experiment for total internal reflection

Total Internal Reflection

We know that light is the form of energy that can undergo various phenomena like reflection, refraction, dispersion, and total internal reflection. It is the phenomenon responsible for optical illusions like a mirage and also the reason for diamonds to shine. In this session, let us know more about the total internal reflection.

What is Total Internal Reflection?

Total internal reflection is defined as:

The phenomenon which occurs when the light rays travel from a more optically denser medium to a less optically denser medium.

Consider the following situation. A ray of light passes from a medium of water to that of air. Light ray will be refracted at the junction separating the two media. Since it passes from a medium of a higher refractive index to that having a lower refractive index, the refracted light ray bends away from the normal. At a specific angle of incidence , the incident ray of light is refracted in such a way that it passes along the surface of the water. This particular angle of incidence is called the critical angle. Here the angle of refraction is 90 degrees. When the angle of incidence is greater than the critical angle, the incident ray is reflected back to the medium. We call this phenomenon total internal reflection.

Formula of Total Internal Reflection

Notations Used In The Total Internal Reflection Formula And Critical Angle

• r is the angle of refraction
• i is the angle of incidence
• n 1 is the refractive index in medium 1
• n 2 is the refractive index in medium 2
• θ is the critical angle

What are the Conditions of Total Internal Reflection?

Following are the two conditions of total internal reflection:

• The light ray moves from a more dense medium to a less dense medium.
• The angle of incidence must be greater than the critical angle.

Related Articles:

• Relation Between Critical Angle And Refractive Index
• What is Reflection of Light?
• Laws Of Reflection

Examples of Total Internal Reflection

Following are the examples of total internal reflection:

When the incident ray falls on every face of the diamond such that the angle formed, the ray is greater than the critical angle. The critical value of the diamond is 23°. This condition is responsible for the total internal reflection in a diamond which makes it shine.

It is an optical illusion that is responsible for the appearance of the water layer at short distances in a desert or on the road. Mirage is an example of total internal reflection which occurs due to atmospheric refraction.

Watch the video to learn more about Mirage.

Optical Fibre:

When the incident ray falls on the cladding, it suffers total internal reflection as the angle formed by the ray is greater than the critical angle.  Optical fibres  have revolutionised the speed with which signals are transferred, not only across cities but across countries and continents making telecommunication one of the fastest modes of information transfer. Optical fibres are also used in endoscopy.

Solved Examples on Total Internal Reflection

Q1. An optical fibre made up the glass with refractive index n 1 = 1.5 which is surrounded by another glass of refractive index n 2 . Find the refractive index n 2 of the cladding such that the critical angle between the two cladding is 80°.

Critical angle, θ = 80°

Refractive index, n 1 = 1.5

Refractive index n 2  = ?

Using the below formula, we can calculate n 2 :

Q2. Find the refractive index of the medium whose critical angle is 40°.

Critical angle, θ = 40°

Refractive index of the medium, μ = ?

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Frequently Asked Questions – FAQs

What is mirage, what is the formula to find the critical angle.

The formula to find the critical angle is: \(\begin{array}{l}sin \theta =\frac{n_{2}}{n_{1}}\end{array} \) Where, n1 is the refractive index in medium 1 n2 is the refractive index in medium 2 Ө is the critical angle

What is the formula to find the total internal reflection?

Total internal reflection is given by the formula: \(\begin{array}{l}\frac{n_{1}}{n_{2}}=\frac{sin r}{sin i}\end{array} \) Where, r is the angle of refraction i is the angle of incidence n 1 is the refractive index in medium 1 n 2 is the refractive index in medium 2

State true or false: Shining of crack in a glass-vessel is due to total internal reflection.

Give some examples where optical fibre is used.

• Optical fibres are used in telecommunication for information transfer.
• It is also used in endoscopy.

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Figure 1: Total Internal Reflection (TIR) causes light to bend through a stream of water.

Figure 2: a laser is pointed at a soda pop bottle with a hole at the bottom. total internal reflection (tir) is produced in a stream of water due to the water and light barrier., figure 3: the video shows total internal reflection (tir) in a water jet., explanation.

When light crosses into a new medium, some of the light will refract (bend) and some of it will reflect. The angle of refraction is given by Snell's Law:

Where n is the index of refrection, a property unique to each material. This, and the reflection, is shown in the following diagram.

Figure 4: Light refracts and reflects at the barrier between air and water.

Where θ 1 is the angle of incidence, or the angle of the source of light, θ r is the angle of reflection, and θ 2 is the angle of refraction.

This is the limiting case. If n 1 ≥ n 2 , then θ 2 = 90°.

Figure 5: At the critical angle, light travels through the plane where the two media meet.

In this demo light will continually reflect through the stream of water creating total internal reflection (TIR). The stream of water will 'carry' the light though, to the end of the stream.

Figure 6: Total Internal Reflection (TIR) causes light to bend through a stream of water.

Total Internal Reflection is the principle behind fiber optics.

Figure 7: Total Internal Reflection is used to carry light in fiber optics.

