Angle of Incidence

[turophile]

Here's the problem:

Light travels with velocity v[sub:1x3kmcay]1[/sub:1x3kmcay] from a point P above the surface of a lake to a point S on the surface, and from there with velocity v[sub:1x3kmcay]2[/sub:1x3kmcay] to a point Q below the surface. (The angle of incidence above the surface is ?[sub:1x3kmcay]1[/sub:1x3kmcay] and the angle below the surface is ?[sub:1x3kmcay]2[/sub:1x3kmcay].) Show that the time required for the light to travel from P to Q will be a minimum if S is chosen so that (sin ?[sub:1x3kmcay]1[/sub:1x3kmcay])/(sin ?[sub:1x3kmcay]2[/sub:1x3kmcay]) = v[sub:1x3kmcay]1[/sub:1x3kmcay]/v[sub:1x3kmcay]2[/sub:1x3kmcay].

My work so far:

I've drawn a picture of the triangles PAS and SBQ, where A is the point directly above S such that PA is perpendicular to AS, and B is the point directly below S such that QB is perpendicular to SB. I've also marked the angles of incidence on the triangles.

I know that distance (d) is velocity (v) multiplied by time (t), so d = vt and t = d/v. Now I'm stuck coming up with an equation f(t) that I can differentiate with respect to t and find the answer to the problem. I could use some help getting the necessary equation(s) set up.

 

[galactus]

This is called Snell's refraction.

The total time for the light to travel from P to S to Q is

\(\displaystyle t=(\text{time from P to S})+(\text{time from S to Q})=\frac{\sqrt{x^{2}+a^{2}}}{v_{1}}+\frac{\sqrt{(c-x)^{2}+b^{2}}}{v_{2}}\)

\(\displaystyle \frac{dt}{dx}=\frac{x}{v_{1}\sqrt{x^{2}+a^{2}}}-\frac{c-x}{v_{2}\sqrt{(c-x)^{2}+b^{2}}}\)

But, \(\displaystyle \frac{x}{\sqrt{x^{2}+a^{2}}}=sin{\theta}_{1}\)

and \(\displaystyle \frac{c-x}{\sqrt{(c-x)^{2}+b^{2}}}=sin{\theta}_{2}\)

Thus, therefore, and whence \(\displaystyle \frac{dt}{dx}=\frac{sin{\theta}_{1}}{v_{1}}-\frac{sin{\theta}_{2}}{v_{2}}\)

So, \(\displaystyle \frac{dt}{dx}=0\), when \(\displaystyle \frac{sin{\theta}_{1}}{v_{1}}=\frac{sin{\theta}_{2}}{v_{2}}\)

 

[turophile]

Aha! Thanks galactus.