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3.6: Reflection, Refraction, and Dispersion

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Page ID 18457

Tom Weideman
University of California, Davis

As we consider more phenomena associated with light, one of our primary concerns will be the direction that light is traveling. We already know that light, like any wave, travels in a direction perpendicular to its planes of constant phase:

Figure 3.6.1 – Light Waves Travel in Several Directions at Once

So in our wave view of light, we say that the light wave is traveling in many directions at once, but now we are going to change our perspective to that of an observer and a source. When we do that, we narrow down all the possible directions of the light wave motion to a single line, which we call a light ray . This is a directed line that originates at the source of light, and ends at the observer of the light:

Figure 3.6.2 – Source and Observer Define a Ray

When most people encounter the idea of a light ray for the first time, what they think of is a thinly-confined laser beam. This is not what is meant here! The ray has no physical meaning in terms of the confinement of light – we just use it as a simple geometrical device to link a source to an observer. Always keep in mind that the actual physical manifestation of the light is a wave that is usually traveling in many directions at once! Our use of rays will become so ubiquitous that this will be easy to forget.

Consider a point source of light that sends out a spherical wave toward an imaginary flat plane, as in the left diagram below. When the wave reaches this plane, then according to Huygens's principle, we can look at every point on the plane and treat it as a point source for an individual wavelet (center diagram below). These wavelets are not in phase, because they are all travel different distances from the source to the plane, and when they are superposed, we know the result is what we see, which is a continued spherical wave (right diagram below).

Figure 3.6.3 – Spherical Wave Passes Through Imaginary Plane

Now suppose the plane is not imaginary, but instead reflects the wave. Every point on this plane becomes a source of a wavelet, but this time, the wave created by these wavelets is going in the opposite direction. The wavelets have the same relative phases as in the previous case, and they are completely symmetric, so they superpose to give the same total wave as before, with the exception that it is a mirror image of the case of the imaginary plane:

Figure 3.6.4 – Spherical Wave Reflects Off Plane

Thanks to the symmetry of the situation, it's not difficult to see that the reflected wave is identical to a spherical wave that has originated from a point on the opposite side of the reflecting plane, exactly the same distance from the plane as the source, and along the line that runs through the source perpendicular to the surface:

Figure 3.6.5 – Image of a Reflected Wave

Of course, there isn't actually a point light source on the other side of the reflecting plane, it's just that someone looking at the reflected light – no matter where they look from – will see the wave originating from the direction of that point. We call such a point an image of the original source of the light.

Now let's put this result in terms of light rays. To do this, we need a source and an observer, and this case, we will require also that a reflection has taken place. Once again drawing the rays perpendicular to the wave fronts, we get:

Figure 3.6.6 – Law of Reflection

It's clear from the symmetry of the situation that the angle the ray makes with the perpendicular (the horizontal dotted line) to the reflecting plane as it approaches, is the same as the angle it makes after it is reflected. This gives us the law of reflection , which states that the incoming angle ( angle of incidence ) equals the outgoing angle ( angle of reflection ):

\[\theta_i = \theta_r\]

The beauty of introducing rays is that from this point on, we can discuss sources and observers without a complicated reference to the spherical waves and Huygens's principle – we can just use the law of reflection and pure geometry.

We saw that light waves have the capability of changing the direction of the rays associated with it through diffraction. We now consider another way that such a direction change can occur. This process, called refraction , comes about when a wave moves into a new medium. To get to the essence of this phenomenon from Huygens's principle, we don't have a symmetry trick like we did for reflection, so rather than use a point source of the light, we can look at the effect that changing the medium has on a plane wave.

We saw in Figure 3.1.2 how a plane wave propagates according to Huygens's Principle. We can't sketch every one wavelets emerging from the infinite number of points on the wavefront, but we can sketch a few representative wavelets, and if those wavelets have propagated for equal periods of time, then a line tangent to all the wavelets will represent the next wavefront. It's clear that following this procedure for a plane wave will continue the plane wave in the same direction. But now let's imagine that such a plane wave approaches a new medium from an angle, as shown in the figure below. As each point on the wave front comes in contact with the new medium, it becomes a source for a new Huygens wavelet within the medium . These wavelets will travel at a different rate than they traveled in the previous medium (in the figure, the light wave is slowing down in the new medium). This means that the distance the wave in medium #1 travels is farther than it travels in medium #2 during the same time. The effect is a bending of the direction of the plane wave in medium #2 relative to medium #1.

