Which function do we want to minimize?

Question

A ray of light travels from the point $A$ to the point $B$ across the border between two materials. At the first material the speed is $v_1$ and at the second it is $v_2$. Show that the journey is achieved at the least possible time when Snell's law: $$\frac{\sin \theta_1}{\sin \theta_2}=\frac{v_1}{v_2}$$ holds.

To show that do we have to use Lagrange multipliers method??

But which function do we want to minimize??

Answer 1

The total travel time, which is the sum of travel times from $A$ to the surface and fromthe surface to $B$. The two part times are obtained as length divided by speed, the two lengths depend on the angles.

Answer 2

Hint:

Use Fermat's principle.

the path taken between two points by a ray of light is the path that can be traversed in the least time.

And you don't need Lagrange multipliers.

Answer 3

Without Lagrange multipliers

You don't need to use Lagrange multipliers. Snell's law can be derived directly from Fermat's principle of least time, which consists in minimizing the time taken by the ray of light to travel from one point to another. Take a look at the diagram below:

In the first medium, with refractive index $n_1$, light will travel at velocity $v_1=\dfrac{c}{n_1}$. In the second medium, at $v_2=\dfrac{c}{n_2}$. Therefore, the total time from point $P$ to $Q$ is $T(x)=\dfrac{d_1}{v_1}+\dfrac{d_2}{v_2}$.

To find $d_1$ and $d_2$, the distances traveled in media $1$ and $2$ respectively, you can simply use the Pythagorean theorem: $$d_1=\sqrt{x^2+a^2},\quad d_2=\sqrt{(l-x)^2+b^2}$$ Thus: $$T(x)=\dfrac{\sqrt{x^2+a^2}}{v_1}+\dfrac{\sqrt{(l-x)^2+b^2}}{v_2}$$ Applying Fermat's principle, you then take minimize $T(x)$: $$\dfrac{dT}{dx}=\dfrac{x}{v_1\sqrt{x^2+a^2}}+\dfrac{x-l}{\sqrt{(l-x)^2+b^2}}=0$$ Using trigonometry: $$\dfrac{x}{\sqrt{x^2+a^2}}=\sin\theta_1,\quad \dfrac{l-x}{\sqrt{(1-x)^2+b^2}}=\sin\theta_2$$ Therefore: $$\dfrac{dT}{dx}=0\Longleftrightarrow \dfrac{\sin\theta_1}{v_1}=\dfrac{\sin\theta_2}{v_2}$$ Using the aforementioned relation between the speed of light in a given medium and the corresponding refractive index, we recover Snell's law: $$\dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{n_2}{n_1}$$

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With Lagrange multipliers

We try to minimize the time taken by the ray of light to travel from $P$ to $Q$, which we can express as: $$c\cdot T(\theta_1,\theta_2)=\dfrac{an_1}{\cos\theta_1}+\dfrac{bn_2}{\cos\theta_2}$$ (You can check that this is equivalent to $T(x)$ in the previous part). We also derive the following constraint, from the fixed horizontal distance between $P$ and $Q$: $$l=x+(l-x)=a\tan\theta_1+b\tan\theta_2$$ We can therefore write down the Lagrangian: $$L(\theta_1,\theta_2,\lambda)=\dfrac{an_1}{\cos\theta_1}+\dfrac{bn_2}{\cos\theta_2}+\lambda(a\tan\theta_1+b\tan\theta_2-l)$$ And we solve the Lagrange equations for $\lambda$: $$\dfrac{\partial L}{\partial\theta_1}=0\Longrightarrow\dfrac{an_1\sin\theta_1+a\lambda}{\cos^2\theta_1}=0\Longrightarrow -\lambda=n_1\sin\theta_1$$ $$\dfrac{\partial L}{\partial\theta_2}=0\Longrightarrow\dfrac{bn_2\sin\theta_2+b\lambda}{\cos^2\theta_2}=0\Longrightarrow -\lambda=n_2\sin\theta_2$$ And indeed, combining the two results above yields Snell's law: $$\dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{n_2}{n_1}$$

Answer 4

Let the total horizontal distance travelled be $a$. The light travels in the first medium a distance $d_{1}$ at an angle $\theta _{1}$ to the vertical. Then if the horizontal distance traveled in this step is $x$ and the vertical distance traveled is $h$, the time it takes to reach the border is

$t_{1}=\frac{\left ( x^{2}+h^{2} \right )^{1/2}}{v_{1}}$.

From there, the light enters the second medium has velocity $v_{2}$, makes an angle $\theta _{2}$ to the vertical, and travels a distance $d_{2}$. The horizontal distance travelled is $a-x$ and the vertical distance travelled is $y$, so that

$t_{2}=\frac{\left ( (a-x)^{2}+y^{2} \right )^{1/2}}{v_{2}}$.

The trick here is to observe that we may assume that only $x$ varies. i.e. that $h$ and $y$ are constant.

Then, put $t=t_{1}+t_{2}$ and minimize this function as a function of the single variable $x$. There is some messy algebra involved and you will use the relations

$\sin \theta _{1}=\frac{x}{d_{1}}$

and

$\sin \theta _{2}=\frac{a-x}{d_{2}}$.