Refraction through a nonuniform medium

[darkfall13]

1. The problem statement, all variables and given/known data

A plane electromagnetic wave refracts in a nonuniform medium. The ray trajectory is known,
y = a \left[ 1 - \left( \frac{x}{2a} \right)^2 \right]
where a=const.

a) Plot y(x)

b) From the slope of the curve find the angle \theta \left( y \right) which the k-vector makes with respect to the surface normal (parallel to the y-axis)

c) Using Snell's law, find the refractive index as a function of height, n\left(y\right)

2. Relevant equations

n_i sin \left( \theta_i \right) = n_t sin \left( \theta_t \right)

3. The attempt at a solution

a) simple downward facing parabola with y-intercept at a, x-intercepts at -2a and 2a

b) It hinted at what to do by stating the slope of the curve which is the derivative of the trajectory.

\frac{dy}{dx} = \frac{d}{dx} a \left[ 1 - \left( \frac{x}{2a} \right)^2 \right]
= a \left[ -2 \left( \frac{x}{2a} \right) \left( \frac{1}{2a} \right) \right] = \frac{-x}{2a}

The slope is thought of as "rise over run" so if we're looking for the angle with respect to the y-axis the triangle formed from the tangential to the trajectory and the y-axis the sides of the triangle given from the slope is adjacent and opposite.

cot \theta = \frac{-x}{2a}
\theta \left( y \right) = tan \frac{-x}{2a} = -tan \frac{x}{2a}

c) This is where I'm stuck, Snell's law is given in the earlier section. I can foresee the need to go to infinitesimal levels as the index of refraction changes over the minute values of y through the trajectory but I cannot see how to begin this task.

[gabbagabbahey]

b) It hinted at what to do by stating the slope of the curve which is the derivative of the trajectory.

\frac{dy}{dx} = \frac{d}{dx} a \left[ 1 - \left( \frac{x}{2a} \right)^2 \right]
= a \left[ -2 \left( \frac{x}{2a} \right) \left( \frac{1}{2a} \right) \right] = \frac{-x}{2a}

The slope is thought of as "rise over run" so if we're looking for the angle with respect to the y-axis the triangle formed from the tangential to the trajectory and the y-axis the sides of the triangle given from the slope is adjacent and opposite.

cot \theta = \frac{-x}{2a}

Good...

[quote] \theta \left( y \right) = tan \frac{-x}{2a} = -tan \frac{x}{2a}

Huh?:confused: First,The cotangent is defined as the reciprocal of the tangent (multiplicative inverse), not its functional inverse. Second, you are supposed to express \theta as a function of y, not x.

[darkfall13]

Ha thank you for the catch, I haven't been completely here lately.

\cot \theta = \frac{-x}{2a}

\theta \left( \textbf{x} \right) = \cot^{-1} \left( \frac{-x}{2a} \right)

= \cot^{-1} \left( - \frac{2a \sqrt{1-\frac{y}{a} } }{2a} \right) = - \cot^{-1} \left( \sqrt{1-y/a} \right) = \theta \left( y \right)