# Assume light travels from point a to b. Find the total time to get from a to b

1. Dec 8, 2011

### NWeid1

1. The problem statement, all variables and given/known data
Suppose that light travels from point A to point B as shown in the figure/ Assume that the velocity of light above the boundary line is v1 and the velocity above the boundary line is v2. Find the total time T(x) to get from point A to point B, Write out the equation T'(x) = 0, replace the square roots using the sines of the angles in the figure and derive Snell's Law $$\frac{sinθ_1}{sinθ_2} = \frac{v_1}{v_2}$$

2. Relevant equations

3. The attempt at a solution
Here is the work I've done so far.

$$T(x) = \frac{\sqrt{1+x^2}}{v_1} + (-\frac{\sqrt{1+(2-x)^2}}{v_2})$$
$$T'(x) = \frac{1}{v_1\sqrt{(1+x^2)}} - \frac{1(x-2)}{v_2(\sqrt{(1+(2-x)^2)}} = 0$$
$$T'(x) = \frac{1}{v_1\sqrt{(1+x)^2}} = \frac{x-2}{v_2\sqrt{1+(2-x)^2}}$$

And I know I have to get

$$\frac{sinθ_1}{sinθ_2} = \frac{v_1}{v_2}$$

and I replaced the fractions with sines but the (x-2) on the right side of the equation messes up the formula, so now I'm confused.

Also!!! Is my latex working for you guys? What did I do wrong? lol

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Last edited: Dec 8, 2011
2. Dec 8, 2011

You confused the parser by using bv code (all those and tags) inside the $$tag. 3. Dec 8, 2011 ### Simon Bridge I think you have forgotten to find x so that T(x) is a minima. (assuming x is the point the light hits the interface, A=(0,1) and B=(2,-1)?) 4. Dec 8, 2011 ### NWeid1 Wouldn't making T'(x) = 0 be finding a minima? 5. Dec 8, 2011 ### NWeid1 And what is the right way to type up this function then? haha 6. Dec 8, 2011 ### NWeid1 Oh okay, so finding where the function = 0 would be finding the min. so before i use the sine functions I should solve for x? 7. Dec 8, 2011 ### Simon Bridge Sorry, misread .. you did do that step. You have a pic with two triangles, and the normal drawn in and the angles? Then, assuming A=(0,1) and B=(2,-1): [tex]\sin(\theta_1) = \frac{x}{\sqrt{1+x^2}}$$

you do it for $\sin(\theta_2)$

Aside:
Sub and superscripts in LaTeX are like this: x_1^{23}would be $x_1^{23}$
For more than one character you have to enclose them in curly brackets.

Last edited: Dec 8, 2011
8. Dec 8, 2011

### NWeid1

Also, I am soooooo stupid. There is a picture I forgot to attach. Wow, sorry anyone who was confused by this lol. The pic is up now.

9. Dec 8, 2011

### Simon Bridge

No worries, I stuffed up the sine in prev post.
You should be able to find the other one now.

In your last two equations, how come (x-2) instead of (2-x)? How come you get a 1 in the numerator of the first term in T' and not an x?

In the last equation, as written, does not make sense ... presumably you mean:

$$T'(x) = 0 \Rightarrow \frac{x}{v_1\sqrt{(1+x)^2}} = \frac{2-x}{v_2\sqrt{1+(2-x)^2}}$$

Last edited: Dec 8, 2011
10. Dec 8, 2011

### NWeid1

Why would

$$sin(\theta_1) = \frac{x}{\sqrt{1+x^2}}$$

and not just

$$sin(\theta_1) = \frac{1}{\sqrt{1+x^2}}$$ ?

11. Dec 8, 2011

### Simon Bridge

Look at the diagram.

By definition: $\sin \theta = \frac{O}{H}$
The hypotenuse has length $\sqrt{1+x^2}$ because the opposite side has length $x$.

(the angle is pointing the other way to how you are used to - $1/\sqrt{1+x^2} = \cos \theta$.)

12. Dec 8, 2011

### NWeid1

Ooooh ok I got it. So

$$\sin(\theta_2) = \frac{2-x}{\sqrt{1+(2-x)^2}}$$

? or would it be negative

13. Dec 8, 2011

### NWeid1

Ok no I got it it's $$\sin(\theta_2) = \frac{2-x}{\sqrt{1+(2-x)^2}}$$

so then I get $$\frac{\sin{\theta_1}}{v_1} = \frac{\sin{\theta_2}}{v_2}$$
which means we would get
$$\frac{sin{\theta_1}}{sin{\theta_2}} = \frac{v_1}{v_2}$$

But now I am confused after reading it again and again...What am I exactly trying to find? The total time, right? So how do I use this formula it asked me to get to find the total time? Lol, I'm lost right now

14. Dec 8, 2011

### Simon Bridge

Weren't you supposed to derive Snell's Law?
Isn't this what you have?

Well done!

An observation and a lesson:

In general, light always takes the path of least time. This is just a specific example.

Get the geometry right, everything else takes care of itself.

15. Dec 8, 2011

### NWeid1

Ok, so, how can I show that this will be the minimum time to travel from A to B?

16. Dec 8, 2011

### Simon Bridge

What did you find T'(x) for?

17. Dec 9, 2011

### HallsofIvy

Staff Emeritus
a minimum. "Minima" is the plural of "minimum".

18. Dec 9, 2011

### NWeid1

Haha sorry, didn't mean to break the calculus rules...-.-