# Geometric optics

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Topics

• What shape, if any, of a mirror or lens focuses light to a point?
• What shape, if any, of a lens focuses light to a point?

## Introduction

The theory of optics has one of the longest histories of study in the physical sciences. Lenses have been used for more than 6000 years. Over that time, we observed and developed theories to explain how light bounces off a mirror (reflection) and turns when passing into a new medium (refraction).

One of the first surviving natural philosophies of light was Heron of Alexandria’s. Heron proposed that sight was based on rays emitted by our eyes, that traveled instantaneously. Despite being wrong about the origins of light, he was correct about the properties of mirrors. He argued that because light must travel the shortest, most efficient path between two points, the angle at which a ray comes into a mirror must be the same as the angle at which it comes off the mirror.

Experiments with the properties of light are some of the simplest for us to perform these days, and were not too hard in the past either. With a laser pointer and a protractor, we can check Heron’s theory of reflection. Set up a coordinate system with the mirror at the origin, let $$\theta _ m$$ be the angle between the laser’s direction of pointing before it hits the mirror and the line made by the mirror’s (assumable flat) surface. Let $$\theta _ r$$ be the angle of between the laser’s old direction and new direction. say more? add a data table that we can fit with a regression model

When put concisely, the observed relationship between the incoming and out-going light is called the Law of Reflection in optics, which states that light reflects off a mirror so that the angle of incidence is always equal to the angle of reflection.

The refraction of light is not as simple. We can perform a similar set of experiments to see how light’s direction changes depending on its angle of incidence, but it turns out that the relationship changes if the light is entering glass or water or clear plastic. say more

A second governing principle of optics is Snell’s Law or the Law of Refraction, which states that along a smooth boundary between two media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant that depends on the physical properties of the two media the ray is transitioning between – the ratio of the indices of refraction of each material.

The law of reflection and law of refraction special case of what’s called Fermat’s principle: “Light follows the path of least time between two points.” They can also be derived from Huygens principle: “Every point reached by a wavefront of light becomes the source of a new spherically symmetric wave of light”.

• Hobbes’ particle theory of light (supported by Newton) vs Huygens’ wave theory
• Fresnel, Arago spot, and Poisson’s prediction
• Eventual revitalization by Einstein’s paper on the photoelectric effect.

Today, based on our modern understanding of light as part wave, we say instead that light is observed to follow paths of stationary action (which might be a minimum or maximum of the distance traveled).

In addition, we will also have to make use of trigonometry, vector algebra, and matrix arithmetic to formulate the problem and derive a differential equation.

We will now apply math to address two classical questions. First we consider what shape mirror is needed to focus parallel rays to a point. Second what shape a single lens interface must take to internally focus parallel incident rates. In the problems we have already considered, mechanical considerations (e.g. the Kentucky do-nothing) and coordinate geometry have been enough to obtain complete solutions. But optics, these alone are not enough. We must employ some general governing principles about the redirection of light. These governing principles must be distilled from empirical observation.

## A focusing mirror

Suppose we would like to build a mirror that reflects parallel light rays to a single focal point. What shape should that mirror be? There are a number of ways to successfully approach this problem. The descriptions below illustrate two. The first method relies on vectors and matrix-algebra descriptions of certain linear transformations. The second uses some slick interpretations of complex variables. simplify the calculation.

### Using vector geometry

Take the focus at $$(0,0)$$. Start a curve $$y(x)$$ at $$y(0) = -q$$. At every point, $$(x, y(x))$$. The basic law of optics says that when light bounces off of a mirror, the angle of incidence equals the angle of reflection.

The reflection matrix $\mathrm{R} = \mathrm{I} - 2 \frac{v v^T}{v^T v}$ is such that the part perpendicular to $$v$$ stays the same, but the part parallel to $$v$$ is reflected. We can show that this matrix does what we say it will by breaking any vector $$x$$ up into parts parallel to $$v$$ and orthogonal to $$v$$.

Tangent vector $$(1, dy/dx)$$. Let’s define the normal vector $$v= (-dy/dx, 1)$$.

