# Refraction

## Introduction

Figure 1. Pencil in glass of water.

Refraction is the bending of light rays when they cross an interface between two materials with different refractive indices. For example, when light passes from air into the water at an angle, the light ray will bend toward the normal to the surface. It is a familiar effect to anyone that has ever looked a the broken image of a pencil in a glass of water.

The law relating the refraction of the light was discovered by Dutchman, Willibrord Snellius in 1621, although the law is now called Snell's law.

Flash 1. The refraction of light. Use the slider to change the angle of incidence and the text-box to change the refractive index of n2.

In the Flash animation above, a light ray is incident at an angle θ1 from the normal on an interface between two materials, in this case, the upper medium is air and the lower interface is glass.

A fraction of the incident light is reflected from the surface at the angle of incidence and the remaining light enters the second medium. The light that travels through the second medium is slowed and this causes the angle to reduce and the light ray moves toward the normal.

The degree to which the angle of the transmited ray changes is dependent on the difference in refractive indices at the interface. We can be more quantitative.

n1 sin θ1 = n2 sin θ2 (1)

Where n1 and n2 are the refractive indices of medium 1 and medium 2, respectively and θ1 and θ2 are the angle of incidence and angle of refraction, respectively. Thus if we know, the angle of incidence and the refractive indices of both media, we can predict the angle of refraction. Try changing the refractive index of the lower medium by altering the value for n2 and the angle of incidence using the slider.

This assumes that the refractive index of medium 1 is less than that of medium 2. According to the principle of reversibility, the path of the light in travelling from medium 1 to medium 2 is the same as it would be if the light were traveling in the opposite direction from medium 2 to medium 1. So when the light is traveling from a medium with a high refractive index to a medium of lower refractive index, the refraction causes the ray of light to be bent away from the normal. In addition, if the angle of incidence is increased, the angle of refraction is also increased with the angle of the refracted light increasing still futher.

## The Critical Angle and Total Internal Reflection

At a certain angle of incidence, known as the critical angle the refracted ray will be at 90° from the normal. No light is refracted. It is all reflected back into the higher refractive index material (n2). This is known as total internal reflection. The critical angle can be found by substituting &theta2 = &pi/2 into (1).

θc = sin-1(n2/n1) (2)

Try altering the value of n2. The reflection and transmission coefficients are calculated. Diamonds have a refractive index of 2.417. One of the reasons why diamonds are so sparkling, is that the difference in refractive index of the air and diamond leads to a small critical angle. Light that is transmitted through the top boundary with a small angle of incidence is travels with a little reflection into the diamond. However, the cut of the diamond is such that the light will undergoe total internal reflection at a facet boundary, therefore the light is trapped until it can escape by meeting an diamond/air boundary at an angle of incidence which is less than the critical angle.

Figure 2. Total internal reflection
in a brilliant-cut diamond

When the refractive index of the second medium is very close to the first medium, the light path is not changed significantly for most angles of incidence. The amount of light reflected is minimised. This finds applications in refractive index matching fluids. Refractive index matching fluids are liquids used to reduce the difference in refractive index between two media to reduce back-reflection. Such fluids are used particularly when splicing two optical fibres together where it is important to minimise the loss at the interfaces of the fibre.

The losses are given by -10 Log10( (n2-n1)2/(n2+n1)2).(3)