# The Path of Light Using Calculus of Variations

• pelmel92
In summary: Since we know that cosθ1 = x/y, we can simplify this equation to:x + y'√(1 - (x/y)^2) = constantFinally, we can rearrange this equation to get our desired equation for the path of the light y(x):y(x) = x(1 + constant/y')^2This is a hyperbolic function, as desired. I hope this helps to clarify the problem and guide you in finding the solution. Good luck!
pelmel92

## Homework Statement

A rectangular solid of height h
increases in density as its height
increases, so the index of refraction of
the solid increases with height
according to:
n(y) = 1.20(2y + 1)
where y is the distance, in meters,
from the origin (see diagram). A beam
of light traveling in air (n = 1.00) in the
x-y plane strikes the bottom of the
tank at the origin, making an angle of
incidence with the normal of θ1. Assume:
• n varies only with y, not with x or z.
• The light travels in the x-y plane.
• The block is wide, so the light leaves through the top and not through a side.

a) Use Fermat’s principle and Euler’s equation to find the equation for the path of the
light y(x) in the tank. There will be two undetermined constants.
HINT: If the equations become too complicated (and they will), remember we learned
some alternative ways to solve calculus of variations problems.
HINT: The answer should come out as a hyperbolic function.

## Homework Equations

Euler's equations for extremum: f - y'(∂f/∂y') = constant
n=c/v

## The Attempt at a Solution

I'm looking to use Fermat's least time principle, I've tried to find the minimum of an integrated functional equal to time t.
Currently I have it as:

t = ∫ ds/v = ∫(1.20/c)(2y+1)√(1+y'^2)dx

which would make my functional:

f=(2y+1)√(1+y'^2)

and using the Euler equations above I have from that:

(2y+1)(1+y'^2)^.5 - y'^2(2y+1)(1+y'^2)^-.5 = (2y+1)(1+y'^2)^-.5 = constant

and that's where I'm stuck...

(2y+1)(1+y'^2)^-.5 = constant seems like it should be easy enough to integrate with dy and dx somehow, but try as I might I'm totally stumped.

Somehow I'm supposed to get a hyperbolic function for y(x), which leads me to believe I'm approaching this whole problem the wrong way.

Thank you for your interesting question. I can offer some guidance on how to approach this problem. First, it is important to note that the problem is asking for the path of the light, which is represented by the function y(x). This means that we need to find an equation that relates the height of the solid (y) to the distance traveled by the light (x).

To start, let's consider the Fermat's principle, which states that light will take the path that minimizes the travel time. In this case, the travel time (t) can be expressed as the distance traveled (x) divided by the speed of light in the material (v), which is given by n(y) = c/v. Therefore, we can write the travel time as:

t = x/n(y)

Now, we can use the Euler's equation for extremum, which states that f - y'(∂f/∂y') = constant. In this case, our functional (f) is the travel time (t). So, we can write:

t - y'(∂t/∂y') = constant

Substituting in our expression for t, we have:

x/n(y) - y'(x/n'(y)) = constant

where n'(y) is the derivative of n(y) with respect to y. Simplifying this equation, we get:

x - y'x/n'(y) = constant

Now, let's consider the relationship between the angle of incidence (θ1) and the angle of refraction (θ2). Using Snell's law, we can write:

sinθ1 = n(y)sinθ2

where n(y) is the index of refraction at a given height y. Solving for sinθ2, we have:

sinθ2 = sinθ1/n(y)

Now, using the small angle approximation (sinθ ≈ θ), we can write:

θ2 ≈ θ1/n(y)

Therefore, the derivative of θ2 with respect to y is:

dθ2/dy ≈ -θ1/n'(y)

Substituting this into our equation above, we have:

x + y'θ1 = constant

Now, we can use the trigonometric identity cos^2θ + sin^2θ = 1 to rewrite this equation as:

x + y'√(1

## 1. What is the "Path of Light"?

The Path of Light refers to the trajectory that light takes when traveling from one point to another. It is a fundamental concept in physics and optics that is used to understand how light behaves and interacts with its surroundings.

## 2. What is the Calculus of Variations?

The Calculus of Variations is a branch of mathematics that deals with finding the optimal path or curve for a given function. It involves using calculus techniques to find the minimum or maximum value of a functional, which is a mathematical expression that takes in a function as its input.

## 3. How does the Calculus of Variations relate to the Path of Light?

The Calculus of Variations can be used to find the path of light that minimizes the time it takes to travel from one point to another. This is known as Fermat's principle, and it states that light will always take the path that requires the least amount of time.

## 4. What are some real-world applications of the Path of Light and Calculus of Variations?

The Path of Light and Calculus of Variations have numerous applications in fields such as optics, physics, engineering, and economics. Some examples include designing optical systems, optimizing transportation routes, and determining the most efficient way to harvest crops.

## 5. Are there any limitations to using the Calculus of Variations for analyzing the Path of Light?

While the Calculus of Variations is a powerful tool for finding optimal paths, it does have some limitations. For example, it assumes that light travels in a straight line, which is not always the case in certain materials or environments. Additionally, it does not take into account factors such as diffraction, scattering, or other phenomena that can affect the path of light.

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