For more information see Wikipedia's entry on Fiber Optics. http://en.wikipedia.org/wiki/Optical_fiber .

• First set up the soda bottle by drilling a hole near the bottom of the bottle. Begin with a drill bit that has a diameter which is slightly larger than the diameter of the laser that will be used. We used a 1/4 inch drill bit, however sizes as small as 7/32 inch worked as well.
• First tape the hole and then fill the bottle with water. The cap will prevent leaking because it creates a vacuum in the bottle.
• Stand the soda bottle on top of a stack of books so the hole is facing the bucket. The laser should be placed in a binder clip so it stays on, and then set on a stack of books and papers. The laser should be lined up so that the laser light goes through the soda bottle, and into the center of the hole. See Figure 2 (top left picture) for details.
• Carefully remove the tape and then unscrew the top of the soda bottle. The light should reflect within the stream of water so that you could see at least a few points of reflection. The light should be visible through the entire stream.
• If the reflections of the light isn't clear, it may be necessary to expand the hole by drilling through the existing hole with a larger drill bit. This process may need to be repeated several times.
• This is an messy experiment. Be ready to adjust the bucket which catches the stream of water.
• Also be aware that the stream's curvature will change as the water level decreases. It will bend closer to the bottle, and the bucket may need to be adjusted again. When the water level is a little above the hole there will be no total internal reflection although the stream will continue. Place the cap back on, or put the bottle inside of the bucket.
• Make sure to have lots of paper towels! Towels or rags could be useful too. However, this mess is water, and therefore easy to clean up.
• Some resources suggest putting a drop of food coloring in the bottom of the bucket to match the laser light, giving the appearance that the water has permanently 'trapped' the colored light. This is a magic trick, and may cause students to misunderstand what total internal reflection is.
• http://wildcat.phys.northwestern.edu/vpl/optics/snell.html This site contains an applet which allows the user to change the angles and indices of refraction. It is a more simple and straight version compared to Stony Brook's (the link below).
• http://www.eserc.stonybrook.edu/ProjectJava/snell/ This site is an applet which allows the user to change the angles and indices of refraction in order to observe reflection and refraction. There is a 'simpler version' option.
• http://en.wikipedia.org/wiki/Total_internal_reflection The Wikipedia entry on Total Internal Reflection
• http://hyperphysics.phy-astr.gsu.edu/Hbase/phyopt/totint.html An explanation of total internal reflection with drawings. It includes a calculator which determines the critical angle when the indices of reflection are provided.
• http://www.glenbrook.k12.il.us/gbssci/phys/CLass/refrn/u14l3b.html This is a tutorial on Total Internal Reflection with many drawings to explain the concepts.

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1.5: Total Internal Reflection

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Learning Objectives

By the end of this section, you will be able to:

• Explain the phenomenon of total internal reflection
• Describe the workings and uses of optical fibers
• Analyze the reason for the sparkle of diamonds

A good-quality mirror may reflect more than 90% of the light that falls on it, absorbing the rest. But it would be useful to have a mirror that reflects all of the light that falls on it. Interestingly, we can produce total reflection using an aspect of refraction.

Consider what happens when a ray of light strikes the surface between two materials, as shown in Figure \(\PageIndex{1a}\). Part of the light crosses the boundary and is refracted; the rest is reflected. If, as shown in the figure, the index of refraction for the second medium is less than for the first, the ray bends away from the perpendicular. (Since \(n_1>n_2\), the angle of refraction is greater than the angle of incidence—that is, \(θ_1>θ_2\).) Now imagine what happens as the incident angle increases. This causes \(θ_2\) to increase also. The largest the angle of refraction \(θ_2\) can be is \(90°\), as shown in Figure \(\PageIndex{1b}\).

The critical angle \(θ_c\) for a combination of materials is defined to be the incident angle \(θ_1\) that produces an angle of refraction of \(90°\). That is, \(θ_c\) is the incident angle for which \(θ_2=90°\). If the incident angle \(θ_1\) is greater than the critical angle, as shown in Figure \(\PageIndex{1c}\), then all of the light is reflected back into medium 1, a condition called total internal reflection . (As Figure \(\PageIndex{1}\) shows, the reflected rays obey the law of reflection so that the angle of reflection is equal to the angle of incidence in all three cases.)

Snell’s law states the relationship between angles and indices of refraction. It is given by

\[n_1\sin θ_1=n_2 \sin θ_2. \nonumber \]

When the incident angle equals the critical angle (\(θ_1=θ_c\)), the angle of refraction is \(90°\) (\(θ_2=90°\)). Noting that \(\sin 90°=1\), Snell’s law in this case becomes

\[n_1 \, \sin \, θ_1 = n_2. \nonumber \]

The critical angle \(θ_c\) for a given combination of materials is thus

\[ θ_c = \sin^{−1}\left(\frac{n_2}{n_1}\right)\label{critical} \]

for \(n_1>n_2\).

Total internal reflection occurs for any incident angle greater than the critical angle \(θ_c\), and it can only occur when the second medium has an index of refraction less than the first. Note that this equation is written for a light ray that travels in medium 1 and reflects from medium 2, as shown in Figure \(\PageIndex{1}\).

