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Suppose an incident light from vacuum ($n_1=1.0$) into some media ($n_2=n_1+\mu\; x^2$) as in the figure below.

How to calculate the refracted light path curve in closed form?

enter image description here

Update:

Try to set up ordinary differential equation for the refracted light path per Snell's law.

Suppose the curve is $y=y(x)$;

Since $n_i \sin\theta_i=\text{constant}=n_1\sin\alpha=\sin\alpha$.

For any point $P:(x_0,y(x_0))$ on the path $y(x)$, we have: $$\tan(\theta_P)=\dfrac{\sin\theta_P}{\cos\theta_P}=y'(x)=\dfrac{\rm{d}y}{\rm{d}x},\quad \text{where }\theta_P \text{ is incident / refracted angle}$$

Since $\theta_P$ is always an acute angle, we have:

$$\dfrac{\sin^2\theta_P}{{1-\sin^2\theta_P}}=y'(x)^2\Rightarrow \sin\theta_P=\dfrac{\pm y'(x)}{\sqrt{1+y'(x)^2}}$$

Clearly $n_P\sin\theta_P=\sin\alpha$, where $n_P=1+\mu x^2$, then we have:

$$\left(1+\mu x^2\right)\dfrac{\pm y'(x)}{\sqrt{1+y'(x)^2}}=\sin\alpha\quad\text{with: y(0)=5} ||y'(0)=\tan\alpha$$

Then it becomes how to solve the ODE with a boundary condition. Can the ODE be solved in closed form?

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    $\begingroup$ Did you try integrating over Snell's law? While I have never done the calculation, it seems to me, that one of sines can be replaced by a Taylor series, since the change of angle in a thin layer of the medium should be a small quantity. $\endgroup$
    – CuriousOne
    Sep 10 '14 at 5:34
  • $\begingroup$ I find it is even difficult to set up a differential equation to solve for such reflected ray per Snell's law. $\endgroup$ Sep 10 '14 at 5:36
  • $\begingroup$ It proves to be a failure. I tried to use Snell's law for a small layer of the media $dx$, and then tried to obtain a differential equation based on the relations. It does not work. $\endgroup$ Sep 10 '14 at 5:54
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    $\begingroup$ Try looking up information on GRIN fibers and lenses (Gradient Index) , e.g. en.wikipedia.org/wiki/Gradient-index_optics $\endgroup$ Sep 10 '14 at 11:43
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    $\begingroup$ Just a quick not e that the relationship between Snell's law and Fermat's principle is a subtle one when the index of refraction varies continuously. In general $n(x) \sin(\theta(x))$ is not a conserved quantity. iopscience.iop.org/article/10.1088/0143-0807/37/2/025301/meta $\endgroup$ Mar 22 '19 at 19:47
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This may (or may not) lead to the same answer as CuriousOne's suggestion above, but the most appropriate (and the longest) way of attempting a solution would to be to employ the Fermat's principle. The method's nicely described in the link, but in a nutshell, you would be led to a condition of the type $$\delta \int n ds = 0$$ where this $ds$ can be cast in terms of your 2D co-ordinates. Now, substitute for the spatial dependence of $n$ and arrive at $$\delta \int n(x,y) \sqrt{(1+(dy/dx)^2)} dx = 0$$

This is a sort of an ab-initio approach. I won't be surprised if there's a shorter method (maybe CuriousOne's suggestion.)

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    $\begingroup$ I can't imagine why this wouldn't lead to identical results. Snell's law is an application of Fermat's principle, after all. The only reason I suggested starting with Snell is because it is usually taught before Fermat, if I remember correctly. $\endgroup$
    – CuriousOne
    Sep 10 '14 at 6:02
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    $\begingroup$ Nothing wrong with that. I think what might throw him off about Snell are the two sines, one of which should become a cosine trough the first order series approximates, after which we probably end up with a tangens, which, I believe can be expressed by the square root term in your version... but now I am just guessing. $\endgroup$
    – CuriousOne
    Sep 10 '14 at 6:07
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    $\begingroup$ Googling leads me to this, but please note that I haven't seen the video myself. So, don't blame me if it isn't fine :) $\endgroup$
    – 299792458
    Sep 10 '14 at 6:18
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    $\begingroup$ thank you! I remember in Feynman's lectures on physics, he also mentioned such possibility. PS: I have no access to youtube/facebook/twitter, because we have a very big firewall :( $\endgroup$ Sep 10 '14 at 6:20
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    $\begingroup$ Based on the Wikipedia article on Snell's law it seems to have been around a lot earlier than Fermat's principle... which means that some folks had the right intuition based on geometric arguments, already. I did the first part of the calculation... and yes, the taylor series for one of the sines leads to a tangens term. The bad news is, that it expresses the problem in dn, not dx, which requires another transformation and integration over the inverse of n(x)... which may be a backwards way of calculating the result. $\endgroup$
    – CuriousOne
    Sep 10 '14 at 6:20
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The derivation of the "equations of motion" for the light ray from the Fermat principle is given in the book "Reflections on Relativity", chapter 8.4 "Refractions on Relativity".

We know that the index of refraction $n$ at a point $(x,y)$ equals $c/v$, where $v$ is the velocity of light at that point. Thus, if we parameterize the path by the equations $x = x(u)$ and $y = y(u)$, the "optical path length" from point $A$ to point $B$ (i.e., the time taken by a light beam to traverse the path) is given by the integral

$$L=\int\limits_A^B n\,\mathrm{d}s=\int\limits_A^Bn\sqrt{\dot{x}^2+\dot{y}^2}\,\mathrm{d}u$$

where dots signify derivatives with respect to the parameter $u$. To make this integral an extremum, let $f$ denote the integrand function

$$f(x,y,\dot{x},\dot{y})=n(x,y)\sqrt{\dot{x}^2+\dot{y}^2}$$

Then the Euler equations (introduced in Section 5.4) are

\begin{align} \frac{\partial n}{\partial x}=\frac{\mathrm d}{\mathrm{d}u}\left(\frac{\partial f}{\partial \dot x}\right) && {} && \frac{\partial n}{\partial y}=\frac{\mathrm d}{\mathrm{d}u}\left(\frac{\partial f}{\partial \dot y}\right) \end{align}

which gives

\begin{align} \frac{\partial n}{\partial x}\sqrt{\dot{x}^2+\dot{y}^2}=\frac{\mathrm d}{\mathrm{d}u}\left[\frac{n\dot x}{\sqrt{\dot{x}^2+\dot{y}^2}}\right] && {} && \frac{\partial n}{\partial y}\sqrt{\dot{x}^2+\dot{y}^2}=\frac{\mathrm d}{\mathrm{d}u}\left[\frac{n\dot y}{\sqrt{\dot{x}^2+\dot{y}^2}}\right] \end{align}

Now, if we define our parameter $u$ as the spatial path length $s$, then we have $\dot{x}^2+\dot{y}^2=1,$ and so the above equations reduce to

$$\frac{\partial n}{\partial x}=\frac{\mathrm d}{\mathrm{d}s}\left(n\frac{\mathrm{d} x}{\mathrm{d} s}\right)\tag{1a}$$ $$\frac{\partial n}{\partial y}=\frac{\mathrm d}{\mathrm{d}s}\left(n\frac{\mathrm{d} y}{\mathrm{d} s}\right)\tag{1b} $$

These are the "equations of motion" for a photon in a heterogeneous medium, as they are usually formulated, in terms of the spatial path parameter $s$.

Now, solving these equations for $x(s)$ and $y(s)$, you'll get your ray curve in your medium $n(x,y)$.

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