If we take a ray from our focus and bounce it off of our mirror, we want to reflected ray $$(x, y(x))$$ to be vertical, hence perpendicular to $$(1,0)$$. So, $\begin{bmatrix} 1 & 0\end{bmatrix} \mathrm{R} \begin{bmatrix} x\\y\end{bmatrix} = 0,$ $\begin{bmatrix} 1 & 0\end{bmatrix} \left( \mathrm{I} - 2 \frac{v v^T}{v^T v} \right) \begin{bmatrix} x\\y\end{bmatrix} = 0,$ $\begin{bmatrix} 1 & 0\end{bmatrix} \left( \begin{bmatrix} 1&0\\0&1 \end{bmatrix} - \frac{2}{(y^\prime)^2 + 1} \begin{bmatrix} (y^\prime)^2 & -y^\prime \\-y^\prime & 1 \end{bmatrix} \right) \begin{bmatrix} x\\y\end{bmatrix} = 0.$

We can perform the calculation using the sympy package for symbolic computations in python.

[Show code]
v = Matrix([[-y(x).diff(x)],[1]])
R = 2 * (v * v.T)/(v.T * v)[0,0] - eye(2)
(Matrix([[1,0]]) * R * Matrix([[-x],[q-y(x)]]))[0,0].factor()

The matrices reduce to the scalar nonlinear ordinary differential equation $-\frac{ \left(x \left[ \left(\frac{d y}{d x} \right)^{2} - 1\right] - 2 y \frac{d y}{d x} \right) }{\left(\frac{dy}{dx} \right)^{2} + 1} = 0$ or equivalently, just $x \left(\frac{d y}{d x} \right)^{2} - 2 y \frac{d y}{d x} - x = 0$

We’ll call this the mirror equation. Just from looking at the mirror equation, we can learn things. The mirror equation is first order, nonlinear, and quadratic in the first-derivative term. It is a kind of equation you’ll probably never see in the introductory classes on differential equations, and so the solution methods we’re most familiar with won’t be of much help.

While the mirror equation lacks unique solutions, it is still very solvable. If you recognize that the mirror equation is a homogeneous equation of the form $F(dy/dx, y/x) = 0,$ you might discover that a substitution of the form $$u(x):= y(x)/x$$ makes the equation separable and integrable in close form. My preferred approach (because it will always give you atleast part of a solution) is to guess that our solution has a power series $y(x) = c _ 0 + c _ 1 x + c _ 2 x^2 + c _ 3 x^3 \ldots.$ We expect our solution to be symmetric accross the y-axis, so we can drop the odd terms ($$0 = c _ 1 = c _ 3 = \ldots$$). If we now substitute into the mirror equation and match terms in the power series, we find our power series is a solution as long as $$-1/4 = c _ 0 c _ 2$$, $$0 = c_4 = c_6 = \ldots$$. Thus, solutions are parabolas of the form $y(x) = \frac{x^2}{4f} - f$ where $$f$$ is the focal length of the mirror (the distance from the focus to the bottom of the mirror). If the parabola passes below the focus ($$f > 0$$), then it is convex and reflects light from above to the focus. If the parabola passes above the focus ($$f < 0$$), then it is concave and reflects light from below to the focus.

### A complex variables approach

The mirror equation is much much easier to derive if we have a good understanding of complex variables. Every complex number $$z$$ can be interpreted as both a vector ($$z = x+yi$$) and as a linear transformation representing a rotation-with-dilation ($$z = r e^{i \theta}$$). The following approach is a special case of techniques generally referred to as geometric algebra, a powerful set of techniques largely ignored in most university curricula.

Let $$w = x+y(x)i$$ be the position vector, and $$t = 1+y^\prime i$$ be a vector tangent to the mirror at that location, where $$y' = dy/dx$$. The tangent vector $$t$$ should be such that we rotate it by the angle of incidence, the resulting vector is verticle. A rotation from $$w$$ to $$t$$ is $t w^{-1} = \frac{r_t e^{i\theta_t}}{r_w e^{i\theta_w}} = \frac{r_t}{r_w} e^{i(\theta_t - \theta_w)}$ If we then apply that rotation to the vector $$t$$, the new vector will point in the direction of the reflected ray. If we want that ray to be vertical, that the real part will be zero, so $\operatorname{Re}\left( (t w^{-1}) t \right) = 0$ or $\begin{gather*} \operatorname{Re}\left( \frac{t^2}{w} \right) = 0 \\ \operatorname{Re}\left( \frac{(1 + y^\prime i)^2}{x + yi} \right) = 0 \\ x (y^\prime)^2 - 2 y y^\prime -x = 0 \end{gather*}$