Figure 3.6.7 – Huygens's Principle Refracts a Plane Wave

The amount that the direction of the light ray changes when the wave enters a new medium depends upon how much the wave slows down or speeds up upon changing media. In other words, it depends upon the indices of refraction of the two media. We can actually calculate this effect by freezing the figure above and looking at some triangles:

Figure 3.6.8 – The Geometry of Refraction

$refraction_geometry.png$

We are looking at what happens to a wavefront when it passes from position $A$ to position $B$. The left side of the wave front is traveling within medium #2, during the same time period that the right side is traveling through medium #1. The rays are by definition perpendicular to the wavefronts, and we have defined the angles the rays make with the perpendicular in each medium as $\theta_1$ and $\theta_2$. Before we do any of the math at all, we immediately note:

Light passing from a faster medium into a slower medium bends toward the perpendicular, and light passing from a slower medium to a faster medium bends away from the perpendicular.

While the second of these conclusions is not expressed in our figure, it's not hard to see that it must be true, if we just imagine the wavefronts in the figure moving up to the left from medium #2 to medium #1.

Now for the math. We have two right triangles (yellow and orange) with a common hypotenuse of length we have called $L$. The distance between wavefronts in the upper medium is the speed of the wave there ($\frac{c}{n_1}$) multiplied by the time spent propagating, while the distance measured within the lower medium is calculated the same way, with a different speed ($\frac{c}{n_2}$). The angle $\theta_1$ (shown on the right side of the diagram) is clearly the complement of the acute angle on the right-hand-side of the yellow triangle, which makes it equal to the acute angle on the left-hand-side of the yellow triangle. We therefore have:

\[\sin\theta_1=\dfrac{\left(\frac{c}{n_1}\right)t}{L}\]

Similarly we find for $\theta_2$:

\[\sin\theta_2=\dfrac{\left(\frac{c}{n_2}\right)t}{L}\]

Dividing these two equations results in $c$ and $L$ dropping out, leaving:

\[n_1\sin\theta_1 = n_2\sin\theta_2\]

This relationship between the rays of a light wave which changes media is called the law of refraction , or Snell's law . While this works in either direction of light propagation, for reasons that will be clear next, it is generally accepted that the "1" subscript applies to the medium where the light is coming from, and the "2" subscript the medium that the light is going into.

Total Internal Reflection

It was noted above that light which passes from a slower medium to a faster one bends away from the perpendicular. What happens then if the incoming angle is made larger and larger (obviously it can't be more than $90^o$)? For example, suppose we have $n_1=2.0$, $\theta_1=45^o$, and $n_2=1.0$. Plugging these values into Snell's law gives:

\[\sin\theta_2 = \frac{n_1}{n_2}\sin\theta_1 = 2.0\cdot \sin 45^o = 1.4 \]

The sine function can never exceed 1, so there is no solution to this. This means that the light incident at this angle cannot be transmitted into the new medium . Every time light strikes a new medium some can be transmitted, and some reflected, so this result tells us that all of it must be reflected back into the medium in which it started. This phenomenon is called total internal reflection . The angle at which all of this first blows up is the one where the outgoing angle equals $90^o$ (the outgoing light refracts parallel to the surface between the two media). This angle is called the critical angle , and is computed by choosing the outgoing angle to be $90^o$:

\[n_1\sin\theta_c = n_2 \sin 90^o \;\;\;\Rightarrow\;\;\; \theta_c =\sin^{-1}\left(\dfrac{n_2}{n_1}\right)\]

Figure 3.6.9 – Partial and Total Internal Reflections By Incident Angle

Note that there is at least partial reflection (obeying the law of reflection) every time the light hits the surface, but all of the light along that ray is only reflected when the ray's angle exceeds the critical angle.

Note that when light is coming from one medium to another, unless that light is a plane wave, it will be moving in many directions at once. Only the portions of the light wave with rays that equal or exceed the critical angle are not transmitted into the new medium. So the word "total" in "total internal reflection" to express the fraction of light at a specific angle that is reflected back, not necessarily the fraction of all the light that is reflected back.

Example $\PageIndex{1}$

The diagram to the right shows the path of a ray of monochromatic light as it hits the surfaces between four different media (only the primary ray is considered – partial reflections are ignored). Order the four media according to the magnitudes of their indices of refraction.

We know from Snell’s Law that when light passes from a higher index to a lower one, it bends away from the perpendicular, so we immediately have $n_1>n_2>n_3$. For the ray to reflect back from the fourth medium, it has to be a total internal reflection (we are only considering primary rays, so this is not a partial reflection), which can only occur when light is going from a higher index of refraction to a lower one, so $n_3>n_4$.

What determines the index of refraction for a medium is a very complicated problem in E&M, but there is one easily-observable fact: The amount that a ray bends as it enters a new medium is dependent upon the light’s frequency. Specifically, the higher the frequency of the light, the more it bends – it essentially experiences a higher index of refraction when its frequency is higher. This phenomenon is most evident when white light is shone through a refracting object. The most iconic example of this is white light through a prism.

