A silicon optical fibre with a core diameter large enough has a core refractive index of 1.50 and a cladding refractive index 1.47. Determine(i) the critical angle at the core cladding interface,(ii) the numerical aperture for the fibre(iii) the acceptance angle in air for the fibre.

Here,\${ \mu }_{ 1 }=1.50;{ \mu }_{ 2 }=1.47;\$\${ \mu }_{ 0 }=1\$(i) Critical angle \${ \theta }_{ c }\$ at the core-cladding interface is given by \${ \theta }_{ c }={ sin }^{ -1 }\dfrac { 1.47 }{ 1.50 } =78.{ 5 }^{ 0 }\$(ii) Numerical aperture,\$NA={ \left( { { \mu }_{ 1 } }^{ 2 }-{ { \mu }_{ 2 } }^{ 2 } \right) }^{ \frac { 1 }{ 2 } }\$ \$={ \left[ { \left( 1.50 \right) }^{ 2 }-{ \left( 1.47 \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }={ \left( 2.25-2.16 \right) }^{ 1/2 }=0.30\$(iii) Acceptance angle \${ \theta }_{ a }={ sin }^{ -1 }\left( NA \right) \$\$={ sin }^{ -1 }\left( 0.30 \right) ={ 17.4 }^{ 0 }\$

The refractive index of the material of a prism is \$\sqrt{2}\$ and the angle of the prism is \$30^o\$. One of the two refracting surface of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is?

Using \$ A = r_{1} + r_{2}\$\$ 30 = r_{1} + 0\$so \$r_{1} = 30\$Using Snell's Law, \$ 1 \times sin i = \mu \times sin r\$\$ 1 \times sin i = \sqrt{2} \times sin 30 = \sqrt{2} 3/2\$\$sin i = 1/ \sqrt{2}\$\$ i = 45^{o}\$option 4 is correct

Mention three applications of total internal reflection of light.

Total internal reflection can be applied in the following:1. Telecommunication systems2. Automotive rain sensors and windscreen wipers3. Optical fingerprinting devices

What are the necessary conditions for total internal reflection to take place?

For total internal reflection to take place (i) light must travel from a denser medium to a rarer medium and (ii) the angle of incidence inside the denser medium must be greater than the critical angle. i.e., \$\angle i > \angle c\$.

A transparent solid cylindrical rod has a refractive index of \$\displaystyle \dfrac{2}{\sqrt{3}}\$. It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle \$\theta\$ for which the light ray grazes along the wall of the rod is -

\$ \displaystyle I\times \sin90 = \frac{2}{\sqrt{3}}\sin(90-\alpha)\$\$\Rightarrow \displaystyle \cos\alpha=\frac{\sqrt{3}}{2}\$So, \$\displaystyle \sin\alpha=\sqrt{1-\frac{3}{4}}=\frac{1}{2}\$Now, \$ 1\displaystyle \times\sin\theta=\frac{2}{\sqrt{3}}\sin\alpha=\displaystyle \frac{2}{\sqrt{3}}\times\frac{1}{2}=\frac{1}{\sqrt{3}}\$\$\Rightarrow \theta =\displaystyle \sin^{-1}\frac{1}{\sqrt{3}}\$

What is total internal reflection?

Total internal reflection is a phenomenon of reflection of ray back to the same medium when passing from denser medium to rarer medium in a such away that angle of incidence greater than its critical angle.

A diver at a depth of \$12m\$ in water \$\left( \mu =\cfrac { 4 }{ 3 } \right) \$ sees the sky in a cone of semi-vertical angle of :

Given : \$\mu = \dfrac{4}{3}\$ and \$\mu_{air} = 1 \$The ray emerges out of water making an angle \$90^o\$ with the water surface, i.e \$r = 90^o\$ Using Snell's law of refraction at point A:\$\Rightarrow \mu sin \theta = \mu_{air} sin r\$\$\Rightarrow \$\$\dfrac{4}{3} \times sin\theta = 1\times sin90^o\$\$\Rightarrow \dfrac{4}{3} \times sin\theta = 1\times 1\$ \$\Rightarrow \$ \$sin\theta =\dfrac{3}{4}\$\$\Rightarrow \$ \$\theta = sin^{-1 } \bigg( \dfrac{3}{4} \bigg)\$

In the figure shown, for an angle of incidence 45, at the top surface, what is the minimum refractive index needed for total internal reflection at vertical face AD?