Example \(\PageIndex{1}\): Determining a Critical Angle

What is the critical angle for light traveling in a polystyrene (a type of plastic) pipe surrounded by air? The index of refraction for polystyrene is 1.49.

The index of refraction of air can be taken to be 1.00, as before. Thus, the condition that the second medium (air) has an index of refraction less than the first (plastic) is satisfied, and we can use the equation

\[θ_c=\sin^{−1}\left(\frac{n_2}{n_1}\right) \nonumber \]

to find the critical angle \(θ_c\), where \(n_2=1.00\) and \(n_1=1.49\).

Substituting the identified values gives

\[\begin{align} θ_c &= \sin^{−1}\left(\frac{1.00}{1.49}\right) \nonumber \\[4pt] &= \sin^{−1}(0.671) \nonumber \\[4pt] &= 42.2°. \nonumber \end{align} \nonumber \]

Significance

This result means that any ray of light inside the plastic that strikes the surface at an angle greater than 42.2° is totally reflected. This makes the inside surface of the clear plastic a perfect mirror for such rays, without any need for the silvering used on common mirrors. Different combinations of materials have different critical angles, but any combination with \(n_1>n_2\) can produce total internal reflection. The same calculation as made here shows that the critical angle for a ray going from water to air is 48.6°, whereas that from diamond to air is 24.4°, and that from flint glass to crown glass is 66.3°.

Exercise \(\PageIndex{1}\)

At the surface between air and water, light rays can go from air to water and from water to air. For which ray is there no possibility of total internal reflection?

air to water, because the condition that the second medium must have a smaller index of refraction is not satisfied

In the photo that opens this chapter, the image of a swimmer underwater is captured by a camera that is also underwater. The swimmer in the upper half of the photograph, apparently facing upward, is, in fact, a reflected image of the swimmer below. The circular ripple near the photograph’s center is actually on the water surface. The undisturbed water surrounding it makes a good reflecting surface when viewed from below, thanks to total internal reflection. However, at the very top edge of this photograph, rays from below strike the surface with incident angles less than the critical angle, allowing the camera to capture a view of activities on the pool deck above water.

Fiber Optics: Endoscopes to Telephones

Fiber optics is one application of total internal reflection that is in wide use. In communications, it is used to transmit telephone, internet, and cable TV signals. Fiber optics employs the transmission of light down fibers of plastic or glass. Because the fibers are thin, light entering one is likely to strike the inside surface at an angle greater than the critical angle and, thus, be totally reflected (Figure \(\PageIndex{2}\)). The index of refraction outside the fiber must be smaller than inside. In fact, most fibers have a varying refractive index to allow more light to be guided along the fiber through total internal refraction. Rays are reflected around corners as shown, making the fibers into tiny light pipes.

Bundles of fibers can be used to transmit an image without a lens, as illustrated in Figure \(\PageIndex{3}\). The output of a device called an endoscope is shown in Figure \(\PageIndex{1b}\). Endoscopes are used to explore the interior of the body through its natural orifices or minor incisions. Light is transmitted down one fiber bundle to illuminate internal parts, and the reflected light is transmitted back out through another bundle to be observed.

Fiber optics has revolutionized surgical techniques and observations within the body, with a host of medical diagnostic and therapeutic uses. Surgery can be performed, such as arthroscopic surgery on a knee or shoulder joint, employing cutting tools attached to and observed with the endoscope. Samples can also be obtained, such as by lassoing an intestinal polyp for external examination. The flexibility of the fiber optic bundle allows doctors to navigate it around small and difficult-to-reach regions in the body, such as the intestines, the heart, blood vessels, and joints. Transmission of an intense laser beam to burn away obstructing plaques in major arteries, as well as delivering light to activate chemotherapy drugs, are becoming commonplace. Optical fibers have in fact enabled microsurgery and remote surgery where the incisions are small and the surgeon’s fingers do not need to touch the diseased tissue.

Optical fibers in bundles are surrounded by a cladding material that has a lower index of refraction than the core (Figure \(\PageIndex{4}\)). The cladding prevents light from being transmitted between fibers in a bundle. Without cladding, light could pass between fibers in contact, since their indices of refraction are identical. Since no light gets into the cladding (there is total internal reflection back into the core), none can be transmitted between clad fibers that are in contact with one another. Instead, the light is propagated along the length of the fiber, minimizing the loss of signal and ensuring that a quality image is formed at the other end. The cladding and an additional protective layer make optical fibers durable as well as flexible.

Special tiny lenses that can be attached to the ends of bundles of fibers have been designed and fabricated. Light emerging from a fiber bundle can be focused through such a lens, imaging a tiny spot. In some cases, the spot can be scanned, allowing quality imaging of a region inside the body. Special minute optical filters inserted at the end of the fiber bundle have the capacity to image the interior of organs located tens of microns below the surface without cutting the surface—an area known as nonintrusive diagnostics. This is particularly useful for determining the extent of cancers in the stomach and bowel.

