20.4 C
New York
Friday, September 24, 2021

Buy now

Topic 2 – The Wave Nature of Light

Some Revision

In this section we will consider light to be a wave carrying energy that travels in straight lines at a speed of 3 × 108 m/s in air.  When light hits an object it is:

  • Reflected
  • Transmitted
  •  Absorbed.


You should be familiar with reflection which you did in early secondary school.  All reflection depends on the smoothness of the surface.  If the surface is smooth, then parallel rays are reflected parallel (specular reflection).  If the surface is rough, then parallel rays are scattered.  However each individual ray obeys the law of reflection.

High quality mirrors are silvered on the front to prevent multiple images.  Here is a ray of light striking a mirror at an angle:

We observe the following:

  • The angle of incidence = angle of reflection, q1 = q2
  • The reflected ray is in the same plane as the incident ray.
  • All angles are measured from the normal, a line running at 90o to the surface.

The image in a mirror has these features:

  • It is the same distance behind the mirror line as the object is in front.
  • It is the same size as the object.
  •  It is the right way up (erect).
  •  It is laterally inverted, i.e. left and right swapped over.
  •  It is virtual, meaning that the rays cannot be projected onto a screen.  The image is not really behind the mirror; it just appears to be.

  Curved mirrors obey the same rules of reflection as flat (plane) mirrors do.


Light rays are bent when they travel from a medium of one optical density into another, for example from air to glass.  In air light travels at 3 × 108 m/s, while in glass its speed is 2 × 108 m/s.  We say that glass in an optically denser medium than air.  We can see that the direction of the ray is abruptly changed, or deviated as it passes the boundary:

We should note the following:

  •  The ray bends in towards the normal as the ray enters the glass.  The angle of incidence is less than the angle of refraction.
  • On leaving the glass the light regains its original speed.  The emergent ray bends away from the normal.  In this case the angle of incidence is less than the angle of refraction.
  • The path of the emergent ray is parallel to the path of the undeviated ray (where the ray would have gone if there hadn’t been a piece of glass in the way).  This only happens when the sides of the object are parallel.  It wouldn’t happen in a prism.
  •   If the boundary is indistinct, then refraction is gradual with the change occurring over a long distance.  Earthquake waves refract in curves due to the gradually changing density of rocks.

Different wavelengths of light are refracted by different amounts; blue light is refracted the most, red light the least.

Question 1.  Can you complete these diagrams showing the refracted ray and the emergent ray?

Snell’s Law

It was the Dutch physicist Snell, who in 1621 established the quantitative relationship between the angle of incidence and the angle of refraction.  If we double the angle of incidence, we do NOT double the angle of refraction

The refractive index given the physics code n and has no units.  We can also describe the absolute refractive index as the ratio of the speed of light in a vacuum to the speed of light in an optical medium.  Glass has a refractive index of 1.5, so the speed of light is 3 × 108 m/s ÷ 1.5 = 2 × 108 m/s.

  Before we carry on, we need to be sure of what these terms mean:

  • The absolute refractive index is the ratio compared with the refractive index of a vacuum.  (n for a vacuum = 1.000;)
  • The relative refractive index is the ratio of the absolute refractive index of one material compared to that of another, for example from water to glass.

  From this we can write:

                                    n1 = c/c1



  We can rearrange each equation and combine the two to write:

                                    n1c1 = c = n2c2

  So we can write:

                                                n2 = c1

                                                n1    c2

 Question 2.  What is the speed of light in glass, refractive index 1.5, when the speed of light in air is 3.0 x 10m/s?

The ratio n1/n2 is the relative refractive index, which we can also define as the ratio of the sines of the angles of incidence and refraction:

                                                n2 = c1  = sinq1

                                                n1    c2    sinq2

We can write this formula in a more useful form:

The ratio n1/n2 is the relative refractive index, which we can also write as 1n2. This means the refractive index of light going in from material 1 to 2.

Question 3 What does 2n1 mean?

Question 4 A ray of light strikes an air-glass boundary at an incident angle of 30 o.  If the refractive index of glass is 1.5, what is the angle of refraction?

