Differential Calculus: From Practice to Theory

Another example of this second type is problem Problem 3.3.1. Recall that we showed that light bounces off of a mirror in such a manner that the angle of incidence equals the angle of reflection. We did this by reasoning geometrically from the figure below to show that the minimal path must satisfy the condition \(\alpha=\gamma\text{.}\)

The geometric argument was actually pretty easy to understand once we found the right trick. But that’s the difficulty. The geometric argument required the very counter–intuitive trick of pretending that the light passed through the mirror instead of reflecting off of it. Calculus provides a more systematic approach, one that does not require a stroke–of–genius level of insight to solve each new problem. Let’s re–solve this problem, this time using the Calculus tools we’ve learned.

We will need an objective function so we modify the sketch above to suit the needs of Calculus. In particular, the relevant quantities need names as shown below.

We assume that \(a\) (the height of the light source), \(b\) (the height of the observer’s eye), and \(c\) (the distance between the light source and the observer) are constants while \(x, y, z\text{,}\) and \(w\) are variables.

Differentiating equation (9.13) gives \(\dx{L} = \dx{z}+\dx{w}\text{.}\)

We began the derivation of Snell’s Law in Section 3.7, using the following sketch and assuming that the velocity of light in Medium 1 is \(v_1\text{,}\) and the velocity of light in Medium 2 is \(v_2\text{.}\)

Notice that this is an optimization problem. The objective function we need to minimize is \(T(x)\) and there are several constraints. For example, \(x+y=c\) is one. Another is \(a^2+x^2=z^2\text{.}\) There is at least one more.

It is tempting to compute \(\dfdx{T}{x}\) and proceed to solve \(\dfdx{T}{x}=0\) as we’ve done before. Computing \(\dfdx{T}{x}\) is a bit challenging but it can be done if you proceed carefully.

The real issue is solving \(\dfdx{T}{x}=0\) for \(x\text{.}\) Unfortunately, this is still rather difficult to accomplish. Fortunately, we are not really interested in the value of \(x\) or \(T\) when \(T(x)\) is minimal. Instead, we want to show that the time traveled along the path from \(A\) to \(C\) to \(B\) is minimal when \(\frac{\sin(\alpha)}{v_1}=\frac{\sin(\beta)}{v_2}\text{.}\) This is the second type of optimization problem. We’ll proceed carefully.

What Part 9.3.3.d means physically is that when you look upward from beneath the surface of a pond the entire view of the world above the surface is compressed into a circle at the top of a cone with central angle \(\beta\text{.}\) Attempts to look beyond that angle will result in you seeing a (probably dark) reflection from the bottom of the pond as no light that enters from above and outside this circle will reach your eye at point \(P\text{.}\) This phenomenon is called Snell’s Window and is illustrated in the image below.

The solution to the following problem will rely on a derivation similar to the one we used to prove Snell’s Law.

Look carefully at the double rainbows in the pictures below. What do you notice?

To tackle these questions the first thing we’ll need to do is build a mathematical model of the problem. As you will see developing the model is probably the most challenging part.

When a ray of light enters a raindrop it gets both refracted and reflected, so we will need to use all that we have learned about how light refracts and reflects. Clearly Snell’s Law will play a prominent role. To keep things simple we assume the drop is spherical.

By the Principle of Local Linearity when light strikes a point on a curved surface it will refract (if the surface is transparent) or reflect (if the surface is reflective) as if the surface is actually the line tangent to the surface. Since we’re assuming for simplicity that our raindrops are spherical this means that the light will reflect or refract as if it is striking the tangent line of the circle in the diagram at the points \(A\text{,}\) \(B\text{,}\) or \(C\) where the light enters the droplet, reflects inside the droplet, and where it exits the droplet, respectively. All of the angles in the diagram above are measured between the light beam (in red) and the extended radii (blue, dashed lines) at points \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)

Actually, only a portion of the light is refracted. The rest is reflected off the exterior of the raindrop, and is no longer of concern to us. Once inside the raindrop, it reflects off of the back of the raindrop (again only a portion is reflected) so that the angle of incidence equals the angle of reflection. A portion of the light then exits the raindrop, and once again the angles are governed by Snell’s Law.

The total amount of (clockwise) rotation in this process is called the deviation angle, denoted \(D(\alpha)\text{.}\) This is shown in the sketch above.

From this graph it appears that the minimum of the curve is approximately at the point \((1.03,2.41)\text{.}\) So the minimum value of \(D(\alpha)\) is about \(2.41 \text{ radians }\approx138^\circ\) and it occurs when the angle at which the light enters the raindrop is \(\alpha\approx1.03 \text{ radians }\approx59.4^\circ\text{.}\) Notice (from the enlargement of the graph near that minimum) that because the graph of \(D(\alpha)\) is nearly horizontal (this is Fermat’s Theorem 9.1.1 coming into play again) for values of \(\alpha\) near \(1.03\) the amount of deviation is roughly the same.

This is pertinent because it means that the lower the deviation angle the more concentrated that light rays exiting the raindrop. So the light rays entering the raindrop at approximately \(59^\circ\) have the highest concentration when exiting the raindrop. At other values of \(\alpha\) the light is more diffused. It is these concentrated light rays that we see as a rainbow.

So you will see a rainbow when the air is filled with water droplets, for example immediately after a rain, and the sun is behind you. Relative to the sun’s rays the angle from your eye to the top of the rainbow will be about \(42^\circ\text{.}\) Every droplet reflects and refracts each ray of light entering it. The aggregation of all of those rays is the rainbow.

Of course, rainbows will manifest under other circumstances too, but we have not analyzed those.

The color spread in a rainbow occurs because the index of refraction \(k\text{,}\) changes with the frequency (color) of the light. This can be seen by shining a beam of light through a triangular prism or, by listening to rock music from the 1970s.

Of course we can only do so much by reasoning from graphs and pictures. If we want to nail things down very precisely (we do), we will eventually need to look at the problem analytically by minimizing an objective function. Fortunately, we already have our objective function. It is \(D(\alpha)\text{.}\) Our constraint comes from Snell’s Law, \(\sin(\alpha)=k\sin(\beta)\text{.}\) We said that the general parameter \(k\) would depend on the frequency of the light and we’ve observed that for water and air \(k=\frac43\text{.}\) Rather than graphing the function for different values of \(k\) we’ll just leave it as an arbitrary constant while we calculate. Afterward we can substitute specific values as needed.

Normally to solve such a system, we would use one of the equations to eliminate one of the variables in the other equation. We could do that here but it would require some very messy expressions involving inverse trigonometric functions. Instead there is a trick that will be cleaner.

The values of \(R(\alpha)\) you obtained in the table account for the various color bands that appear in the rainbow as sunlight hits droplets at various angles.

In a double rainbow the primary rainbow is created from light entering the upper portion of the droplet and exiting from the bottom. If light is bright enough then a secondary rainbow is created from light entering the lower portion and exiting the upper portion of the droplet as in the diagram below.

For the secondary rainbow the deviation angle \(D(\alpha)\) is the total amount of counterclockwise rotation and the rainbow angle would be \(R(\alpha)=D(\alpha)-\pi\text{.}\) Note that the extra reflection explains why the secondary rainbow is not as bright as the primary.