Rohan Joshi
5 min readJul 21, 2021

Hello! In today’s article, we’ll be looking at the concept of refraction from an interesting perspective. As we all know, refraction is the change in the direction of travel of a wave when it moves from one medium to another. The refraction of light is probably the most well known example.

In this article, we’ll be exploring a different kind of refraction, and the situation will offer an interesting analogy to the same concept. I’m talking about…a particle moving through a potential barrier. While it’s not perfect, it is quite similar, and differs in one major way. Here’s the basic setup.

Here’s what this image means. Consider a particle of mass m moving in the x-y plane, with an initial speed v1, at angle θ1 to the normal to the y-axis. Its initial kinetic energy is K. Now the space is such that left of the y axis, the potential energy is zero, but to the right has a constant value U.

This is the graph of the above function U(x)
A 3D graph

As a particle crosses from left to right, it experiences a force due to a change in potential energy, and we can find out the magnitude and direction of its final velocity v2 and θ2 using the conservation of energy and momentum.

The first thing to realise is that the potential energy varies only in the x direction, so the force will be directed in the opposite direction as that of the gradient, that is in the negative x direction. This means that the y component of momentum, that is the component parallel to the potential boundary, is conserved. As for the speed, we can use conservation of energy to find out. So if we look at figure 1, we realize

Now the equation we derived above looks a lot like Snell’s law for refraction, but with one big difference-if the speed of a particle decreases as it crosses the potential barrier, the angle of ‘refraction’ θ2 is greater than the angle of incidence θ1. While this is a big deal, it doesn’t exactly hinder our comparison of the motion of a particle to optics. So let’s use conservation of energy now.

So that’s it! The two highlighted equations show the final speed and angle of the particle as it crosses through the barrier. As you can see, if θ1=0 degrees, then the particle will pass through undeflected. Another important topic now is the ‘critical angle’, the angle at which a particle will be refracted at 90 degrees to the normal, or parallel to the interface. It can be derived as follows:

The expression highlighted in green is the ‘critical angle’. If a particle approaches the interface at this angle, then the angle of ‘refraction’ will be 90 degrees.


Now comes the interesting part. What if the angle of approach is greater than the critical angle? That presents a problem because equation (2) breaks down for angle greater than θcritical. So what do we do? We simply look at an analogous situation. Instead of looking like a simple step function, what if U(x) was continuously increasing for x>0? Maybe (for the sake of simplicity) U was proportional to x? Well in that case, there would be a constant force acting in the negative x direction. But we know what that is! If a particle approaches the interface at an angle, then the it’ll simple be projectile motion, and the path will be parabolic!

A potential directly proportional to x can be thought of as multiple ‘steps’ of potential

Now we use what we learnt above. Moving through from one potential energy level to another can be thought of as continuous ‘refraction’ of the particle’s motion. So, comparing to the parabolic path, we can deduce that as the angle of ‘refraction’ gets closer and closer to the critical angle, the particle essentially begins to take a U-turn and reflects off the maximum energy level, the same way a parabola curves back down. So, if we assume this holds for our extreme case scenario where the potential well isn’t continuous, but a step function, then if a particle approaches at the critical angle, it simple reflects back at the same angle of incidence, with the same speed, due to the conservation of energy. The particle never actually gains the potential energy U, it just reflects, so kinetic energy is conserved.

If the angle of approach is greater than the critical angle, or the particle doesn’t have enough energy, it simply reflects off the barrier as shown.

Now, what if our particle simple doesn’t have enough energy to cross the barrier? What if it isn’t moving fast enough? Well, we simple use the same logic as above. If a particle is thrown in a gravitational field at a speed v and angle θ, then the maximum height reached, and the maximum potential energy gained, will depend on its initial kinetic energy. As soon as the maximum potential energy and height is reached, the particle falls back down. Using the same principle, a particle which doesn’t have enough energy to cross the barrier cannot move to the other side, the positive x axis, and instead is again reflected, the same scenario as above.

So that’s it for today! The combination of motion and optics is quite bizarre, but it is still a fun exploration! Many discoveries have been made in physics by simply applying a principle from a disparate topic into another! And while this topic didn’t exactly produce Nobel prize winning results, it was still interesting-at least to me. Bye!