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

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.15Consider 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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.15To 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. \[t = \int\limits_{A}^{B} \frac{ ds }{ v }~~~~\text{where}~~~~ v=\frac{ c }{ n }\] \[ds = \sqrt{dx^2+dy^2} \implies \sqrt{1+y'^2} = f\]\[\frac{ \partial f }{ \partial y } = 0 \] \[\frac{ \partial f }{ \partial y' } = \frac{ 1 }{ 2 }\frac{ 2y' }{ \sqrt{1+y'^2} }= \frac{ y' }{ \sqrt{1+y'^2} }=C\] \[y'^2 = C(1+y'^2) = C+Cy'^2 \implies C = (1C)y'^2 \] \[y'^2 = \frac{ C }{ 1C } \implies constant\] 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.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.15Now 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.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.15dw:1440133766815:dw hence we have \[t = \frac{ n_1 }{ c }\sqrt{h_{A}^2+x^2}+\frac{ n_2 }{ c }\sqrt{(lx)^2+h_{B}^2}\] 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. \[\frac{ dt }{ dx } = 0\] \[\frac{ d }{ dx }\left( \frac{ n_1 }{ c } \sqrt{h_{A}^2+x^2}+\frac{ n_2 }{ c }\sqrt{(lx)^2+h_{B}^2}\right)=0\] ...simplifying after taking the derivative we get \[\frac{ n_1x }{ c \sqrt{h_{A}^2+x^2} }\frac{ n_2(lx) }{ c \sqrt{(lx)^2+h_{B}^2} }=0\] and using the beautiful triangle I made haha...we have \[\sin \theta_1 = \frac{ x }{ \sqrt{h_{A}^2+x^2} }\] and \[\sin \theta_2 = \frac{ (lx) }{ \sqrt{(lx)^2+h_{B}^2} }\] Setting \[\frac{ n_1 }{ c }\frac{ x }{ \sqrt{h_{A}^2+x^2} } = \frac{ n_2 }{ c }\frac{ (lx) }{ \sqrt{(lx)^2+h_{B}^2} }\] we then get...\[n_1 \sin \theta_1 = n_2 \sin \theta_2 \] hopefully I did it correctly :P well that was fun.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.15@ganeshie8 @IrishBoy123 @Michele_Laino @Empty @nincompoop

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0nicely done!! @Astrophysics

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.15Thanks @Michele_Laino !! :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.15dw:1440139323475:dw I think that got cut off

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1great idea for a thread and very clear explanation the idea that photons have SatNav has always blown me away!!!

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0I 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...

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.15@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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.15I 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 \[\frac{ \partial d }{ \partial y } + \frac{ d }{ dx }\left( \frac{ \partial f }{ \partial y' }\right)=0\]

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Can you give me some video lectures on this? Fermat's principle is very new to me.... Is it needed in quantum physics?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.0How does the light know what path will minimize time before it takes the path?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Good 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.`

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i think a cool example that is close to this behavior is seen when water is passing on an inclined plane

dan815
 one year ago
Best ResponseYou've already chosen the best response.0all to make sure the water at the end is about the same as the water that enter this surface area dw:1440378616487:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.0something similar happens with light too i think

dan815
 one year ago
Best ResponseYou've already chosen the best response.0it 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.0its like this trying to take the shortest path turns into the fastest path transition period

dan815
 one year ago
Best ResponseYou've already chosen the best response.0its 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
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