Remember that each wavelength of light has a slightly different refractive index.  Generally we use yellow light.

The table shows some absolute refractive indices for some common materials:

Medium Refractive Index
Vacuum 1.000
Air 1.003
Ice 1.30
Water 1.33
Glass 1.50
Diamond 2.42

The absolute refractive index will always be always be greater than 1.0 as light cannot travel any faster then it does in a vacuum.  It is possible for particles to travel faster than light in a material (but NOT in a vacuum).  The result of this will be little flashes, the light equivalent of a sonic boom, called Cherenkov radiation.  This is why radioactive materials appear to make water glow.

Now let us have a look at what happens when a ray passes from a dense medium to a less dense medium.  Diagram A shows the ray going from air to glass, while Diagram B shows the ray leaving the glass and going into air.

In Diagram A we have established that:

            1n2 = sinq1


In Diagram B we see:

  •  Ray of light travels faster in the optically less dense medium
  • The ray bends away from the normal.
  • The angle of incidence is less than the angle of refraction.

  The relative refractive index going from medium 2 to medium 1 is 2n1, so we can write:

            2n1 = sinq2


  Therefore 1n2 = 1 ÷ 2n1.  For example, the refractive index from air to water is 1.33, while the refractive index from water to air = 1/1.33 = 0.75

Question 5  What is the relative refractive index when a light ray passes from glass (RI = 1.5) to air?

Question 6 What is the angle of refraction and what is the angle through which the ray is deviated when light passes at an incident angle of 48o into water of refractive index 1.33 from air at a refractive index of 1.00?

Critical Angle

We have seen how a ray of light passing from glass to air bends away from the normal.  If we increase the angle of incidence, the angle of refraction increases more (Diagrams A and B):

Angle of refraction getting bigger:

At a particular value of angle of incidence, the angle of refraction is 90o, as shown in Diagram C.  This particular angle of incidence is called the critical angle.

Above the critical angle we get total internal reflection.  There is no transmission of light at all. See diagram D.

Above the critical angle we get total internal reflection.  There is no transmission of light at all.  We know that:

 n1 sinq1 = n sinq2

  At the critical angle:

n1 sinqc = n sin 90

                                         Ž  n1 sinqc = n2

                                         Ž sinqc = n2


We can write this as:

                                    sinqc =   1


  Since n1/n2 is the relative refractive index 1n2, we can write

Question 7 What is the critical angle of water, of which the refractive index is 1.33?

Question 8  Rays from a point source of light at the bottom of a swimming pool 1.8 m deep strike the water surface and only emerge through a circle of radius r, as shown in the diagram.  If the refractive index between water and air is 1.33, calculate the value of r.

An example of the use of total internal reflection is in optical fibres.  At their simplest, optical fibres are long thin strands of glass that carry light from one end a long distance to the other.  The light can be guided round corners using total internal reflection.

There are two problems:

  • Ordinary glass is not very good.  It is impure and has a high attenuation coefficient, losing most of its intensity after a short distance.  Red and blue light is absorbed, leaving only green.
  • Wide fibres tend to smear the signals as there is a path difference between the rays that go down the middle of the fibre and those that bounce from side to side.

The problems can be overcome by:

  • Using very high purity glass.
  • Using infra-red transmitters.
  • The light is channelled along a very thin central fibre that is cladded with glass of a lower refractive index.  Fibres with a step index have one layer of cladding, while graded index fibres have several layers of cladding.  The best fibres are called monomode fibres, with a channel of no more than 1.25 mm, which is very narrow.  The path difference between the axial ray and the reflected rays is negligible.

  Optical fibres are amazingly flexible and strong.  They mounted in a polythene tube for further strength.  Optical fibres offer many advantages over wires.

Although we have looked at light, all these phenomena can be observed with other forms of waves, e.g.

  • radio waves
  • water waves
  • sound waves

Related Articles


Please enter your comment!
Please enter your name here

Stay Connected

- Advertisement -spot_img

Latest Articles