In another type of application, optical fibers are commonly used to carry signals for telephone conversations and internet communications. Extensive optical fiber cables have been placed on the ocean floor and underground to enable optical communications. Optical fiber communication systems offer several advantages over electrical (copper)-based systems, particularly for long distances. The fibers can be made so transparent that light can travel many kilometers before it becomes dim enough to require amplification—much superior to copper conductors. This property of optical fibers is called low loss. Lasers emit light with characteristics that allow far more conversations in one fiber than are possible with electric signals on a single conductor. This property of optical fibers is called high bandwidth. Optical signals in one fiber do not produce undesirable effects in other adjacent fibers. This property of optical fibers is called reduced crosstalk. We shall explore the unique characteristics of laser radiation in a later chapter.

Corner Reflectors and Diamonds

Corner reflectors are perfectly efficient when the conditions for total internal reflection are satisfied. With common materials, it is easy to obtain a critical angle that is less than 45°. One use of these perfect mirrors is in binoculars, as shown in Figure \(\PageIndex{5}\). Another use is in periscopes found in submarines.

Total internal reflection, coupled with a large index of refraction, explains why diamonds sparkle more than other materials. The critical angle for a diamond-to-air surface is only 24.4° , so when light enters a diamond, it has trouble getting back out (Figure \(\PageIndex{6}\)). Although light freely enters the diamond, it can exit only if it makes an angle less than 24.4°. Facets on diamonds are specifically intended to make this unlikely. Good diamonds are very clear, so that the light makes many internal reflections and is concentrated before exiting—hence the bright sparkle. (Zircon is a natural gemstone that has an exceptionally large index of refraction, but it is not as large as diamond, so it is not as highly prized. Cubic zirconia is manufactured and has an even higher index of refraction (≈ 2.17 ), but it is still less than that of diamond.) The colors you see emerging from a clear diamond are not due to the diamond’s color, which is usually nearly colorless, but result from dispersion . Colored diamonds get their color from structural defects of the crystal lattice and the inclusion of minute quantities of graphite and other materials. The Argyle Mine in Western Australia produces around 90% of the world’s pink, red, champagne, and cognac diamonds, whereas around 50% of the world’s clear diamonds come from central and southern Africa.

Explore refraction and reflection of light between two media with different indices of refraction. Try to make the refracted ray disappear with total internal reflection. Use the protractor tool to measure the critical angle and compare with the prediction from Equation \ref{critical}.

25.4 Total Internal Reflection

Learning objectives.

By the end of this section, you will be able to:

• Explain the phenomenon of total internal reflection.
• Describe the workings and uses of fiber optics.
• Analyze the reason for the sparkle of diamonds.

A good-quality mirror may reflect more than 90% of the light that falls on it, absorbing the rest. But it would be useful to have a mirror that reflects all of the light that falls on it. Interestingly, we can produce total reflection using an aspect of refraction .

Consider what happens when a ray of light strikes the surface between two materials, such as is shown in Figure 25.13 (a). Part of the light crosses the boundary and is refracted; the rest is reflected. If, as shown in the figure, the index of refraction for the second medium is less than for the first, the ray bends away from the perpendicular. (Since n 1 > n 2 n 1 > n 2 , the angle of refraction is greater than the angle of incidence—that is, θ 2 > θ 1 θ 2 > θ 1 .) Now imagine what happens as the incident angle is increased. This causes θ 2 θ 2 to increase also. The largest the angle of refraction θ 2 θ 2 can be is 90º 90º , as shown in Figure 25.13 (b).The critical angle θ c θ c for a combination of materials is defined to be the incident angle θ 1 θ 1 that produces an angle of refraction of 90º 90º . That is, θ c θ c is the incident angle for which θ 2 = 90º θ 2 = 90º . If the incident angle θ 1 θ 1 is greater than the critical angle, as shown in Figure 25.13 (c), then all of the light is reflected back into medium 1, a condition called total internal reflection .

Critical Angle

The incident angle θ 1 θ 1 that produces an angle of refraction of 90º 90º is called the critical angle, θ c θ c .

Snell’s law states the relationship between angles and indices of refraction. It is given by

When the incident angle equals the critical angle ( θ 1 = θ c θ 1 = θ c ), the angle of refraction is 90º 90º ( θ 2 = 90º θ 2 = 90º ). Noting that sin 90º =1 sin 90º =1 , Snell’s law in this case becomes

The critical angle θ c θ c for a given combination of materials is thus

Total internal reflection occurs for any incident angle greater than the critical angle θ c θ c , and it can only occur when the second medium has an index of refraction less than the first. Note the above equation is written for a light ray that travels in medium 1 and reflects from medium 2, as shown in the figure.

Example 25.4

How big is the critical angle here.

What is the critical angle for light traveling in a polystyrene (a type of plastic) pipe surrounded by air?

