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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)

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  1. Astrophysics
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    |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
    • one year ago
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    @ganeshie8 @IrishBoy123 @Michele_Laino @Empty @nincompoop

  6. Michele_Laino
    • one year ago
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    nicely done!! @Astrophysics

  7. Astrophysics
    • one year ago
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    Thanks @Michele_Laino !! :)

  8. Michele_Laino
    • one year ago
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    :)

  9. Astrophysics
    • one year ago
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    |dw:1440139323475:dw| I think that got cut off

  10. IrishBoy123
    • one year ago
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    great idea for a thread and very clear explanation the idea that photons have SatNav has always blown me away!!!

  11. arindameducationusc
    • one year ago
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    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
    • one year ago
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    @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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    How does the light know what path will minimize time before it takes the path?

  16. anonymous
    • one year ago
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    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
    • one year ago
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    i think a cool example that is close to this behavior is seen when water is passing on an inclined plane

  18. dan815
    • one year ago
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    |dw:1440378480870:dw|

  19. dan815
    • one year ago
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    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
    • one year ago
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    something similar happens with light too i think

  21. dan815
    • one year ago
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    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
    • one year ago
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    |dw:1440378687124:dw|

  23. dan815
    • one year ago
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    its like this trying to take the shortest path turns into the fastest path transition period

  24. dan815
    • one year ago
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    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
    • one year ago
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    https://youtu.be/vlxMSz7XRys?t=3m21s

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