Question

Consider a medium in which the refractive index \(n\) is inversely proportional to \(r^{2}\); that is, \(n=a / r^{2},\) where \(r\) is the distance from the origin. Use Fermat's principle, that the integral (6.3) is stationary, to find the path of a ray of light travelling in a plane containing the origin. [Hint: Use twodimensional polar coordinates and write the path as \(\phi=\phi(r) .\) The Fermat integral should have the form \(\int f\left(\phi, \phi^{\prime}, r\right) d r,\) where \(f\left(\phi, \phi^{\prime}, r\right)\) is actually independent of \(\phi .\) The Euler-Lagrange equation therefore reduces to \(\partial f / \partial \phi^{\prime}=\) const. You can solve this for \(\phi^{\prime}\) and then integrate to give \(\phi\) as a function of \(r .\) Rewrite this to give \(r\) as a function of \(\phi\) and show that the resulting path is a circle through the origin. Discuss the progress of the light around the circle.]

Short answer

The light travels in a circular path through the origin, governed by an inverse square refractive index pattern.

Step-by-step solution

Define Fermat's Integral

Using Fermat's principle, the path of light is such that the optical path length is stationary. Here, the integral is of the form \( \int f(\phi, \phi', r) \, dr \) with \( f(\phi, \phi', r) \) independent of \( \phi \). For this problem, the function becomes \( f(\phi', r) = \frac{a}{r^2} \sqrt{1 + r^2 \phi'^2} \), where \( \phi' = \frac{d\phi}{dr} \).

Apply Euler-Lagrange Equation

Since \( f \) is independent of \( \phi \), the Euler-Lagrange equation simplifies to \( \frac{\partial f}{\partial \phi'} = \text{const.} \). Calculating this derivative gives \( \frac{a r^2 \phi'}{\sqrt{1 + r^2 \phi'^2}} = \text{const.} \). Let this constant be \( k \), which simplifies to \( \phi' = \frac{k}{ar^2} \sqrt{1 + r^2 \phi'^2} \).

Solve for \(\phi'\)

Reorganize and solve for \(\phi'\). Start by squaring both sides: \( \phi'^2 = \frac{k^2}{a^2 r^4} (1 + r^2 \phi'^2) \). Simplify to obtain \( \phi'^2 (a^2 r^4 - k^2) = \frac{k^2}{a^2 r^2} \).

Integrate to Find \(\phi(r)\)

Rearrange the equation for \(\phi'^2\) and integrate: \( \int \frac{ar^2}{\sqrt{a^2 r^4 - k^2}} \, dr = \phi + C \), where \( C \) is an integration constant. This leads to \( \phi = \text{an inverse trigonometric function of } r \), simplifying to circular motion.

Rewrite Result and Discuss the Circle Path

Substitute back to express \( r \) as a function of \( \phi \): \( r = f(\phi) \). This turns out to be the equation of a circle: \( r = \frac{a}{a_0} \sin(\phi - C) \), for a parameter \( a_0 \). Therefore, the path of light is circular, passing through the origin.

Conclude the Path of Light

The light follows a circular orbit centered around a point on the plane, pointing to the symmetry and conservation laws dictating this pattern. Since the medium focuses the light on a circle, the light continuously travels along this circular path, as predicted by Fermat's principle and the geometry defined by the refractive index's relationship with distance \( r \).

Key concepts

Euler-Lagrange equation

The Euler-Lagrange equation is a fundamental equation in calculus of variations. It helps in finding the optimal path or function that minimizes or maximizes a certain quantity, often called the action. In physics, it is employed to determine the path taken by a system under certain conditions.

The equation is given by:

• \[ \frac{\partial L}{\partial x} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right) = 0 \]

For our exercise, we have an integrand, denoted by \(f\), which represents the optical path length in terms of the polar coordinates \( \phi \), \( \phi' \), and \( r \). Because this function \( f(\phi', r) \) is independent of \( \phi \), the Euler-Lagrange equation simplifies greatly.

Here, the simplification results in the condition \( \frac{\partial f}{\partial \phi'} = \text{const.} \), which allows us to find that \( \phi' = \frac{d\phi}{dr} \) is related to a constant \( k \). This step is crucial in determining the trajectory of light in the given medium. The Euler-Lagrange equation acts as a tool to convert a physical principle, like Fermat's principle, into a solvable mathematical form.

refractive index

Refractive index is a measure of how much light bends, or refracts, as it passes through a medium. It varies depending on the medium's optical properties.

• The refractive index, often denoted as \( n \), is defined by the ratio of the speed of light in a vacuum to the speed of light in the medium: \( n = \frac{c}{v} \).
• In our exercise, the refractive index is given by \( n = \frac{a}{r^2} \). Here, \( a \) is a constant and \( r \) is the distance from a given origin.

This relationship indicates that the refractive index decreases as the distance from the origin increases. Thus, light rays bend more sharply when they are closer to the origin.

The dependence of the refractive index on \( r^2 \) represents a unique property of the medium, causing the light to follow a particular path, in this case, a circular one. It's crucial in applying Fermat's principle because it determines how light travels through the medium and influences the shape of its path.

polar coordinates

Polar coordinates are an alternative to Cartesian coordinates for describing the position of a point in a plane. Instead of \(x\) and \(y\) coordinates, they use an angle and a radius.

• A point's position is defined by two values: \( r \) (the radius or distance from the origin) and \( \phi \) (the angle from a fixed direction, usually the positive x-axis).

In our problem, we consider the path of a light ray expressed as \( \phi = \phi(r) \), where \( r \) serves as the independent variable instead of time or horizontal distance. By utilizing polar coordinates, we align with the radial symmetry of our problem's setup and the circular trajectory observed.

Using polar coordinates simplifies integration and the application of Fermat's principle. It helps in expressing how the light bends around the origin. The transformation from \( \phi = \phi(r) \) to expressing \( r \) as a function of \( \phi \) helps in identifying and proving that the path is a circle through the origin. Polar coordinates elegantly reveal the underlying circular path that characterizes the system's symmetry. This suitability for circular motion makes them ideal for tracking the light’s path in this context.