The index of refraction for polystyrene is found to be 1.49 in Figure 25.14 , and the index of refraction of air can be taken to be 1.00, as before. Thus, the condition that the second medium (air) has an index of refraction less than the first (plastic) is satisfied, and the equation θ c = sin − 1 n 2 / n 1 θ c = sin − 1 n 2 / n 1 can be used to find the critical angle θ c θ c . Here, then, n 2 = 1 . 00 n 2 = 1 . 00 and n 1 = 1 . 49 n 1 = 1 . 49 .

The critical angle is given by

Substituting the identified values gives

This means that any ray of light inside the plastic that strikes the surface at an angle greater than 42.2º 42.2º will be totally reflected. This will make the inside surface of the clear plastic a perfect mirror for such rays without any need for the silvering used on common mirrors. Different combinations of materials have different critical angles, but any combination with n 1 > n 2 n 1 > n 2 can produce total internal reflection. The same calculation as made here shows that the critical angle for a ray going from water to air is 48 . 6º 48 . 6º , while that from diamond to air is 24 . 4º 24 . 4º , and that from flint glass to crown glass is 66 . 3º 66 . 3º . There is no total reflection for rays going in the other direction—for example, from air to water—since the condition that the second medium must have a smaller index of refraction is not satisfied. A number of interesting applications of total internal reflection follow.

Fiber Optics: Endoscopes to Telephones

Fiber optics is one application of total internal reflection that is in wide use. In communications, it is used to transmit telephone, internet, and cable TV signals. Fiber optics employs the transmission of light down fibers of plastic or glass. Because the fibers are thin, light entering one is likely to strike the inside surface at an angle greater than the critical angle and, thus, be totally reflected (See Figure 25.14 .) The index of refraction outside the fiber must be smaller than inside, a condition that is easily satisfied by coating the outside of the fiber with a material having an appropriate refractive index. In fact, most fibers have a varying refractive index to allow more light to be guided along the fiber through total internal refraction. Rays are reflected around corners as shown, making the fibers into tiny light pipes.

Bundles of fibers can be used to transmit an image without a lens, as illustrated in Figure 25.15 . The output of a device called an endoscope is shown in Figure 25.15 (b). Endoscopes are used to explore the body through various orifices or minor incisions. Light is transmitted down one fiber bundle to illuminate internal parts, and the reflected light is transmitted back out through another to be observed. Surgery can be performed, such as arthroscopic surgery on the knee joint, employing cutting tools attached to and observed with the endoscope. Samples can also be obtained, such as by lassoing an intestinal polyp for external examination.

Fiber optics has revolutionized surgical techniques and observations within the body. There are a host of medical diagnostic and therapeutic uses. The flexibility of the fiber optic bundle allows it to navigate around difficult and small regions in the body, such as the intestines, the heart, blood vessels, and joints. Transmission of an intense laser beam to burn away obstructing plaques in major arteries as well as delivering light to activate chemotherapy drugs are becoming commonplace. Optical fibers have in fact enabled microsurgery and remote surgery where the incisions are small and the surgeon’s fingers do not need to touch the diseased tissue.

Fibers in bundles are surrounded by a cladding material that has a lower index of refraction than the core. (See Figure 25.16 .) The cladding prevents light from being transmitted between fibers in a bundle. Without cladding, light could pass between fibers in contact, since their indices of refraction are identical. Since no light gets into the cladding (there is total internal reflection back into the core), none can be transmitted between clad fibers that are in contact with one another. The cladding prevents light from escaping out of the fiber; instead most of the light is propagated along the length of the fiber, minimizing the loss of signal and ensuring that a quality image is formed at the other end. The cladding and an additional protective layer make optical fibers flexible and durable.

The cladding prevents light from being transmitted between fibers in a bundle.

Special tiny lenses that can be attached to the ends of bundles of fibers are being designed and fabricated. Light emerging from a fiber bundle can be focused and a tiny spot can be imaged. In some cases the spot can be scanned, allowing quality imaging of a region inside the body. Special minute optical filters inserted at the end of the fiber bundle have the capacity to image tens of microns below the surface without cutting the surface—non-intrusive diagnostics. This is particularly useful for determining the extent of cancers in the stomach and bowel.

Most telephone conversations and Internet communications are now carried by laser signals along optical fibers. Extensive optical fiber cables have been placed on the ocean floor and underground to enable optical communications. Optical fiber communication systems offer several advantages over electrical (copper) based systems, particularly for long distances. The fibers can be made so transparent that light can travel many kilometers before it becomes dim enough to require amplification—much superior to copper conductors. This property of optical fibers is called low loss . Lasers emit light with characteristics that allow far more conversations in one fiber than are possible with electric signals on a single conductor. This property of optical fibers is called high bandwidth . Optical signals in one fiber do not produce undesirable effects in other adjacent fibers. This property of optical fibers is called reduced crosstalk . We shall explore the unique characteristics of laser radiation in a later chapter.

