To bridge the gap between Fermat's principle and Snell's law using a graphical animation tool to apply different geometrical configurations.
Light travels along the path that requires the least time.
Light traveling from point \( P \) to point \( Q \) reflects off a mirror so that the angle of incidence equals the angle of reflection.
In the following content, we will see how Fermat's principle implies Snell's law from geometrical and mathematical perspectives.
To prove the implication, we show that if the light travels along the path that requires the least time (Fermat's principle), then we will indeed obtain an angle of incidence that is equal to the angle of reflection (Snell's law).
Click the \(``\)Play\( " \) button to play the animation and observe the speed of the light rays.
The starting and ending position positions is are fixed at \( P(-5,5) \) and \( Q(5,5) \) respectively.
Note that the orange light ray which reflects off the mirror at \( (0,0) \) denotes the shortest path with a distance of \( 14.1 \) units. In this case, the angle of incidence (\( 45^{\circ} \)) equals the angle of reflection (\( 45^{\circ} \)).
The yellow light ray which reflects off the mirror at \((-3,0)\) denotes another path with a distance of \( 14.8 \) units. In this case, the angle of incidence (\( 21.8^{\circ} \)) does not equal the angle of reflection (\( 58^{\circ} \)).
Click the \(``\)Play\( " \) button to play the animation.
Observe that parabolic mirrors focus all parallel light to a single point.
Click the \(``\)Play\( " \) button to play the animation and observe the speed of the light rays.
The starting and ending position positions is are fixed at \( P(2.5,0) \) and \( Q(6,3) \) respectively. The equation of the mirror is given by \( x=0.1y^2.\)
Note that the orange light ray which reflects off the mirror at \( (0.9,3) \) denotes the shortest path with a distance of \( 8.5 \) units. In this case, the angle of incidence (\( 31^{\circ} \)) equals the angle of reflection (\( 31^{\circ} \)).
The yellow light ray which reflects off the mirror at \( (0.4,2) \) denotes another path with a distance of \( 14.8 \) units. In this case, the angle of incidence (\( 21.8^{\circ} \)) does not equal the angle of reflection (\( 31.9^{\circ} \)).
Click the \(``\)Play\( " \) button to play the animation.
Observe that spherical mirrors do not focus all parallel light to a single point.
Now, we will show that Fermat’s principle implies Snell’s law from a mathematical perspective using an example of a curved mirror.
Assume the mirror is in the shape of a graph \( y=f(x) \) and the light travels from \( P(a,b) \) to \( Q(u,v) \) passing through a point \( R(x,f(x)) \)on the mirror. By Pythagorean theorem, the distance travelled is: \begin{equation*} L_P + L_Q \end{equation*} where \begin{equation*} L_P=\sqrt{(x-a)^2+(f(x)-b)^2} \end{equation*} and \begin{equation*} L_Q=\sqrt{(u-x)^2+(v-f(x))^2} .\end{equation*}
Hence, the time required for the light to pass along the path is \begin{equation*} t(x)=\dfrac{1}{c}(\sqrt{(x-a)^2+(f(x)-b)^2}+\sqrt{(u-x)^2+(v-f(x))^2}) \end{equation*} where \( c \) is the speed of light.
To find the minimum time required, we differentiate the above equation on both sides. Then, we have \begin{equation*} c\dfrac{dt}{dx}=\dfrac{(x-a)+(f(x)-b)f'(x)}{L_P} - \dfrac{(u-x)+(v-f(x))f'(x)}{L_Q}. \end{equation*}
The minimum is attained when the derivative is zero, so by setting \( \dfrac{dt}{dx} =0 \), we have the equation \begin{align} \textrm{Equation (1)}: \quad \quad \dfrac{(x-a)+(f(x)-b)f'(x)}{L_P}=\dfrac{(u-x)+(v-f(x))f'(x)}{L_Q}. \end{align}
To see how this equation implies Snell’s law, we consider the angle between the tangent vector \begin{bmatrix} 1 \\ f'(x) \end{bmatrix} with vector \( PR \) \begin{bmatrix} x-a \\ f(x)-b \end{bmatrix} and vector \( RQ \) \begin{bmatrix} u-x \\ v-f(x) \end{bmatrix} respectively.
By the dot product, we have \begin{align} \textrm{Equation (2)}: \quad \quad \cos \, \theta_1=\dfrac{(x-a)+(f(x)-b)f'(x)}{L_P \sqrt{1+(f'(x))^2}} \end{align} and \begin{align} \textrm{Equation (3)}: \quad \quad \cos \, \theta_2=\dfrac{(u-x)+(v-f(x))f'(x)}{L_Q \sqrt{1+(f'(x))^2}}. \end{align}
Substituting Equation (2) and Equation (3) into Equation (1), we have \( \theta_1 = \theta_2 \) and hence the angle of incidence \( i \) equals angle of reflection \( r \). Therefore, we can see that Fermat’s principle implies Snell’s Law.
Can we develop the same solution without using the calculus approach, namely, by using the algebraic approach?