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

[NWeid1]

Homework Statement

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}

Homework Equations

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

 

[Simon Bridge]

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

You confused the parser by using bv code (all those _{and ^{tags) inside the tag.}}

 

[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)?)

 

[NWeid1]

Wouldn't making T'(x) = 0 be finding a minima?

 

[NWeid1]

And what is the right way to type up this function then? haha

 

[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?

 

[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):
\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.

 

[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.

 

[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}}

 

[NWeid1]

Why would

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

and not just

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

 

[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.)

 

[NWeid1]

Ooooh ok I got it. So

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

? or would it be negative

 

[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

 

[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.

 

[NWeid1]

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

 

[Simon Bridge]

What did you find T'(x) for?

 

[HallsofIvy]

Wouldn't making T'(x) = 0 be finding a minima?

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

 

[NWeid1]

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

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