Figure 3.6.10 – Dispersion Through a Prism

Screen Shot 2019-08-25 at 11.15.58 PM.png

The emergence of the fully-separated spectrum of colors from a prism is reminiscent of a rainbow, and in fact rainbows are also a result of dispersion. Unlike the prism depicted above, however, internal reflection is an integral part of the rainbow effect (and in fact prisms can also feature internal reflection).

A droplet of water suspended in the atmosphere is a refracting sphere. White light that enters near the top of the droplet gets dispersed inside the droplet, reflects, and then gets dispersed as it exits the droplet, sending rays of different-colored light in different directions. The diagram below shows this effect for rays of red and blue light for two droplets.

Figure 3.6.11 – Rainbows

A few things to note here:

Notice that the sun always needs to be behind the observer in order to witness a rainbow. That’s why it seems to move as you move, and why reaching the “end of the rainbow” is impossible (unless you can catch a leprechaun).
The reason it is shaped like a bow is that the sun is nearly a point source, so the geometry is symmetric around the line joining the sun and the observer. If you create a “human-made rainbow” with a light and some mist, you can get close to an entire circle (minus whatever light your body blocks out).
The secondary rainbow above the primary one comes from the light that enters the bottom of the droplets, and has two internal reflections. This reversed direction of the light bouncing around inside the droplets results in the colors being reversed (the violet is at the top and the red at the bottom).

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Law of Reflection Problems

Introduction to law of reflection problems.

The law of reflection is an essential concept in optics that describes the behavior of light when it reflects off a surface. It states that the angle of incidence is equal to the angle of reflection, and it applies to all types of surfaces, including mirrors, water, and glass. However, applying this law to real-world scenarios can be challenging, and it requires an understanding of the concepts involved and the ability to identify the relevant variables.

Understanding the Law of Reflection

The law of reflection is based on the principle that light travels in straight lines and that it changes direction when it interacts with a surface. The angle of incidence is the angle between the incoming light ray and the normal, which is a line perpendicular to the surface at the point of incidence. The angle of reflection is the angle between the reflected light ray and the normal. When light reflects off a smooth surface, the angle of incidence equals the angle of reflection, and the reflected rays are parallel to each other.

Common Challenges with Reflection Problems

One of the main challenges with reflection problems is identifying the relevant variables and applying the correct formula. For example, when a light ray is incident on a curved mirror, the angle of incidence and the angle of reflection are not the same, and the reflected rays are not parallel. In this case, the mirror equation and the magnification equation must be used to calculate the image distance and size. Another common challenge is determining the sign conventions for the variables, which can vary depending on the orientation of the object and the mirror.

Example of a Law of Reflection Problem

Suppose a light ray is incident on a plane mirror at an angle of 30 degrees. What is the angle of reflection, and what is the direction of the reflected ray? According to the law of reflection, the angle of reflection is also 30 degrees, and the reflected ray is in the same plane as the incident ray and the normal. Therefore, the reflected ray makes an angle of 30 degrees with the normal, and it is directed away from the mirror. This simple example illustrates the basic principles of the law of reflection, and it can be used as a starting point for more complex reflection problems.

25.2 The Law of Reflection

Whenever we look into a mirror, or squint at sunlight glinting from a lake, we are seeing a reflection. When you look at this page, too, you are seeing light reflected from it. Large telescopes use reflection to form an image of stars and other astronomical objects.

The law of reflection is illustrated in Figure 25.4 , which also shows how the angles are measured relative to the perpendicular to the surface at the point where the light ray strikes. We expect to see reflections from smooth surfaces, but Figure 25.5 illustrates how a rough surface reflects light. Since the light strikes different parts of the surface at different angles, it is reflected in many different directions, or diffused. Diffused light is what allows us to see a sheet of paper from any angle, as illustrated in Figure 25.6 . Many objects, such as people, clothing, leaves, and walls, have rough surfaces and can be seen from all sides. A mirror, on the other hand, has a smooth surface (compared with the wavelength of light) and reflects light at specific angles, as illustrated in Figure 25.7 . When the moon reflects from a lake, as shown in Figure 25.8 , a combination of these effects takes place.

The law of reflection is very simple: The angle of reflection equals the angle of incidence.

The Law of Reflection

The angle of reflection equals the angle of incidence.

