Optics Concept Page - 14

Definition

Total internal reflection

Two essential conditions for total internal reflection are:
1)Light should travel from denser medium to rarer medium.
2)The angle of incidence should be greater than the critical angle.

Definition

Optical fibre and it uses

An optical fiber or optical fibre is a flexible, transparent fiber made by drawing glass (silica) or plastic to a diameter slightly thicker than that of a human hair. Optical fibers are used most often as a means to transmit light between the two ends of the fiber and find wide usage in fiber-optic communications, where they permit transmission over longer distances and at higher bandwidths than wire cables. Fibers are used instead of metal wires because signals travel along them with lesser amounts of loss; in addition, fibers are also immune to electromagnetic interference, a problem from which metal wires suffer excessively. 
The uses of optical fibre are given bellow:
1. Communication - Telephone transmission method uses fibre-optic cables. Optical fibres transmit energy in the form of light pulses. The technology is similar to that of the coaxial cable, except that the optical fibres can handle tens of thousands of conversations simultaneously.
2. Medical uses - Optical fibres are well suited for medical use. They can be made in extremely thin, flexible strands for insertion into the blood vessels, lungs, and other hollow parts of the body. Optical fibres are used in a number of instruments that enable doctors to view internal body parts without having to perform surgery.
3. Simple uses - The simplest application of optical fibres is the transmission of light to locations otherwise hard to reach. Also, bundles of several thousand very thin fibres assembled precisely side by side and optically polished at their ends, can be used to transmit images.

Definition

Total internal reflection in optical fibres

Total internal reflection is a powerful tool 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. The construction of a single optical fibre is shown in the figure.The basic functional structure of an optical fiber consists of an outer protective cladding and an inner core through which light pulses travel. The overall diameter of the fiber is about 125μm$125\mu m$ and that of the core is just about 50μm$50\mu m$. The difference in refractive index of the cladding and the core allows total internal reflection in the same way as happens at an air-water surface show in the figure . 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 over a very long distance (tens of kilometers). Optical fibers are commonly used in telecommunications, because information can be transported over long distances, with minimal loss of data. Another common use can be found in medicine in endoscopes. The field of applied science and engineering concerned with the design and application of optical fibers are called fiber optics.

Diagram

Process of total internal reflection in optical fibres

We can say that the refracted angle θ2${\theta }_2$ is dependent on the ratio of the indices of the two materials n1n2${\displaystyle \frac{n_1}{n_2}}$ as well as the incident angle θ1${\theta }_1$. As a result, by controlling the ratio of the indices, one can control the refracted angle such that all of the light is reflected back from the interface. This is known as total internal reflection and is the method that allows for light to be contained and guided inside of a fiber optic.

Diagram

Process of total internal reflection in diamond

An incident light gets refracted into the diamond crystal. Dispersion causes the light to split into different colours that are seen in the diamond crystal. Due to high refractive index of diamond (around 2.4$2.4$), diamond has a very small critical angle. Thus most of the rays of light go multiple total internal reflections before they refract out of the crystal.

Shortcut

Michelson’s Method for Measuring the Speed of Light

Light from the source passes through a narrow slit.
It is reflected by face A of the octagonal (8 sided) metal prism.
It then travels a distance of a few kilometres and returns to be reflected by face B.
When the prism is stationary, a stationary image of the slit is observed.
The prism is now rotated.
If the prism rotates fast enough, when light returns to the prism, face B is no longer in the right position to reflect it into the observers eye.
The image of the slit disappears.
However, if the speed of rotation is increased, at a certain speed of rotation, the image of the slit reappears.
This is because the time taken for light to go from face A to face B was the same as the time taken by the prism to rotate 1/8th of a revolution.
So, the calculation needed to find the speed of light is remarkably simple:
If the prism completes n rotations per second then
Time of rotation = 1n${\displaystyle \frac{1}{n}}$

Therefore, the time, t taken for the light to cover the distance, s is 18th${{\displaystyle \frac{1}{8}}}^{th}$ of this

t=18n$t={\displaystyle \frac{1}{8n}}$
and so the speed of light c is given by

c=st=8ns$c={\displaystyle \frac{s}{t}}=8ns$

In 1931, Michelson found c = 2.99774×108 ms$\times {10}^8{\displaystyle \frac{m}{s}}$

Result

Factors affecting critical angle

Critical angle is affected by the refractive index of second medium with respect to the first medium. This in turn depends on:

• Wavelength: Critical angle increases with increase in wavelength (least for violet).
• Temperature: Critical angle increases with increase in temperature. 

Formula

Relation between refractive index and critical angle

Let us consider medium 1 −$-$ Incident medium and 2 −$-$ Refractive medium, Then value of critical angle can be derived by Snell's law.
n1sinθi=n2sinθr$n_{1}sin{\theta }_i=n_{2}sin{\theta }_r$
n1sinθcrit=n2sin90∘$n_{1}sin{\theta }_{crit}=n_{2}sin{90}^{\circ }$
sinθcrit=n2n1$sin{\theta }_{crit}={\displaystyle \frac{n_2}{n_1}}$
sinθcrit=1n12$sin{\theta }_{crit}={\displaystyle \frac{1}{n_{12}}}$

n12$n_{12}$ : Refractive index of denser medium 1 with respect to rarer medium 2.

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