00:02
This problem, they asked to demonstrate for ma's principle and show that it reduces to snow's law for light.
00:17
And so the idea is that the path light takes between two points will be the, be the, we'll minimize the time taken.
00:32
So if you have light that wants to go from a to b, it's going to minimize the time that it takes to get from a to b.
00:44
And so now if you just have a single medium where the speed of light is the same, then you're just going to have a straight line.
00:55
But if the light passes from one substance to another where things change, so then you, then you, have a different problem.
01:11
It's not going to necessarily be a straight line.
01:14
And so they want us to show that, again, that we can figure out where this point is here, given the speed of light in these two materials, such that the time it takes to get from here to here is minimized.
01:32
So first we need to look at the distance.
01:35
So we look at this distance here, a to c, and that's square root of a.
01:39
Squared plus x squared.
01:43
So the time that the light takes the light to go from a to c is this distance divided by the velocity of the light in that medium.
01:52
We can take the derivative of that with respect to x so that we can define the path that minimizes it.
02:00
And so with that we wind up with taking the derivative and then using ac again we wind up with the derivative of t with respect to x is x divided by v1 over the distance ac and x over the distance ac is just sign of this angle here so we get this is sign of theta 1 that divided by b1 now we can do the same thing for for the for the distance here and time it takes so that distance i call this y where y is now c this total distance, this total horizontal distance, c minus x.
02:42
So this distance here is squared or b squared plus y squared...