### A comment on method

These two competing methods both lead to the same mirror equation at the end, but illustrate a common trade-off in applied mathematics between discovery and elegance. When we start in on a new problem, we don’t know where to begin, make a few clumsy guesses at first, before the pieces start to fit together, and then suddenly (or not-so-suddenly) we’ve discovered our solution before we really got a good understanding. We could just stop there, but we might also continue to study things – having reached the end, we can look back and see all the false-starts, stumbles, and sidetracks. We can redo everything so it’s clean, efficient, and elegant – the right way to solve the problem. We’d probably never have come up with this elegant solution the first time through, and at the end, we’ve only solved a problem to which we already had the solution. On the other hand, we now know how to solve a whole family of problems, and our deeper understanding is something we can share with others.

Is the extra work worth it? Well, that depends on the situation, and is a question we can always keep in the back of our heads.

### Analysis and solution

Since the mirror equation is quadratic in the first derivative, we can use the quadratic formula to calculate the mirror slope at any given point – if we know a position $$(x,y(x))$$ where the mirror passes, we can determine the slope of the curve at that point by using the quadratic formula. But, the equation has 2 solutions for the slope most of the time! Why? Well, a little thought reveals that this makes perfect sense – there are actually two different mirrors that solve our equations – one that collects light from above, and one that collects light from below! So the ambiguity actually makes good sense and reveals a property of the problem that we had actually overlooked so far. (Another property the mirror equation reveals is symmetry of solutions – see exercises).

While the mirror equation lacks unique solutions, it is still very solvable. One approach is to guess a power series $$y(x) = c _ 0 + c _ 1 x + c _ 2 x^2 + c _ 3 x^3 \ldots$$. We expect our solution to be symmetric across the y-axis, so we can drop the odd terms ($$0 = c _ 1 = c _ 3 = \ldots$$). If we now substitute into the mirror equation and match terms in the power series, we find our power series is a solution as long as $$-1/4 = c _ 0 c _ 2$$, $$0 = c_4 = c_6 = \ldots$$. Thus, solutions are parabolas of the form $y(x) = \frac{x^2}{4f} - f$ where $$f$$ is the focal length of the mirror (the distance from the focus to the bottom of the mirror). If the parabola passes below the focus ($$f > 0$$), then it is convex and reflects light from above to the focus. If the parabola passes above the focus ($$f < 0$$), then it is concave and reflects light from below to the focus.

## Refraction

The problem of lens design is based on refraction rather than reflection and is more difficult than that of mirror design. Snell’s law for refraction states that when a light ray hits an interface between two media, $$n_1 \sin \theta_1 = n_2 \sin \theta_2$$, where $$\theta_1$$ is the angle of incidence from vertical (normal, orthogal to the surface), $$\theta_2$$ is the angle of exodus from vertical, and $$n_1$$ and $$n_2$$ are refractive indices of the two materials.

Assume all rays propagate outward from a focus at $$(0,0)$$ to the lens interface at $$(x,y)$$. On hitting the lens, we want them to be refracted to travel up, in direction $$(0,1)$$.

Let $$y(x)$$ be the equation of our lens interface, with tangent vector $$(1,y')$$ and normal vector $$(-y',1)$$.

To convert Snell’s law into a vector form, we can use trigonometry’s the sine-of-sums identity: $\sin(b + a) = \cos a \sin b + \sin a \cos b$ and the closely related $\sin(b - a) = \cos a \sin b - \sin a \cos b.$ One interpretation of this identity is that if we have a unit vector $$\hat{v} = (\cos a, \sin a)$$ and another unit vector $$\hat{w} = (\cos b, \sin b)$$, the angle between these vectors $$\theta = b-a$$, and the sine of the angle between the vectors is $$\hat{v}_1 \hat{w}_2 - \hat{v}_2 \hat{w}_1$$. Thus, we have a formula for the sine of the angle between any two unit vectors that involves only basic algebra. This generalizes to vectors of arbitrary magnitude if each is first converted to a unit vector.
$\sin \theta = \frac{\vec{v}_1 \vec{w}_2 - \vec{v}_2 \vec{w}_1}{\sqrt{(\vec{v}_1^2+\vec{v}_2^2)(\vec{w}_1^2+\vec{w}_2^2)}}$ or stated more concisely using the wedge product $$\wedge$$, $\sin \theta = \frac{|\vec{v} \wedge \vec{w}|}{| \vec{v} | | \vec{w} |}.$

