You may have noticed when experimenting with ray boxes and glass blocks in the previous
section that sometimes, when you changed the angle of incidence of the light, it was not
refracted out into the air, but was reflected back through the block. When the entire
incident light ray travelling through an optically denser medium is reflected back at
the boundary between that medium and another of lower optical density, instead of
passing through and being refracted, this is called total internal reflection.
As we increase the angle of incidence, we reach a point where the angle of refraction is
\(\text{90}\)\(\text{°}\) and the refracted ray travels along the boundary of the
two media. This angle of incidence is called the critical angle.
Critical angle
The critical angle is the angle of incidence where the angle of refraction is
\(\text{90}\)\(\text{°}\). The light must travel from an optically
more dense medium to an optically less dense medium.
Figure 5.15: When the angle of incidence is equal to the
critical angle, the angle of refraction is equal to
\(\text{90}\)\(\text{°}\).
If the angle of incidence is bigger than this critical angle, the refracted ray will not
emerge from the medium, but will be reflected back into the medium. This is called
total internal reflection.
The conditions for total internal reflection are:
light is travelling from an optically denser medium (higher refractive index) to an
optically less dense medium (lower refractive index).
the angle of incidence is greater than the critical angle.
Figure 5.16: When the angle of incidence is greater than the
critical angle, the light ray is reflected at the boundary of the two media and
total internal reflection occurs.
Each pair of media have their own unique critical angle. For example, the critical angle for
light moving from glass to air is \(\text{42}\)\(\text{°}\), and that of water to
air is \(\text{48,8}\)\(\text{°}\).
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A recommended experiment for informal assessment is also included. This covers
determining the critical angle for light travelling through a rectangular glass
block. Learners will need a rectangular glass block, a
\(\text{360}\)\(\text{°}\) protractor, pencil, paper, ruler and a ray box.
Learners should all get similar results at the end of the experiment.
Calculating the critical angle
Instead of always having to measure the critical angles of different materials, it is
possible to calculate the critical angle at the surface between two media using
Snell's Law. To recap, Snell's Law states:
\[n_1 \sin \theta_1 = n_2 \sin \theta_2\]
where \(n_1\) is the refractive index of material \(\text{1}\), \(n_2\) is the
refractive index of material \(\text{2}\), \(\theta_1\) is the angle of
incidence and \(\theta_2\) is the angle of refraction. For total internal
reflection we know that the angle of incidence is the critical angle. So,
\[\theta_1 = \theta_c.\]
However, we also know that the angle of refraction at the critical angle is
\(\text{90}\)\(\text{°}\). So we have:
\[\theta_2=90^{\circ}.\]
We can then write Snell's Law as:
\[n_1 \sin{\theta_c} = n_2 \sin{90^{\circ}}\]
Remember that for total internal reflection the incident ray is always in the
denser medium!
Worked example 5: Critical angle 1
Given that the refractive indices of air and water are
\(\text{1,00}\) and \(\text{1,33}\) respectively, find the
critical angle.
Determine how to approach the problem
We can use Snell's law to determine the critical angle since we know
that when the angle of incidence equals the critical angle, the
angle of refraction is \(\text{90}\)\(\text{°}\).
The critical angle for light travelling from water to air is
\(\text{48,8}\)\(\text{°}\).
Worked example 6: Critical angle 2
Complete the following ray diagrams to show the path of light in each
situation.
Identify what is given and what is asked
The critical angle for water is \(\text{48,8}\)\(\text{°}\)
We are asked to complete the diagrams.
For incident angles smaller than \(\text{48,8}\)\(\text{°}\)
refraction will occur.
For incident angles greater than \(\text{48,8}\)\(\text{°}\)
total internal reflection will occur.
For incident angles equal to \(\text{48,8}\)\(\text{°}\)
refraction will occur at \(\text{90}\)\(\text{°}\).
The light must travel from a medium with a higher refractive index
(higher optical density) to a medium with lower refractive index
(lower optical density).
Complete the diagrams
Refraction occurs (ray is bent away from the normal).
Total internal reflection occurs.
\(\theta_c = \text{48,8}\text{°}.\)
Refraction towards the normal (air is less dense than water).
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Fibre optics
Total internal reflection is a very useful natural phenomenon since it can be used to
confine light. One of the most common applications of total internal reflection
is in fibre optics. An optical fibre is a thin, transparent fibre,
usually made of glass or plastic, for transmitting light. Optical fibres are
usually thinner than a human hair! The construction of a single optical fibre is
shown in Figure 5.17.
The basic functional structure of an optical fibre consists of an outer protective
cladding and an inner core through which light pulses travel.