Corner Reflectors and Diamonds

A light ray that strikes an object consisting of two mutually perpendicular reflecting surfaces is reflected back exactly parallel to the direction from which it came. This is true whenever the reflecting surfaces are perpendicular, and it is independent of the angle of incidence. Such an object, shown in Figure 25.17 , is called a corner reflector , since the light bounces from its inside corner. Many inexpensive reflector buttons on bicycles, cars, and warning signs have corner reflectors designed to return light in the direction from which it originated. It was more expensive for astronauts to place one on the moon. Laser signals can be bounced from that corner reflector to measure the gradually increasing distance to the moon with great precision.

Corner reflectors are perfectly efficient when the conditions for total internal reflection are satisfied. With common materials, it is easy to obtain a critical angle that is less than 45º 45º . One use of these perfect mirrors is in binoculars, as shown in Figure 25.18 . Another use was in optical periscopes found in older submarines, although modern subs have replaced periscopes and their lenses, mirrors, and prisms with photonic masts that use sensors, lasers, and fiber optics.

The Sparkle of Diamonds

Total internal reflection, coupled with a large index of refraction, explains why diamonds sparkle more than other materials. The critical angle for a diamond-to-air surface is only 24 . 4º 24 . 4º , and so when light enters a diamond, it has trouble getting back out. (See Figure 25.19 .) Although light freely enters the diamond, it can exit only if it makes an angle less than 24 . 4º 24 . 4º . Facets on diamonds are specifically intended to make this unlikely, so that the light can exit only in certain places. Good diamonds are very clear, so that the light makes many internal reflections and is concentrated at the few places it can exit—hence the sparkle. (Zircon is a natural gemstone that has an exceptionally large index of refraction, but not as large as diamond, so it is not as highly prized. Cubic zirconia is manufactured and has an even higher index of refraction ( ≈ 2.17 ≈ 2.17 ), but still less than that of diamond.) The colors you see emerging from a sparkling diamond are not due to the diamond’s color, which is usually nearly colorless. Those colors result from dispersion, the topic of Dispersion: The Rainbow and Prisms . Colored diamonds get their color from structural defects of the crystal lattice and the inclusion of minute quantities of graphite and other materials. The Argyle Mine in Western Australia produces around 90% of the world’s pink, red, champagne, and cognac diamonds, while around 50% of the world’s clear diamonds come from central and southern Africa.

PhET Explorations

Bending light.

Explore bending of light between two media with different indices of refraction. See how changing from air to water to glass changes the bending angle. Play with prisms of different shapes and make rainbows.

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Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units

• Authors: Paul Peter Urone, Roger Hinrichs
• Publisher/website: OpenStax
• Book title: College Physics 2e
• Publication date: Jul 13, 2022
• Location: Houston, Texas
• Book URL: https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units
• Section URL: https://openstax.org/books/college-physics-2e/pages/25-4-total-internal-reflection

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Total Internal Reflection - Complete Toolkit

• To describe how the relative intensity of a reflected ray and refracted ray at a boundary is dependent upon the angle of incidence.  
• To define total internal reflection and to state the conditions that are required in order for a light ray to undergo total internal reflection.  
• To define the critical angle, to derive an equation for the critical angle from Snell's law equation, and to use the equation to calculate the critical angle for the boundary between two media.  

Readings from The Physics Classroom Tutorial

• The Physics Classroom Tutorial, Refraction and the Ray Model of Light Chapter, Lesson 3

Interactive Simulations

Video and Animations

Labs and Investigations

• The Physics Classroom, The Laboratory, R and R Lab Students use laser light and a Lucite hemicylindrical prism to investigate the effect of the angle of incidence upon the relative brightness of the reflected and transmitted ray.  
• The Physics Classroom, The Laboratory, A Critical Lab Students experimentally determine the critical angle for the air-water boundary and the air-Lucite boundary. Link:  http://www.physicsclassroom.com/lab/refrn/RLlabs.html

Demonstration Ideas

Minds On Physics Internet Modules:

• Refraction and Lenses, Ass’t RL5 -  Total Internal Reflection
• Refraction and Lenses, Ass’t RL6 -  TIR and the Critical Angle

Concept Building Exercises:

• The Curriculum Corner, Refraction and Lenses, Total Internal Reflection

Problem-Solving Exercises:

• The Calculator Pad, Refraction and Lenses, Problems #13-#18 Link:  http://www.physicsclassroom.com/calcpad/refrn/problems

Science Reasoning Activities:

• Reflection and Transmission Link: http://www.physicsclassroom.com/reasoning/refraction

Real Life Connections:

• Different patterns may be observed at each of the scales at which a system is studied and can provide evidence for causality in explanation of phenomena.
• Models (e.g., physical, mathematical, computer models) can be used to simulate systems and interactions – including energy, matter, and information flows – between and within systems at different scales. 
• Develop and use models (physical, mathematical, and computer models) to predict the behavior of a system and design systems to do specific tasks.
• Algebraic thinking is used to examine scientific data and predict the effect of a change in one variable on another.
• Analyze data using tools, technologies, and/or models (e.g., computational, mathematical) in order to make valid and reliable scientific claims.
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• Use a model to provide mechanistic accounts of phenomena.
• Plan and conduct an investigation individually and collaboratively to produce data to serve as the basis for evidence, and in the design: decide on types, how much, and accuracy of data needed to produce reliable measurements.
• Construct an explanation that includes qualitative or quantitative relationships between variables that predict phenomena.
• Construct and review an explanation based on valid and reliable evidence obtained from a variety of sources (including students’ own investigations, models, simulations, and peer review) and the assumption that theories and laws that describe the natural world operate today as they did in the past and will continue to do so in the future.
• Make and defend a claim based on evidence about the natural world that reflects scientific knowledge and student-generated evidence. 

