# Application of Differential Equations to Optics and the Brachistochrone Problem

Differential Equations is indeed one of the most important chapter in mathematics. It has a wide range of application in day to day life also. In this article, we have shown you the application of differential equations to optics. Apart from the application of differential equations to optics, we have also discussed the famous brachistochrone problem in this article. Basically, you will see the application of differential equations to optics and the brachistochrone problem.

 Example: 1 Difficulty: Difficult

(a) Light travels in medium $I$ with velocity ${v_1}$ and a denser medium $II$ with velocity ${v_2}$. It has to travel from the point $A(0,a)$ to the point ${\rm{B(c, - b)}}$, as shown below:

Assuming that Fermat’s principle of least time, which tells us that light travels from one path to another along the path requiring the shortest time, is true, find the relation between ${\alpha _{\;1}}$ and ${\alpha _{\;2}}$.

(b) Suppose that instead of two, there are three mediums, with velocities ${v_1},\;{v_2}$ and $\;{v_3}$, and the corresponding angles the light ray makes with the vertical in the different materials being ${\alpha _1},\;{\alpha _2}$ and ${\alpha _3}$. Generalize the result of the previous part to this case.

(c) Suppose that there is a medium with a continuously increasing optical density, so that light takes a ‘smooth’ path, as shown below:

 Generalize the results of the previous two parts to this case.

(d) A wire in the form of a continuous curve joins the point $A(0,0)$ to the point $B(a, - b)$. A bead travels on the wire under the influence of gravity:

What should be the shape of this wire so that the bead traverses the path $A \to B$ in the least amount of time? You can apply Fermat’s principle of least time to this problem.

 Solution1-(a)

The total time required for the journey is

 $T = \dfrac{{\sqrt {{a^2} + {x^2}} }}{{{v_1}}} + \dfrac{{\sqrt {{b^2} + {{\left( {c - x} \right)}^2}} }}{{{v_2}}}$

We have to find that value of $x$ for which $T$ is minimum. Thus,

 $\dfrac{{dT}}{{dx}} = 0 \,\,\,\,\,\, \Rightarrow \,\,\,\,\, \dfrac{x}{{{v_1}\sqrt {{a^2} + {x^2}} }} = \dfrac{{c - x}}{{{v_2}\sqrt {{b^2} + {{\left( {c - x} \right)}^2}} }}$ $\Rightarrow \,\,\,\, \dfrac{{\sin {\alpha _1}}}{{{v_1}}} = \dfrac{{\sin {\alpha _2}}}{{{v_2}}}$

The reader may recall that this is the Snell’s law of redfraction.

 Solution1-(b)
 $\dfrac{{\sin {\alpha _1}}}{{{v_1}}} = \dfrac{{\sin {\alpha _2}}}{{{v_2}}} = \dfrac{{\sin {\alpha _3}}}{{{v_3}}}$
 Solution1-(c)
 $\dfrac{{\sin \alpha }}{v} = {\rm{constant}}$
 Solution1-(d)

From conservation of energy, we have $v = \sqrt {2gy}$.

Also, using Fermat’s principle of least time, we can say that

 $\dfrac{{\sin \alpha }}{v} = {\rm{constant}} = k\left( {{\rm{say}}} \right)$

Additionally, $y' = \tan \beta$

Now,

 $\sin \alpha = \cos \beta = \dfrac{1}{{\sqrt {1 + {{\tan }^2}\beta } }} = \dfrac{1}{{\sqrt {1 + {{\left( {y'} \right)}^2}} }}$ $\Rightarrow \,\,\,\, \dfrac{{\sin \alpha }}{v} = \dfrac{1}{{\sqrt {2gy} \sqrt {1 + {{\left( {y'} \right)}^2}} }} = k$ $\Rightarrow \,\,\,\, y\left( {1 + {{\left( {y'} \right)}^2}} \right) = c$${\rm{another\,constant}}$

This is the differential equation of the path we wish to determine. Separating variables, we have

 $dx = \sqrt {\dfrac{y}{{c - y}}} \;dy$

To solve this differential equation, we use the substitution $\dfrac{y}{{c - y}} \to {\tan ^2}\phi .$ This gives

 $y = c{\sin ^2}\phi ,\;dy = 2c\sin \phi \cos \phi \;d\phi$ $\Rightarrow \,\,\,\, dx = 2c{\sin ^2}\phi \;d\phi = c\left( {1 - \cos 2\phi } \right)d\phi$ $\Rightarrow \,\,\,\, x = \dfrac{c}{2}\left( {2\phi - \sin 2\phi } \right) + \lambda$

Since the curve passes through $(0,0),\,\lambda = 0.$ Thus,

 $x = \dfrac{c}{2}\left( {2\phi - \sin 2\phi } \right),\;y = c{\sin ^2}\phi = \dfrac{c}{2}\left( {1 - \cos 2\phi } \right)$

These relations can be written more simply by using $\dfrac{c}{2} \to p$ and $2\phi \to \theta :$

${x = p\left( {\theta - \sin \theta } \right),\;y = p\left( {1 - \cos \theta } \right)}$

The value of $p$ can be obtained by using the fact that the curve passes through $B\left( {a, - b} \right).$ This is the equation of a cycloid, the curve generated by a point on the circumference of a rolling disc.