Fermat s Principle. Fermat s Principle states that a ray of light in a medium will follow the path which takes the least amount of time.

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1 Homework

2 Fermat s Principle Fermat s Principle states that a ray of light in a medium will follow the path which takes the least amount of time.

3 Solution: The traversal time for the path is T = where ds is an element of arclenght. Using 2 ds v ds = + y! 2 dx We have T[y] = x 2 x + y! 2 v(x) dx

4 Solution. Euler s equation is f y d dx f y " = with f ( y, y!;x) = + y! 2 / v(x) Since f / y = we have with a a constant. f y " = y" v(x) +! y 2 = a

5 Solution. From the figure we see that y =dy/dx=tanθ on each side, so equation above we have y! = v(x) + y! 2 Simplifying, we obtain Snell s law. tanθ v + tan 2 θ = tanθ 2 v 2 + tan 2 θ 2

6 Solution. From We have y! v(x) + y! 2 y! = av = a a 2 v 2 y! = When x=, v=c and dy/dx=, a=/(2) /2 c. So we have y! = + 2x / l y(x) = v / c 2 v 2 / c 2 dx x = l # + 2x / l & + 2x / l $% '(

7 Light reflection on a smooth plane y v(x) + y 2 = a tan θ + tan 2 θ = tan θ 2 + tan 2 θ 2 cosθ = cosθ 2 θ = θ 2

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9 Light reflection on a smooth spherical surface y! v(x) + y! 2 = a x P = Rsinθ, y P = Rcosθ y! AP = Rcosθ R Rsinθ, y! PB = Rcosθ R + Rsinθ + y! 2 = 2 / (± sinθ) sinθ = + sinθ sinθ = sinθ θ =

10 Light reflection on a smooth spherical surface l(θ) = 2Rsin[( π 2 θ) / 2]+ 2Rsin[(π 2 +θ) / 2] l(θ) = 4Rsin π 4 cosθ 2 l(θ) = 2R + cosθ θ = l max

11 Light reflection on a smooth spherical surface l = 2R 2 2R 2 cos( π 2 θ) l = 2R 2 2R 2 cos( π 2 +θ) l(θ) = 2R( sinθ + + sinθ ) l 2 = 2R 2 (2 + 2 (+ sinθ)( sinθ)) l 2 = 4R 2 (+ cosθ) θ = l max

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13 Fermat s Principle In its simplest form it consists in the assertion that a ray of light between two points in space always follows the path along which its travel time is less than along any of the other paths connecting the points. In its more rigorous formulation Fermat s principle is a variational principle asserting that a real ray of light from one point to another follows the path along which its travel time either is an extremum in comparison with the travel times along the other paths joining the points or is the same for all paths. In other words, the optical path of a ray may be not only less than but also greater than or equal to the other possible paths connecting the points.

14 Fermat s Principle and Brachistochrone Problem T = ds = v(x) + y' 2 v(x) dx y' v + y' 2 = a v(x) = 2gx

15 3.6 Cycloid pendulum (from Greiner p36). A mass point glides without friction on a cycloid, which is given by x = a(! " sin!) and y = a(+ cos!) (with! "! 2# ). Determine (a) the Lagrangian and (b) the equation of motion. (c) Solve the equation of motion.

16 3.6 Cycloid pendulum (from Greiner p36). A mass point glides without friction on a cycloid, which is given by x = a(! " sin!) and y = a(+ cos!) (with! "! 2# ). Determine (a) the Lagrangian and (b) the equation of motion. (c) Solve the equation of motion. (a) L = 2 m( x2 + y 2 ) mgy = ma 2 ( cosθ) θ 2 mga(+ cosθ) (b) ( cosθ) θ + ( 2 sinθ ) θ 2 g 2a sinθ = (c) u = cos θ 2 u = C cos u 2 (4a u + gu) = u + g 4a u = g 4a t + C 2 sin g 4a t

17 The Brachistochrone T = + y' 2 2gx dx d dx f y' = f y' = (4ag) /2 y' 2 x(+ y' 2 ) = 2a y = xdx (2ax x 2 ) /2 (x,y ) y x = a( cosθ) y = a(θ sinθ) x ds v (x,y )

18 The Brachistochrone T = + y' 2 2gx dx x = a( cosθ) y = a(θ sinθ) y θ θ 2du " u 2 u = 2arctan u u 2 u $ 2 2 # $ u 2 u 2 x % ' &' u π

19 Example : The Brachistochrone T = + y' 2 2gx dx x = a( cosθ) y = a(θ sinθ) T = dx 2 + dy 2 2gx dx = asinθdθ dy = a( cosθ)dθ = (asinθ) 2 + a 2 ( cosθ) 2 2ga( cosθ) dθ = a g dθ = a g θ

20 Example : The Brachistochrone T = = + c2 2g + y' 2 2gx dx x 2 dx y = cx = θ sinθ cosθ x x = a( cosθ) y = a(θ sinθ) = 2(+ c2 ) g x 2 y = 2a g ( cosθ ) 2 + (θ sinθ ) 2 ( cosθ ) x

21 The Brachistochrone T = 2a g ( cosθ ) 2 + (θ sinθ ) 2 ( cosθ ) T = a g θ T π = a g 4 + π 2 y x T 3π /2 = a g 2 + 2(3π 2 +)2

22 The Brachistochrone y E = 2 mv2 mgx = mgx T = = dx 2 + dy 2 2g(x x ) (asinθ) 2 + a 2 ( cosθ) 2 dθ 2ga(cosθ cosθ) x x = a( cosθ) y = a(θ sinθ) = a g cosθ dθ cosθ cosθ T π = a g π, <θ < π

23 The Brachistochrone u = cos θ 2 cosθ θ cosθ cosθ dθ = θ = θ θ 2du! u 2 u = 2arctan u u 2 u # 2 2 "# u 2 u 2 θ θ = 2(arctan[] arctan[ ]) = 2( π / 2) = π sin(θ / 2)dθ cos 2 (θ / 2) cos 2 (θ / 2) $ & %& u

24 Huygens Pendulum Clock