Now, if we rewrite Snell’s law using this identity and the three vectors $$(x,y)$$, $$(-y',1)$$, and $$(0,1)$$, to obtain a differential equation for the lens surface. After substituting and simplifying, we find $\left(\frac{n_1}{n_2}\right) \left( \frac{1}{y'} + \frac{y}{x}\right) = \sqrt{1 + \left(\frac{y}{x}\right)^2}$ or in a form free of division and square roots, $\left(\frac{n_1}{n_2}\right)^2 \left( x + y y'\right)^2 = y'^2 \left( x^2 + y^2\right)$

### Analysis and solution

From the first form, we have that solutions are scale-invariant, as expected – the slope along every point of the line y= m x must be the same. Solutions also exhibit mirror symmetry across the y-axis. This equation again does not fall in the scope of our standard methods, but the solution curves can be shown to be conic sections. The general solution, letting $$k = n_1/n_2$$, $x^{2} + \left(C + y\right)^{2} \left(1 - 1/k^2\right) = C^{2} \left(k^{2} - 1\right)$ $\frac{k^2}{k^2-1} \left(\frac{x}{C}\right)^{2} + \left(1 + \frac{y}{C}\right)^{2} = k^2$ where $$C$$ is the integration constant. When we solve for $$y$$ in terms of $$x$$, $y(x) = \frac{ \left[ C \left(k^{2} - 1\right) \pm k \sqrt{\left(1-k^2\right) \left(C^2- C^2 k^2 + x^2\right)}\right]}{k^{2} - 1}$

# Exercises

1. Show by substitution that if $$Y(x)$$ is a solution to the mirror equation, then so is $$-Y(x)$$.

2. Show by substitution that if $$Y(x)$$ is a solution to the mirror equation, then so is $$Y(-x)$$. (Hint: First solve the mirror equation for $$y'$$, then substitute.)

3. If $$z = x + iy$$, then the conjugate $$\overline{z} = x - iy$$. Using conjugation, the mirror equation can be written in the alternative form $\operatorname{Re}\left( (t \overline{w}) t \right) = 0,$ which is slightly more efficient to analyze because it does not require any division. Provide a verbal geometric explanation of why replacing the inverse with the conjugate leads to the same equation.

4. A close approximation of Vitello’s data relating angle of incidence to angle of refraction for light passing from air to glass is given below.

1. Use least squares to fit a curve of the form $$\theta_2 = a_1 \theta_1 + a_3 \theta_1^3$$ to Vitello’s data.
2. Solve Snell’s Law for the angle of refraction $$\theta_2$$ as a function of the angle of incidence $$\theta_1$$, and McLauren-series the function to fourth order.
3. Using your previous results, estimate the refractive index of glass relative to air. Explain your reasoning.

[ Data : hide , shown as table , shown as CSV shown as QR ]

# angle-of-incidence (deg),angle-of-refraction (deg)
#
# Witelo's (aka Vitello) experimental data on the refraction angle at
# the air/glass boundary from *Optica*, extracted from
# http://opticalengineering.spiedigitallibrary.org/data/journals/optice/23437/121704_1_4.png by g3data
# all angles are measured in degrees
#
# Warning: These numbers have not been double-checked against the original manuscript.  Apply with caution.
#
79.97,42.11
70.07,38.58
60.17,34.56
50.12,30.04
40.07,25.04
30.02,19.54
20.11,13.54
10.05,07.06
00.00,00.00


1. Find the shape of a lens that focuses light from 1 unit to the left of the lens center to one unit to the right of the lens center.

2. We have shown above that a perfectly focusing mirror should be shaped like a circular paraboloid (a parabola extended to three dimensions by rotation). However, it is much easier to make a spherical mirror (by rotating a circle) than a circular paraboloid (by rotating a parabola). If the aperture (explained?) radius of the mirror is small relative to the focal length of the mirror, we might make a spherical mirror that does almost as well as a circular paraboloid.

1. Find the equation for a circle that focuses light to the origin and is the best approximation to the parabola $$y = b - x^2/4b$$ when b is large. (Hint: Match the curvature of the parabola at it’s minimum to that of the circle.)

2. (Hard) While the spherical lens is a good approximation when the appeture is small, it is not perfect, and introduces some error known as spherical abberation. Spherical abberation can be removed by including a lens to focus light before it reaches the mirror. Find a differential equation for a lens that will remove spherical abboration.

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