# Homework Help: Using Snell's Law for Brachistochrone Project

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1. May 8, 2017

### zengodspeed

Hi all,

I'm after a little guidance for I do not know what is going wrong.

I understand that for Johann Bernoulli's proof of the brachistochrone problem he used refraction of light and Fermat's principle of least time.

I have decided to do a project on the subject, in which I am dividing up a vertical displacement into media of differing refractive index as a function of velocity (for this example, lets say 10m into 10 sections 1m in height). With this I am finding the velocity of an object at each point by of crossover:

v = sqrt(2gy)

and then plugging the values in velocity into Snell's law to find the angle at which the path changes at each change in medium.

Now, for the first first change (at Δy = 1) the equation works out to be

sinΘ2 = (v2sinΘ1)/(v1)

But as the velocity is ever increasing at each point is always

sinΘ2 > 1

Which obviously cannot be right.

So where is it I am going wrong?

Thanks in advance.

Last edited by a moderator: May 8, 2017
2. May 8, 2017

### TSny

Welcome to PF!
Can you show how the equation sinΘ2 = (v2sinΘ1)/(v1) implies sinΘ2 > 1?

3. May 9, 2017

### zengodspeed

For a decreasing index of refraction velocity will continually increase, therefore we know that:

v1<v2

I have realised where my problem lies. At the beginning, when the object first starts to move, the path will be tangent to the vertical, which implies that:

sinΘ1 = 0

I cannot work with this number, so must I begin my values from a later point?

Thanks

4. May 9, 2017

### TSny

sinθ/v = c where c is a constant. Different values of c correspond to different paths. So, you need to pick a value for c to get a definite path.

When you are finding an approximate solution by using a finite number of horizontal layers, I think there are different ways you can approximate the path.

You could choose the path to be vertical for the first horizontal layer. Then find θ for the second layer using sinθ/v = c where v is the speed at the beginning of the second layer. Then proceed to the third layer, etc.

Or, for each layer, you could take v to be the speed at the center of the layer. Then, for the first layer, you would have a nonzero v and you would choose θ for the first layer to satisfy sinθ/v = c. In this case, the path is not vertical in the first layer. (But for thin layers, it should be approximately vertical.)

5. May 9, 2017

### zengodspeed

That's great, thanks for all the help!

I have solved for the intervals I want and have found the slope for each section.

This is a little off topic but do you know of any graphing software where I would be able to input the slope values and the y coordinates to make a sketch of the curve?

Thanks in advance.

6. May 9, 2017

### TSny

I don't know of such software. However, you can use your slope and y values to calculate (x,y) points on the curve at the boundaries of the layers. Then you can plot these (x, y) points.

7. May 9, 2017

### zengodspeed

Okay thanks, it seems that's the way to go.

Thanks for all the help.

8. May 9, 2017

### TSny

OK. Have fun with your project.

9. May 12, 2017

### zengodspeed

Thanks for the help again, it's going great.

Just one more question. We have the graph now, and we need to find a way to check the validity of our results.
Our first thoughts were to find the equation of the curve (being a cycloid) and check for the error margin of our results. However, the math seems too intensive for us at present (currently studying single var calculus). I was wondering if you have any other ideas of how to go about checking for validity?

thanks

10. May 12, 2017

### TSny

You can express the Snell's law formula in the form $\frac{\sin \theta}{\sqrt y} = C$ for some constant $C$.
The cycloid may be written in parametric forms as $x = R(\phi - \sin \phi)$, $y = R(1-\cos \phi)$.

Determine the relation between $C$ and $R$. Then, for a choice of $C$ you can plot your Snell's law approximation and the cycloid on the same graph to give a nice visual comparison. You can see how the graphs compare for different numbers of layers that you choose for the Snell's law approximation.

You could also compare a specific point $(x_s, y_s)$ of the Snell's law approximation with the corresponding point on the cycloid. Consider the point on the cycloid that has a y value, $y_c$, equal to $y_s$. Solve $y_s = R(1-\cos \phi)$ for $\phi$ and then find the value of $x_c$ on the cycloid using $x_c = R(\phi - \sin \phi)$. Then you can compare $x_c$ with $x_s$.

11. May 12, 2017

### zengodspeed

Hi, thanks again for the help.

We have been working at the problem this afternoon but we don't understand how to relate $C$ and $R$. We have managed to find a parametric equation that looks very much like our curve by playing around with the variables, so we have a decent estimate of the magnitude of the equation but it is not at all precise. Could you please explain the relation a little more in depth?

Thanks

12. May 12, 2017

### TSny

To relate $R$ and $C$, note that the maximum value that $y$ can attain must be the same for the Snell's law formula and the cycloid. Use the Snell's law formula to relate $C$ to $y_{max}$. Use the cycloid formula to relate $R$ to $y_{max}$.

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