## Astrophysics one year ago Fun question that requires some knowledge of calculus of variations, thought I'd derive. (Fermat's principle, Snell's law)

1. Astrophysics

Consider light passing from one medium with index of refraction $$n_1$$ into another medium with index of refraction $$n_2$$ (Will draw figure). Use Fermat's principle to minimize time, and derive the law of refraction: $$n_1 \sin \theta_1 = n_2 \sin \theta_2$$ |dw:1440132161535:dw|

2. Astrophysics

To minimize time, we know light travels in a straight line, so we have $t = \frac{ s }{ v }$ and we have $v = \frac{ c }{ n }$ where c is the speed of light in a vacuum and v is the speed of light in a medium and n is the index of refraction. $t = \int\limits_{A}^{B} \frac{ ds }{ v }~~~~\text{where}~~~~ v=\frac{ c }{ n }$ $ds = \sqrt{dx^2+dy^2} \implies \sqrt{1+y'^2} = f$$\frac{ \partial f }{ \partial y } = 0$ $\frac{ \partial f }{ \partial y' } = \frac{ 1 }{ 2 }\frac{ 2y' }{ \sqrt{1+y'^2} }= \frac{ y' }{ \sqrt{1+y'^2} }=C$ $y'^2 = C(1+y'^2) = C+Cy'^2 \implies C = (1-C)y'^2$ $y'^2 = \frac{ C }{ 1-C } \implies constant$ letting y' = m we then have $y=mx+b$ which we conclude is the quickest way for light to travel is a straight line, it's alright if you don't follow the math entirely I will try explaining it later haha just wanted to get this point across.

3. Astrophysics

Now the funner stuff, deriving Snell's law $$n_1 \sin \theta_1 = n_2 \sin \theta_2$$ |dw:1440133592210:dw| I know there's a lot going on in this drawing and it's not the greatest but I hope it will get the point across haha.

4. Astrophysics

|dw:1440133766815:dw| hence we have $t = \frac{ n_1 }{ c }\sqrt{h_{A}^2+x^2}+\frac{ n_2 }{ c }\sqrt{(l-x)^2+h_{B}^2}$ as the light is going through a medium the velocity must be $\frac{ c }{ n_1 }~~~\text{and}~~~\frac{ c }{ n_2 }$ now we take the derivative and set it equal to 0. $\frac{ dt }{ dx } = 0$ $\frac{ d }{ dx }\left( \frac{ n_1 }{ c } \sqrt{h_{A}^2+x^2}+\frac{ n_2 }{ c }\sqrt{(l-x)^2+h_{B}^2}\right)=0$ ...simplifying after taking the derivative we get $\frac{ n_1x }{ c \sqrt{h_{A}^2+x^2} }-\frac{ n_2(l-x) }{ c \sqrt{(l-x)^2+h_{B}^2} }=0$ and using the beautiful triangle I made haha...we have $\sin \theta_1 = \frac{ x }{ \sqrt{h_{A}^2+x^2} }$ and $\sin \theta_2 = \frac{ (l-x) }{ \sqrt{(l-x)^2+h_{B}^2} }$ Setting $\frac{ n_1 }{ c }\frac{ x }{ \sqrt{h_{A}^2+x^2} } = \frac{ n_2 }{ c }\frac{ (l-x) }{ \sqrt{(l-x)^2+h_{B}^2} }$ we then get...$n_1 \sin \theta_1 = n_2 \sin \theta_2$ hopefully I did it correctly :P well that was fun.

5. Astrophysics

@ganeshie8 @IrishBoy123 @Michele_Laino @Empty @nincompoop

6. Michele_Laino

nicely done!! @Astrophysics

7. Astrophysics

Thanks @Michele_Laino !! :)

8. Michele_Laino

:)

9. Astrophysics

|dw:1440139323475:dw| I think that got cut off

10. IrishBoy123

great idea for a thread and very clear explanation the idea that photons have SatNav has always blown me away!!!

11. arindameducationusc

I couldn't understand from ds= sqrt(dx^2+dx^2) sqrt(1+y^2)=f Can you explain me @Astrophysics? But nice job for the derivation...

12. Astrophysics

@arindameducationusc $ds = \sqrt{(dx)^2+(dy)^2} \implies \sqrt{(dx)^2\left( 1+\frac{ (dy)^2 }{ (dx)^2 } \right)} \implies \sqrt{1+\left( \frac{ dy }{ dx } \right)^2}dx$ $\implies ds= \sqrt{1+y'^2} dx$ it's really just the arc length

13. Astrophysics

I kind of skipped some steps but essentially when we are done integrating $\frac{ n }{ c } \int\limits_{x_1}^{x_2} \sqrt{1+y'^2}dx \implies f = \sqrt{1+y'^2}$ and we have $\frac{ \partial d }{ \partial y } + \frac{ d }{ dx }\left( \frac{ \partial f }{ \partial y' }\right)=0$

14. arindameducationusc

Can you give me some video lectures on this? Fermat's principle is very new to me.... Is it needed in quantum physics?

15. Kainui

How does the light know what path will minimize time before it takes the path?

16. anonymous

Good question, from the Fermat's principle we have a ray of light prefers the path such that there are other paths, arbitrarily nearby on either side, along which the ray would take almost exactly the same time to traverse.

17. dan815

i think a cool example that is close to this behavior is seen when water is passing on an inclined plane

18. dan815

|dw:1440378480870:dw|

19. dan815

all to make sure the water at the end is about the same as the water that enter this surface area |dw:1440378616487:dw|

20. dan815

something similar happens with light too i think

21. dan815

it doesnt initially just take that path u see according to snells law, there is a very small time increment where it eventually goes to that path, just like water waves

22. dan815

|dw:1440378687124:dw|

23. dan815

its like this trying to take the shortest path turns into the fastest path transition period

24. dan815

its not really considered in stuff like lens law, but this stuff is considered when you are trying to quantum physics experiments and light capturing stuff

25. nincompoop