Ishaan Booker

2022-07-22

The velocity of light above a hot surface decreases with the height from that surface. The velocity is given by $v={v}_{0}\left(\frac{1-y}{\alpha }\right)$ where $y$ is the vertical distance above the starting position. Show that the path taken by light from (0,0) to $\left(2X,0\right)$ is the arc of a circle centeblack at $\left(X,\alpha \right)$. Note that the origin is chosen to be well above the ground, so that $y<0$ is allowable.

Attempt at a solution: After working through from the velocity relation given in the question, we get the following ODE: $\frac{1}{{v}_{0}\left(\frac{1+y}{\alpha }\right)\sqrt{1+{y}^{2}}}=-C$ where $C$ is a constant.

Steven Bates

The question concerns the light path in a vertically inhomogeneous medium with the velocity $v\left(y\right)$ supposed linearly dependent on the height $y\left(x\right)$.

Here
$v={v}_{0}\left(1-\frac{y}{\alpha }\right)$
From Fermat's principle one obtains the ode
$\frac{1}{v\sqrt{1+{y}^{\prime 2}}}=\text{const}$
Here one solves the "boundary value problem"
$\left\{\begin{array}{l}{y}^{\prime }=\frac{\sqrt{{\alpha }^{2}-{C}^{2}{v}_{0}^{2}\phantom{\rule{thinmathspace}{0ex}}\left(y-\alpha {\right)}^{2}}}{C{v}_{0}\phantom{\rule{thinmathspace}{0ex}}\left(y-\alpha \right)}\\ \\ \\ y\left(0\right)=0\\ \\ y\left(2X\right)=0\end{array}$
The ode is solved by separation of variables remembering that $\left(\sqrt{u}{\right)}^{\prime }={u}^{\prime }/\left(2\sqrt{u}\right)$.
One obtains
$\sqrt{{\alpha }^{2}-{C}^{2}{v}_{0}^{2}\phantom{\rule{thinmathspace}{0ex}}\left(y-\alpha {\right)}^{2}}=C{v}_{0}\left(x+B\right)$
with
$\left\{\begin{array}{l}B=-X\\ \\ C=\frac{\alpha }{{v}_{0}\sqrt{{X}^{2}+{\alpha }^{2}}}\end{array}$
Finally the equation of the circle
$\left(x-X{\right)}^{2}+\left(y-\alpha {\right)}^{2}={X}^{2}+{\alpha }^{2}$

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