SICM Exercise 1.03
Solution to exercise 1.3 of Structure and Interpretation of Classical Mechanics by Gerald Jay Sussman and Jack Wisdom.
Fermat optics \(\require{gensymb} \)
🛈 Note
Fermat observed that the laws of reflection and refraction could be accounted for by the following facts: Light travels in a straight line in any particular medium with a velocity that depends upon the medium. The path taken by a ray from a source to a destination through any sequence of media is a path of least total time, compared to neighboring paths. Show that these facts imply the laws of reflection and refraction.
Reflection
Let the reflective surface be the \(x\) axis of the coordinate system with the reflection point at \((x, 0)\). The points have the coordinates \((a_x, a_y)\) and \((b_x, b_y)\). The tangents of the angles are
\[\tan(\alpha) = \frac{a_x + x}{a_y}\]\[\tan(\beta) = \frac{b_x - x}{b_y}\]The distance traveled by the light is
\[S = \sqrt{{{(a_x + x)}^2 + a_y^2}} + \sqrt{{(b_x - x)}^2 + b_y^2}\]To find the minimum we take the derivative
\[S' = \frac{a_x + x}{\sqrt{{{(a_x + x)}^2 + a_y^2}}} - \frac{b_x - x}{\sqrt{{(b_x - x)}^2 + b_y^2}} = 0\]\[{(a_x + x)}^2\left({(b_x - x)}^2 + b_y^2\right) = {(b_x - x)}^2\left({(a_x + x)}^2 + a_y^2\right)\]\[\frac{{(b_x - x)}^2}{b_y^2} = \frac{{(a_x + x)}^2}{a_x^2}\]\[tan^2(\beta) = tan^2(\alpha)\]\[\beta = \alpha\]\( \alpha = \beta + k \pi \) would also be a solution, but since the sum of the two angles must be less than \( 180 \degree \), we only have a single solution.
Refraction
Let the boundary between the two media be the \(x\) axis of the coordinate system. The refraction point is at \((x, 0)\). The points have the coordinates \((a_x, a_y)\) and \((b_x, b_y)\). The tangents of the angles are
\[\tan(\alpha) = \frac{a_x + x}{a_y}\]\[\tan(\beta) = \frac{b_x - x}{b_y}\]The distance travelled by the light in the first and the second medium is
\[S_1 = \sqrt{{(a_x + x)}^2 + a_y^2}\]\[S_2 = \sqrt{{(b_x - x)}^2 + b_y^2}\]Thus the total time is
\[T = \frac{S_1}{v_1} + \frac{S_2}{v_2}\]Taking the derivative of this expression
\[DT = \frac{1}{v_1}DS_1 + \frac{1}{v_2}DS_2 = 0\]\[DT = \frac{a_x + x}{v_1 \sqrt{{(a_x + x)}^2 + a_y^2}} - \frac{b_x - x}{v_2 \sqrt{{(b_x - x)}^2 + b_y^2}} = 0\]\[v_2^2{(a_x + x)}^2\left({(b_x - x)}^2 + b_y^2\right) = v_1^2{(b_x - x)}^2\left({(a_x + x)}^2 + a_y^2\right)\]\[v_2^2 \left(1 + \frac{b_y^2}{{(b_x - x)}^2}\right) = v_1^2\left(1 + \frac{a_y^2}{{(a_x + x)}^2}\right)\]\[v_2^2\left(1 + \cot^2(\beta)\right) = v_1^2\left(1 + \cot^2(\alpha)\right)\]Using the trigonometric identity \(1 + \cot^2(\theta) = \csc^2(\theta)\)
\[\frac{v_2^2}{\sin^2(\beta)} = \frac{v_1^2}{\sin^2(\alpha)}\]\[\frac{v_1}{v_2} = \frac{\sin(\alpha)}{\sin(\beta)}\]