Total Internal Reflection and Optical Fibers

Total Internal Reflection (T.I.R.)

When light ray travels from denser to rarer medium, it bends away from the normal. If the angle of incidence is increased, the angle of refraction also increases. At a particular value of angle, the refracted ray subtends $90^{\circ }$ angle with the normal, this angle of incidence is known as critical angle $\left({\theta }_C\right)$. If angle of incidence increases further, the ray comes back to the same medium. This phenomenon is known as total internal reflection.

Using Snell’s Law

${\mu }_D\sin {\theta }_C={\mu }_R\sin 90^{\circ }$

$\sin {\theta }_C=\frac{{\mu }_R}{{\mu }_D}$

$\sin {\theta }_C=\frac{{\mu }_R}{{\mu }_D}=\frac{v_D}{v_R}=\frac{{\lambda }_D}{{\lambda }_R}$

Conclusion:

For T.I.R light must travel from denser to rarer medium(for example,glass to air,water to air,glass to water etc)

$i<{\theta }_C:$ Refraction

$i={\theta }_C$ : Grazing Emergence

$i>{\theta }_C$ : TIR

$\sin {\theta }_C=\frac{{\mu }_R}{{\mu }_D}$

If rarer medium is air ,$\sin {\theta }_C=\frac{1}{\mu }$

Note:

(1) From the ray diagram as shown in figure we conclude that in case of refraction maximum angle from normal is ${\theta }_C$ in denser medium and maximum angle from normal is $90^{\circ }$ in rarer medium.

(2) In case of total internal reflection, as all (i.e. 100%) incident light is reflected back into the same medium there is no loss of intensity while in case of reflection from mirror or refraction from lenses there is some loss of intensity as the entire light cannot be reflected or refracted. Due to this reason, images formed by TIR are much brighter those than formed by mirrors or lenses.

1.0Conditions for Total Internal Reflection

For Total Internal Reflection to occur, two crucial conditions must be met:

1. Light must travel from a denser medium to a rarer medium. The refractive index of the first medium must be greater than the refractive index of the second medium $(n_1>n_2)$.
2. The angle of incidence$\text{ (θі) }$in the denser medium must be greater than the critical angle $(\theta \mathrm{i}>\theta \mathrm{c})$. If the angle of incidence is less than the critical angle, some light will be refracted and some will be reflected.

2.0The Critical Angle $({\theta }_C)$

The critical angle$({\theta }_C)$ is the specific angle of incidence in the denser medium for which the angle of refraction in the rarer medium is exactly 90^{\circ}. At this angle, the refracted ray grazes the boundary between the two media. If the angle of incidence is increased even slightly beyond the critical angle, TIR will occur.

Derivation of the Critical Angle

The critical angle can be derived using Snell's Law of Refraction. Snell's Law states:

$n_1\sin {\theta }_i=n_2\sin {\theta }_r$

where:

• $n_1$ is the refractive index of the denser medium.
• $n_2$ is the refractive index of the rarer medium.
• ${\theta }_i$ is the angle of incidence.
• ${\theta }_r$ is the angle of refraction.

By the definition of the critical angle, when ${\theta }_i={\theta }_c$, the angle of refraction ${\theta }_r=90^{\circ }$

Substituting these values into Snell's Law:

$n_1\sin {\theta }_c=n_2\sin \left(90^{\circ }\right)$

$n_1\sin {\theta }_c=n_2(1)$

Therefore, the formula for the critical angle is:

$\sin {\theta }_c=\frac{n_2}{n_1}$

This formula shows that the critical angle depends solely on the refractive indices of the two media.