Total internal reflection experiments and optical fibres

There are some great demonstrations of total internal reflection occurring in classrooms throughout the world and here are some of them.

Tank of water

The simplest experiment to show total internal reflection is to fill a perspex tank with water. Angle a directed ray of light to the underneath of the surface and total internal reflection can be seen (it is easier to see with the lights off).

Large solid plastic coil

If your school physics lab has a coil of large solid plastic there is a very clear demonstration of total internal reflection that can be seen.

If you shine a laser light down the one end of the coil you will see the path of the laser light all the way through the solid tube even though it is coiled.

Optical fibre cables work on exactly the same principle and directed laser lights are used to communicate electronic signals in the form of rays of light. This relies entirely on total internal reflection.

Fibre optic cables are made of a plastic and glass polymers with cables which can get as thin as a human hair.

Benefits of fibre optic cables include

• They can carry far more information over the same cross-sectional area than copper wire.
• Optical fibres can carry light rays over long distances with little losses compared to copper wires.
• Optical fibres are not affected by electromagnetic losses which are a big problem with metal wires.

NOTE: Quite often the optical fibre cable is covered in a transparent protective medium. That medium is always of a lower index of refraction, otherwise total internal reflection doesn't work.

Large bottle of water

Take a large plastic bottle, drill a hole in the side and fill the hole with a cork. Fill the bottle with water, shine a laser through the bottle and pull out the cork. Direct the laser light into the stream of water, as shown below.

The laser light becomes trapped in the stream of water. With the denser medium (water) against air and the beam of light angled at more than the critical angle, we get total internal reflection.

Optical fibres

Of course, if you have the resources you could also observe how light images can be bent around corners.

If you have a fibre optic cable made up of thousands of strands of only a hair's width, you can see around corners if the cable is bent.

You can observe how images can bend around corners.

Collection of Physics Experiments

Total internal reflection in a stream of water, experiment number : 1765, goal of experiment.

This experiment demonstrates a total internal reflection and shows the principle of optical fibres.

Refraction of light at the interface of two media is described by Snell’s law

wherre \(\alpha\) is the angle of incidence, \(\beta\) angle of refraction and n 1 , n 2 are refractive indices of the media. In order to observe the total reflection, the light must pass from an optically denser medium to an optically thinner medium. The propagating wave refracts away from the normal and with increasing angle of incidence the angle of refraction increases. When we reach the critical angle, the refraction angle reaches the largest possible value of 90°. The critical angle is therefore the largest angle of incidence at which refraction occurs. At larger angles of incidence the refraction does not occur and the ray is reflected at the interface. We call this a total internal reflection. On this phenomenon optical fibres are based. In the figure below there is a ray passing in an optical fibre.

Laser pointer, plastic bottle, milk, container for water, nail.

• Using the nail, make a hole in the bottle about 5 cm above the bottom.
• Place the bottle on an elevated place and fill it with water.
• Place the container under the bottle, so that it catches the water spilling out from it.
• Point the laser beam horizontally through the bottle, so that it comes through to the hole.
• Observe the beam in the spilling water.

Sample result

Video below illustrates the experiment performation.

Technical notes

• For easier penetration of the bottle with a nail, heat the nail above a candle.
• The hole should be wider that the width of the laser beam, so that no unwanted refraction occurs. Above that, the laser beam can pass through the stream of water and you should be able to see some points of refraction, which visualises the demonstrated phenomenon.
• When filling the bottle, it is recommended to seal the hole and the bottle so that the water does not spill until the experiment is prepared.
• The effect can be improved by adding a few drops of milk into the water, which makes the laser beam more visible.

Pedagogical notes

• In this experiment, not only the reflection of the laser beam occurs. There is also a refraction of the beam out from the water stream; otherwise we would not be able to see the beam.

Wait, isn't water supposed to be transparent?

When can you see through the surface, and when does it reflect?

Total internal reflection

Although clean water is transparent, it is not always possible to look through a water surface.

Let your students find out under what conditions the air-water interface lets light pass, and when it reflects the light.

Finding the conditions under which total internal reflection takes place.

Fill a transparent plastic cup or glass with water and hold it just a bit higher than your eyes. Stick your fingertip a centimeter into the water.

1. Why can't you see all of your finger, but only your fingertip and its reflection?

2. Look at the glass from different sides, including top and bottom. Under what conditions can you see through the water surface, and when does it reflect?

Does it make a difference if the other side of the water surface is very bright or dark? › No.

Do you note a difference between the reflections seen from above and from below the water surface? › Reflections seen from above are weaker than those observed from below.

