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 $v={v}_{0}(\frac{1-y}{\alpha})$ where $y$ is the vertical distance above the starting position. Show that the path taken by light from (0,0) to $(2X,0)$ is the arc of a circle centeblack at $(X,\alpha )$. 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: $\frac{1}{{v}_{0}(\frac{1+y}{\alpha})\sqrt{1+{y}^{2}}}}=-C$ where $C$ is a constant.

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

Steven Bates

Beginner2022-07-23Added 15 answers

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

Here

$v={v}_{0}(1-\frac{y}{\alpha})$

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"

$\{\begin{array}{l}{y}^{\prime}={\displaystyle \frac{\sqrt{{\alpha}^{2}-{C}^{2}{v}_{0}^{2}\phantom{\rule{thinmathspace}{0ex}}(y-\alpha {)}^{2}}}{C{v}_{0}\phantom{\rule{thinmathspace}{0ex}}(y-\alpha )}}\\ \\ \\ y(0)=0\\ \\ y(2X)=0\end{array}$

The ode is solved by separation of variables remembering that $(\sqrt{u}{)}^{\prime}={u}^{\prime}/(2\sqrt{u})$.

One obtains

$\sqrt{{\alpha}^{2}-{C}^{2}{v}_{0}^{2}\phantom{\rule{thinmathspace}{0ex}}(y-\alpha {)}^{2}}=C{v}_{0}(x+B)$

with

$\{\begin{array}{l}B=-X\\ \\ C={\displaystyle \frac{\alpha}{{v}_{0}\sqrt{{X}^{2}+{\alpha}^{2}}}}\end{array}$

Finally the equation of the circle

$(x-X{)}^{2}+(y-\alpha {)}^{2}={X}^{2}+{\alpha}^{2}$

Here

$v={v}_{0}(1-\frac{y}{\alpha})$

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"

$\{\begin{array}{l}{y}^{\prime}={\displaystyle \frac{\sqrt{{\alpha}^{2}-{C}^{2}{v}_{0}^{2}\phantom{\rule{thinmathspace}{0ex}}(y-\alpha {)}^{2}}}{C{v}_{0}\phantom{\rule{thinmathspace}{0ex}}(y-\alpha )}}\\ \\ \\ y(0)=0\\ \\ y(2X)=0\end{array}$

The ode is solved by separation of variables remembering that $(\sqrt{u}{)}^{\prime}={u}^{\prime}/(2\sqrt{u})$.

One obtains

$\sqrt{{\alpha}^{2}-{C}^{2}{v}_{0}^{2}\phantom{\rule{thinmathspace}{0ex}}(y-\alpha {)}^{2}}=C{v}_{0}(x+B)$

with

$\{\begin{array}{l}B=-X\\ \\ C={\displaystyle \frac{\alpha}{{v}_{0}\sqrt{{X}^{2}+{\alpha}^{2}}}}\end{array}$

Finally the equation of the circle

$(x-X{)}^{2}+(y-\alpha {)}^{2}={X}^{2}+{\alpha}^{2}$

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