frac 1 q = \frac {4(b^2-ad)} {ac}$$,

which in the paraxial approximation ($$b\rightarrow0$$) indeed becomes $$\frac {-4d} c = \frac 1 f$$. What you basically derived is a correction to the approximation in terms of $$b$$.

• So isn't there a method to calculate the height of the picture? I plugged $a=2.5, b=0.3, c=12, d=-1$ (which the object is between the mirror and the focus), and with my derived coordinates I found the distance of the picture to the mirror to be about $12.9$ and the height of the picture to be about $-1.55$ (the mius sign shows it's upside-down). While when the object is between the mirror and focus the distance of the picture to the mirror should be negative (showing it's virtual) and the height should be positive. But when I choose $b=0$ the distance becomes $-15$ and the height a $0$. Commented Jul 23, 2023 at 21:39
• I also plotted the graphs of $12y^2-x=0$ and the reflecting lights (using my findings) in desmos. Commented Jul 23, 2023 at 21:43
• $-15$ is the true distance, but its problem is that the height of the picture is $0$, while the object has a height. Commented Jul 23, 2023 at 22:10

Here is a for a spherical mirror. Maybe someone else can figure out how to extend this to a parabolic mirror.

I take a spherical mirror of radius $$R$$ centered at $$(0,0)$$, and an image of height $$h$$ with its base at distance $$s$$ from the center of the mirror, so that the tip of the object is at $$(R-s,h)$$.

The ray parallel to the optical axis intersects the mirror at $$(\sqrt{R^2-h^2},h)$$, and is incident at angle $$\theta = \arcsin(h/R)$$ from the surface normal. The slope of the outgoing ray is $$\tan2\theta$$, so the equation for the outgoing ray is \begin{align} y = h + \tan2\theta\left(x-\sqrt{R^2 - h^2}\right) \end{align}

The ray incident at the center of the mirror $$(R,0)$$ has slope $$-h/(R-s)$$. The outgoing ray has the opposite slope, so its equation is \begin{align} y = \frac{h}{R-s}(x-R) \end{align} Equating these and solving for $$x$$ to locate the tip of the image, we get an awful mess, which I hesitate to include (but do for the sake of completeness): \begin{align} x = \frac{-s \sqrt{R^2-h^2} \tan \left(2 \sin ^{-1}\left(\frac{h}{R}\right)\right)+R \sqrt{R^2-h^2} \tan \left(2 \sin ^{-1}\left(\frac{h}{R}\right)\right)-2 h R+h s}{-s \tan \left(2 \sin ^{-1}\left(\frac{h}{R}\right)\right)+R \tan \left(2 \sin ^{-1}\left(\frac{h}{R}\right)\right)-h} \end{align}

To carefully employ the paraxial approximation $$h\ll R$$, let's expand to first order in $$h/R$$. To first order, we have \begin{align} \sqrt{R^2 - h^2} &\approx R\\ \tan \left(2 \sin^{-1}\left(\frac{h}{R}\right)\right) &\approx \frac{2h}{R} \end{align} so \begin{align} x&\approx \frac{-2hs+2hR-2hR + hs}{-2hs/R+2h -h}\\ &=\frac{-sR}{-2s+R}\\ &= \frac{s(R/2)}{s - (R/2)} \end{align} Rearranging, we find that for a spherical mirror with radius $$R$$ and in the paraxial approximation, the object and image positions $$s$$ and $$x$$, respectively, obey \begin{align} \frac{1}{s} + \frac{1}{x} = \frac{2}{R} \end{align} which is the standard form of the mirror equation.