Is light passing from water to air refracted at the surface towards the normal or away from it? › It is refracted away from it.

What happens to a light beam that, according to Snell's law, would be refracted more than 90˚ from the normal? › It is reflected at the surface.

It is easy to see through a water surface from above, even if some weak reflections reduce the contrast. As long as you look along an axis close to the surface normal, it is also possible to see through the water surface from below. However, for observation angles larger than arcsin( n air / n water ) ≈ 48˚ to the surface normal, there is no light visible from above the surface, because it has been refracted to smaller angles. Instead, the observer sees a reflection – as if the surface was a mirror.

It has to be noted though, that due to the second refraction on the glass surface, your students will probably see a total internal reflection whenever they look from the side.

Fingerprints

Magnifying glass near and far

Finding the mirror image

The true colors of white

Does a mirror really invert sides?

Total Internal Reflection ( CIE IGCSE Physics )

Revision note.

Total Internal Reflection

• This phenomenon is called total internal reflection
• Total internal reflection (TIR) occurs when:

The angle of incidence is greater than the critical angle and the incident material is denser than the second material

• The angle of incidence > the critical angle
• The incident material is denser than the second material

Critical angle and total internal reflection

• Optical fibres e.g. endoscopes
• Prisms e.g. periscopes
• They are also used in safety reflectors for bicycles and cars, as well as posts marking the side or edge of roads
• It consists of two right-angled prisms

Reflection of light through a periscope

• The light totally internally reflects in both prisms

Single and double reflection through right-angled prisms

If asked to name the phenomena make sure you give the whole name – total internal reflection

Remember: total internal reflection occurs when light travels from a denser material to less dense material and ALL of the light is reflected.

If asked to give an example of a use of total internal reflection, first state the name of the object that causes the reflection (e.g. a right-angled prism) and then name the device in which it is used (e.g. a periscope)

Critical Angle

• As the angle of incidence is increased, the angle of refraction also increases until it gets closer to 90°
• At this point, the angle of incidence is known as the critical angle   c

As the angle of incidence increases it will eventually surplus the critical angle and lead to total internal reflection of the light

• This is total internal reflection

Worked example

Step 1: Draw the reflected angle at the glass-liquid boundary

• When a light ray is reflected, the angle of incidence = angle of reflection
• Therefore, the angle of incidence (or reflection) is 90° – 25° = 65°

Step 2: Draw the refracted angle at the glass-air boundary

• At the glass-air boundary, the light ray refracts away from the normal
• Due to the reflection, the light rays are symmetrical to the other side

Step 3: Calculate the critical angle

• The question states the ray is “totally internally reflected for the first time” meaning that this is the lowest angle at which TIR occurs
• Therefore, 65° is the critical angle

If you are asked to explain what is meant by the critical angle in an exam, you can be sure to gain full marks by drawing and labelling the same diagram above (showing the three semi-circular blocks)

Refractive Index & Critical Angle Equation

• The critical angle, c , of a material is related to its refractive index , n
• The relationship between the two quantities is given by the equation:

• This can also be rearranged to calculate the refractive index, n:

• The larger the refractive index of a material, the smaller the critical angle
• Light rays inside a material with a high refractive index are more likely to be totally internally reflected

Opals and diamonds are transparent stones used in jewellery. Jewellers shape the stones so that light is reflected inside.Compare the critical angles of opal and diamond and explain which stone would appear to sparkle more.

The refractive index of opal is about 1.5

The refractive index of diamond is about 2.4

Step 1: List the known quantities

• Refractive index of opal, n o = 1.5
• Refractive index of diamond, n d = 2.4

Step 2: Write out the equation relating critical angle and refractive index

Step 3: Calculate the critical angle of opal (c o )

sin(c o ) = 1 ÷ 1.5 = 0.6667

c o = sin –1 (0.6667) = 41.8 = 42°

Step 4: Calculate the critical angle of diamond (c d )

sin(c d ) = 1 ÷ 2.4 = 0.4167

c d = sin –1 (0.4167) = 24.6 = 25°

Step 5: Compare the two values and write a conclusion

• Total internal reflection occurs when the angle of incidence of light is larger than the critical angle (i>c)
• In opal, total internal reflection will occur for angles of incidence between 42° and 90°
• The critical angle of diamond is lower than the critical angle of opal (c o >c d )
• This means light rays will be totally internally reflected in diamond over a larger range of angles (25° to 90°)
• Therefore, more total internal reflection will occur in diamond hence it will appear to sparkle more than the opal

When calculating the value of the critical angle using the above equation:

• First use the refractive index, n , to find sin( c )
• Then use the inverse sin function (sin –1 ) to find the value of c

Optical Fibres

• Communications
• Decorative lamps
• Light travelling down an optical fibre is totally internally reflected each time it hits the edge of the fibre

Optical fibres utilise total internal reflection for communications

• Optical fibres are also used in medicine in order to see within the human body

Endoscopes utilise total internal reflection to see inside a patient's body

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Author: Katie M

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

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Total Internal Reflection

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