To find:Minimum R.Iso that TIRtakes place at AD=?For TIR\$\boxed{90-t=\theta_c}\$Applying snesll's Law at Ab\$u_1\sin i=u_2\sin r\$\$1\times \sin 45=u\sin r\$\$\boxed{u\sin t=\dfrac{1}{\sqrt{2}}}\$............(1)Applying snell's law at AD\$u\sin (90-t)=1\times \sin 90\$\$\boxed{u=\dfrac{1}{\cos r}}\$............(2)From eq (1) and (2)\$\boxed{\tan r=\dfrac{1}{\sqrt{2}}}\$from eq (1)\$u.\dfrac{1}{\sqrt{3}}=\dfrac{1}{\sqrt{2}}\Rightarrow \boxed{u=\sqrt{\dfrac{3}{2}}}\$

(a) Figure shows a cross-section of a light pipe made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.(b) What is the answer if there is no outer covering of the pipe?

(a)\$\mu_1=1.68,\mu_2=1.44\$\$\mu_1/\mu_2=1/sini_1\Rightarrow i_1=59^o\$For critical angle, total internal reflection takes place when \$i \gt i_1\$maximum angle of reflection is \$90-i_1=31^o=r_{max}\$Let \$i_{max}\$ be the maximum angle of incidence.\$\mu_1=sini_{max}/sinr_{max}\Rightarrow i_{max}=60^o\$Thus all the ray between \$0^o\lt i\lt60^o\$ will suffer TIR.(b)Now, \$\mu_1=1\$ (air)\$sini/sinr=\mu_2/\mu_1\Rightarrow r=36.5^o\$\$i=90-36.5=53.5^o\$Since \$i\gt r\$, all incident ray will suffer TIR

Optical Fibres | Definition, Examples, Diagrams

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

), 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.

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

125 μ m

and that of the core is just about

50 μ 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 $θ_{2}$ is dependent on the ratio of the indices of the two materials $\frac{n _{1}}{n _{2}}$ as well as the incident angle $θ_{1}$ . 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.

Optical Fibres

diagram

Process of total internal reflection in diamond

definition

Optical fibre and it uses

definition

Total internal reflection in optical fibres

diagram

Process of total internal reflection in optical fibres

REVISE WITH CONCEPTS

Total Internal Reflection and Critical angle

Applications of Total Internal Reflection

Quick Summary With Stories

Optical Fibers

A silicon optical fibre with a core diameter large enough has a core refractive index of 1.50 and a cladding refractive index 1.47. Determine
(i) the critical angle at the core cladding interface,
(ii) the numerical aperture for the fibre
(iii) the acceptance angle in air for the fibre.

Mention three applications of total internal reflection of light.

What are the necessary conditions for total internal reflection to take place?

A transparent solid cylindrical rod has a refractive index of $\frac{2}{3}$ . It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle $θ$ for which the light ray grazes along the wall of the rod is -

What is total internal reflection?

Define critical angle. Explain total internal reflection using a neat diagram.

A diver at a depth of $12 m$ in water $(μ = \frac{4}{3})$ sees the sky in a cone of semi-vertical angle of :

In the figure shown, for an angle of incidence 45, at the top surface, what is the minimum refractive index needed for total internal reflection at vertical face AD?

diagram

Process of total internal reflection in diamond

definition

Optical fibre and it uses

definition

Total internal reflection in optical fibres

diagram

Process of total internal reflection in optical fibres

REVISE WITH CONCEPTS

Total Internal Reflection and Critical angle

Applications of Total Internal Reflection

Quick Summary With Stories

Optical Fibers

A silicon optical fibre with a core diameter large enough has a core refractive index of 1.50 and a cladding refractive index 1.47. Determine (i) the critical angle at the core cladding interface, (ii) the numerical aperture for the fibre (iii) the acceptance angle in air for the fibre.

Mention three applications of total internal reflection of light.

What are the necessary conditions for total internal reflection to take place?

A transparent solid cylindrical rod has a refractive index of 3​2​. It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle θ for which the light ray grazes along the wall of the rod is -

What is total internal reflection?

Define critical angle. Explain total internal reflection using a neat diagram.

A diver at a depth of 12m in water (μ=34​) sees the sky in a cone of semi-vertical angle of :

In the figure shown, for an angle of incidence 45, at the top surface, what is the minimum refractive index needed for total internal reflection at vertical face AD?

A silicon optical fibre with a core diameter large enough has a core refractive index of 1.50 and a cladding refractive index 1.47. Determine
(i) the critical angle at the core cladding interface,
(ii) the numerical aperture for the fibre
(iii) the acceptance angle in air for the fibre.

A transparent solid cylindrical rod has a refractive index of $\frac{2}{3}$ . It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle $θ$ for which the light ray grazes along the wall of the rod is -

A diver at a depth of $12 m$ in water $(μ = \frac{4}{3})$ sees the sky in a cone of semi-vertical angle of :