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Key Words Reflection, Refraction, Refractive Index, Snell's Law, Critical Angle, Total Internal Reflection. |
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:

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.
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.
| Can you complete these diagrams showing the refracted ray and the emergent ray? |
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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.
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.
n1
= c/c1
and
n2 = c/c2
n1c1 = c = n2c2
n2 = c1
n1 c2
| What is the speed of light in glass, refractive index 1.5, when the speed of light in air is 3.0 x 108 m/s? |
The
ratio n2/n1 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

The ratio n2/n1 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.
| What does 2n1 mean? | ||
| 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
sinq2
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.
2n1 = sinq2
sinq1
| What is the relative refractive index when a light ray passes from glass (RI = 1.5) to air? | ||
| 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
= n2 sinq2
n2
sinqc
= n1 sin 90
Þ
n2
sinqc
= n1
Þ
sinqc
= n1
n2
We can write this as:
sinqc
=
1
n2/n1

| What is the critical angle of water, of which the refractive index is 1.33? | ||
| 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.
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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

| Presentation | Snell's Law | ||
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Now try Topic Quiz 2 |
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