3.0Total Internal Reflection in Everyday Life

Total Internal Reflection is a common phenomenon with many fascinating examples:

• Diamonds: Diamonds sparkle brilliantly because of their very high refractive index $(\mathrm{n}\approx 2.42)$, which gives them a very small critical angle $\left(\approx 24.4^{\circ }\right)$. When light enters a diamond, it undergoes multiple total internal reflections, reflecting the light out of the facets and giving it a fiery appearance.
• Mirage: A mirage is an optical illusion caused by the TIR of light from the sky, which creates a shimmering pool-like reflection on a hot road or in a desert. The air near the ground is hot and less dense, while the air above is cooler and denser.

Prisms in Binoculars and Periscopes: Prisms are often used instead of mirrors in optical instruments. For example, a $45^{\circ }-90^{\circ }-45^{\circ }$prism can be used to reflect light at a $90^{\circ }or180^{\circ }$ angle with almost no loss of intensity, as the light undergoes TIR.

4.0What are Optical Fibers?

An optical fiber is a thin, flexible, transparent fiber made of glass or plastic that acts as a waveguide for light. It is designed to transmit light over long distances with minimal loss of intensity. The working principle of optical fibers is entirely based on Total Internal Reflection.

A typical optical fiber consists of three main parts:

1. Core: The central, light-carrying part of the fiber. It is made of a material with a high refractive index $(n_1)$.
2. Cladding: A layer surrounding the core. It is made of a material with a lower refractive index $(n_2)$, where $n_2<n_1$​.
3. Protective Jacket: An outer coating that protects the core and cladding from damage and moisture.

Mathematical Description in Optical Fibre

Snell’s Law at air/core interface, $1\times \sin i={\mu }_1\times \sin r$

$\sin r=\frac{\sin i}{{\mu }_1}\Rightarrow \sin {\theta }_C=\frac{{\mu }_2}{{\mu }_1}$

For TIR in optical fibre, $90-r>{\theta }_C$

Take sin on both side $\sin 90-r>\sin {\theta }_C$

$\cos r>\sin {\theta }_C$

$\sqrt{1-{\sin }^{2}r}>\sin {\theta }_C$

$\sqrt{1-\frac{{\sin }^{2}i}{{\mu }_1^2}}>\frac{{\mu }_2}{{\mu }_1}$

${\mu }_1^2-{\sin }^{2}i>{\mu }_2^2$

$\sin i<\sqrt{{\mu }_1^2-{\mu }_2^2}$

For maximum value

$\sin i=\sqrt{{\mu }_1^2-{\mu }_2^2}$ here i is called maximum acceptance angle.

5.0Principle and Working of Optical Fibers

The working principle of optical fibers is straightforward and brilliant. A light signal is launched into the core of the optical fiber. It strikes the interface between the core and the cladding. Because the core has a higher refractive index than the cladding, and the light is engineered to strike the boundary at an angle greater than the critical angle, it undergoes Total Internal Reflection.

The light ray is completely reflected back into the core, continuing its path down the fiber. As it travels, it undergoes a series of successive total internal reflections, bouncing off the core-cladding boundary and propagating along the length of the fiber with virtually no loss of energy. This allows the signal to be transmitted over very long distances with high fidelity.

Types of Optical Fibers

Optical fibers are primarily classified into two types based on their refractive index profile and the mode of light propagation.

1. Single-Mode Fiber: This type has a very small core diameter (around $8-10\mu \mathrm{m}$). It allows only one mode (or path) of light to travel down the fiber. This minimizes signal distortion and is ideal for high-speed, long-distance data transmission (e.g., telecommunications).
2. Multi-Mode Fiber: This type has a larger core diameter (typically $50-100\mu \mathrm{m}$) and allows multiple modes or paths of light to travel. There are two sub-types:
3. • Step-Index: The refractive index of the core is uniform. Different modes travel at different speeds, leading to signals spreading over long distances.
   • Graded-Index: The refractive index of the core gradually decreases from the center towards the cladding. This allows different modes to travel at roughly the same average speed, reducing signal dispersion and making it suitable for medium-distance communication.