The overall diameter of the fibre is about \(\text{125}\) \(\text{μm}\)
(\(\text{125} \times \text{10}^{-\text{6}}\) \(\text{m}\)) and that of the core
is just about \(\text{50}\) \(\text{μm}\) (\(\text{50} \times
\text{10}^{-\text{6}}\) \(\text{m}\)). The difference in refractive index of the
cladding and the core allows total internal reflection to occur in the same way
as happens at an air-water surface. If light is incident on a cable end with an
angle of incidence greater than the critical angle then the light will remain
trapped inside the glass strand. In this way, light travels very quickly down
the length of the cable.
Figure 5.17: Structure of a single optical fibre.
Fibre Optics in Telecommunications
Optical fibres are most common in telecommunications, because information can
be transported over long distances, with minimal loss of data. This
gives optical fibres an advantage over conventional cables.
Signals are transmitted from one end of the fibre to another in the form of
laser pulses. A single strand of fibre optic cable is capable of
handling over \(\text{3 000}\) transmissions at the same time which
is a huge improvement over the conventional co-axial cables. Multiple
signal transmission is achieved by sending individual light pulses at
slightly different angles. For example if one of the pulses makes a
\(\text{72,23}\)\(\text{°}\) angle of incidence then a separate
pulse can be sent at an angle of \(\text{72,26}\)\(\text{°}\)! The
transmitted signal is received almost instantaneously at the other end
of the cable since the information coded onto the laser travels at the
speed of light! During transmission over long distances repeater
stations are used to amplify the signal which has weakened
by the time it reaches the station. The amplified signals are then
relayed towards
their destination and may encounter several other repeater stations on
the way.
Fibre optics in medicine
Optic fibres are used in medicine in endoscopes.
Endoscopy means to look inside and refers to looking inside
the human body for diagnosing medical conditions.
The main part of an endoscope is the optical fibre. Light is shone down the
optical fibre and a medical doctor can use the endoscope to look inside
the body of a patient. Endoscopes can be used to examine the inside of a
patient's stomach, by inserting the endoscope down the patient's throat.
Endoscopes also allow minimally invasive surgery. This means that a person
can be diagnosed and treated through a small incision (cut). This has
advantages over open surgery because endoscopy is quicker and cheaper
and the patient recovers more quickly. The alternative is open surgery
which is expensive, requires more time and is more traumatic for the
patient.
Total internal reflection and fibre optics
Textbook Exercise 5.5
Describe total internal reflection by using a diagram and
referring to the conditions that must be satisfied for
total internal reflection to occur.
If the angle of incidence is bigger than the critical angle,
the refracted ray will not emerge from the medium, but
will be reflected back into the medium. This is called
total internal reflection.
The critical angle occurs when the angle of incidence where
the angle of refraction is
\(\text{90}\)\(\text{°}\). The light must travel
from an optically more dense medium to an optically less
dense medium.
The conditions for total internal reflection are the the
light is travelling from an optically denser medium
(higher refractive index) to an optically less dense
medium (lower refractive index) and that the angle of
incidence is greater than the critical angle.
Representing this on a diagram gives:
Define what is meant by the critical angle when
referring to total internal reflection. Include a ray
diagram to explain the concept.
The critical angle occurs when the angle of incidence where
the angle of refraction is
\(\text{90}\)\(\text{°}\). The light must travel
from an optically more dense medium to an optically less
dense medium.
Will light travelling from diamond to silicon ever undergo
total internal reflection?
Diamond (index of refraction is about 3) is less optically
dense than silicon (index of refraction is about 4) and
so total internal reflection cannot occur.
Will light travelling from sapphire to diamond undergo total
internal reflection?
Sapphire (index of refraction is \(\text{1,77}\)) is less
optically dense than diamond (index of refraction is
about 3) and so total internal reflection cannot occur.
What is the critical angle for light travelling from air to
acetone?
Light travelling from diamond to water strikes the interface
with an angle of incidence of
\(\text{86}\)\(\text{°}\) as shown in the picture.
Calculate the critical angle to determine whether the
light be totally internally reflected and so be trapped
within the diamond.
A glass slab is inserted in a tank of water. If the
refractive index of water is \(\text{1,33}\) and that of
glass is \(\text{1,5}\), find the critical angle.
Note that the light must be travelling from the glass into
the water for total internal reflection to occur.
A diamond ring is placed in a container full of glycerin. If
the critical angle is found to be
\(\text{37,4}\)\(\text{°}\) and the refractive
index of glycerin is given to be \(\text{1,47}\), find
the refractive index of diamond.
An optical fibre is made up of a core of refractive index
\(\text{1,9}\), while the refractive index of the
cladding is \(\text{1,5}\). Calculate the maximum angle
which a light pulse can make with the wall of the core.
NOTE: The question does not ask for the angle of
incidence but for the angle made by the ray with the
wall of the core, which will be equal to
\(\text{90}\)\(\text{°}\) − angle of
incidence.