6.0Advantages and Applications of Optical Fibers

Optical fibers have revolutionized the telecommunications and medical industries due to their unique advantages:

• High Bandwidth: They can transmit a vast amount of data simultaneously, making them the backbone of modern internet and telecommunication networks.
• Low Signal Loss: The loss of light intensity over long distances is remarkably low, allowing for transmission over many kilometers without the need for frequent signal amplification.
• Immunity to Electromagnetic Interference: Since they transmit light, optical fibers are not affected by electromagnetic interference (EMI) from power lines or other sources, ensuring signal security and quality.
• Small Size and Lightweight: They are much thinner and lighter than traditional copper cables, making them easier to install and handle.
• Security: The signals cannot be "tapped" without physically breaking the fiber, which is easily detectable.

Applications:

• Telecommunications: High-speed internet, telephone lines, and cable television.
• Medical Endoscopy: Medical endoscopes use optical fibers to transmit images from inside the body to a monitor, allowing for minimally invasive surgery and diagnosis.
• Lighting and Decoration: Used for decorative lighting and for transmitting light to inaccessible areas.
• Sensors: Used in various sensors to measure temperature, pressure, and strain.

7.0Solved JEE-Level Problems

Illustration-1: Figure shows a cross section of a 'light pipe' made of a glass fibre of refractive index $\sqrt{2}$. The outer covering of the pipe is made of a material of refractive index $\sqrt{\frac{3}{2}}$. What is the maximum angle of the incident ray with the axis of the pipe for which total internal reflections inside the pipe take place?

Solution:

$\sin i<\sqrt{{\mu }_1^2-{\mu }_2^2}$

$\sin i<\sqrt{(\sqrt{2}{)}^2-{\left(\sqrt{\frac{3}{2}}\right)}^2}$

$\sin i<\frac{1}{\sqrt{2}}\Rightarrow i<45^{\circ }$

$i_{\max }=45^{\circ }$

Illustration-2: Find the angle of refraction in a medium ($\mu =2$) if light is incident in vacuum, making angle equal to twice the critical angle.

Solution: Since the incident light is in a rarer medium. Total Internal Reflection can not take place.

${\theta }_C={\sin }^{-1}\frac{1}{\mu }=30^{\circ }\left(\therefore 2{\theta }_C=60^{\circ }\right)$

Applying Snell’s Law

$1\sin 60^{\circ }=2\sin r\Rightarrow \sin r=\frac{\sqrt{3}}{4}$

$\Rightarrow r={\sin }^{-1}\left(\frac{\sqrt{3}}{4}\right)$

Illustration-3:What should be the value of angle $\theta$ so that light entering normally through the surface AC of a prism (n=3/2) does not cross the second refracting surface AB.

Solution: Light ray will pass the surface AC without bending since it is incident normally. Suppose it strikes the surface AB at an angle of incidence i.

$i=90-\theta$

For the required condition :

$90^{\circ }-\theta >{\theta }_C$

$\sin \left(90^{\circ }-\theta \right)>\sin {\theta }_C$

$\cos {\theta }_C>\sin {\theta }_C=\frac{1}{3/2}=\frac{2}{3}$

$\theta <{\cos }^{-1}\frac{2}{3}$

Illustration 4: What should be the value of the refractive index n of a glass rod placed in air, so that the light entering through the flat surface of the rod does not cross the curved surface of the rod?

Solution: It is required that all possible $r^{\prime }$ should be more than a critical angle. This will be automatically fulfilled if minimum $r^{\prime }$ is more than a critical angle ...(A)

Angle $r^{\prime }$ is minimum r is maximum i.e. ${\theta }_C$

Therefore,the minimum value of  $r^{\prime }$ is $90-{\theta }_C$

From condition (A):

$90^{\circ }-{\theta }_C>{\theta }_C\text{ or }{\theta }_C<45^{\circ }$

$\sin {\theta }_C<\sin 45^{\circ };\frac{1}{n}<\frac{1}{\sqrt{2}}$

Or $n>\sqrt{2}$