When we see ourselves in a mirror, it appears that our image is actually behind the mirror. This is illustrated in Figure 25.9 . We see the light coming from a direction determined by the law of reflection. The angles are such that our image is exactly the same distance behind the mirror as we stand away from the mirror. If the mirror is on the wall of a room, the images in it are all behind the mirror, which can make the room seem bigger. Although these mirror images make objects appear to be where they cannot be (like behind a solid wall), the images are not figments of our imagination. Mirror images can be photographed and videotaped by instruments and look just as they do with our eyes (optical instruments themselves). The precise manner in which images are formed by mirrors and lenses will be treated in later sections of this chapter.

Take-Home Experiment: Law of Reflection

Take a piece of paper and shine a flashlight at an angle at the paper, as shown in Figure 25.6 . Now shine the flashlight at a mirror at an angle. Do your observations confirm the predictions in Figure 25.6 and Figure 25.7 ? Shine the flashlight on various surfaces and determine whether the reflected light is diffuse or not. You can choose a shiny metallic lid of a pot or your skin. Using the mirror and flashlight, can you confirm the law of reflection? You will need to draw lines on a piece of paper showing the incident and reflected rays. (This part works even better if you use a laser pencil.)

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Chapter 25 Geometric Optics

25.4 Total Internal Reflection

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 1 (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 [latex]{n_1 > n_2}[/latex], the angle of refraction is greater than the angle of incidence—that is, [latex]{\theta _1 > \theta _2}[/latex].) Now imagine what happens as the incident angle is increased. This causes [latex]{\theta _2}[/latex] to increase also. The largest the angle of refraction [latex]{\theta _2}[/latex] can be is [latex]{90 ^{\circ}}[/latex], as shown in Figure 1 (b).The critical angle [latex]{\theta _c}[/latex] for a combination of materials is defined to be the incident angle [latex]{\theta _1}[/latex] that produces an angle of refraction of [latex]{90 ^{\circ}}[/latex]. That is, [latex]{\theta _c}[/latex] is the incident angle for which [latex]{\theta _2 = 90^{\circ}}[/latex]. If the incident angle [latex]{\theta _1}[/latex] is greater than the critical angle, as shown in Figure 1 (c), then all of the light is reflected back into medium 1, a condition called total internal reflection .

Critical Angle

The incident angle [latex]{\theta _1}[/latex] that produces an angle of refraction of [latex]{90 ^{\circ}}[/latex] is called the critical angle, [latex]{\theta _c}[/latex].

$In the first figure, an incident ray at an angle theta 1 with a perpendicular line drawn at the point of incidence travels from n1 to n2. The incident ray suffers both refraction and reflection. The angle of refraction is theta 2. In the second figure, as theta 1 is increased, the angle of refraction theta 2 becomes 90 degrees and the angle of reflection corresponding to 90 degrees is theta c. In the third figure, theta c greater than theta i, total internal reflection takes place and instead of refraction, reflection takes place and the light ray travels back into medium n1.$

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

When the incident angle equals the critical angle ([latex]{\theta _1 = \theta _c}[/latex]), the angle of refraction is [latex]{90^{\circ}}[/latex] ([latex]{\theta _2 = 90^{\circ}}[/latex]). Noting that [latex]{\text{sin} \; 90^{\circ} = 1}[/latex], Snell’s law in this case becomes

The critical angle [latex]{\theta _c}[/latex] for a given combination of materials is thus

Total internal reflection occurs for any incident angle greater than the critical angle [latex]{\theta _c}[/latex], 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 1: 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 2 , 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 [latex]{\theta _c = \text{sin}^{-1} \;(n_2 / n_1)}[/latex] can be used to find the critical angle [latex]{\theta _c}[/latex]. Here, then, [latex]{n_2 = 1.00}[/latex] and [latex]{n_1 = 1.49}[/latex].

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 [latex]{42.2^{\circ}}[/latex] 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 [latex]{n_1 > n_2}[/latex] 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 [latex]{48.6^{\circ}}[/latex], while that from diamond to air is [latex]{24.4^{\circ}}[/latex], and that from flint glass to crown glass is [latex]{66.3^{\circ}}[/latex]. 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 2 .) 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.

Light ray enters an S-shaped tube and undergoes multiple reflections, finally emerging through the other end.

Bundles of fibers can be used to transmit an image without a lens, as illustrated in Figure 3 . The output of a device called an endoscope is shown in Figure 3 (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.

Picture (a) shows how an image A is transmitted through a bundle of parallel fibers. Picture (b) shows an endoscope image.

Fibers in bundles are surrounded by a cladding material that has a lower index of refraction than the core. (See Figure 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. 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.

$The image shows a bundle fiber with a medium of refractive index n sub 1 inside surrounded by a medium n sub 2. Medium n sub 2 is made up of cladding material and n sub 1 is the core.$

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 5 , 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.

Picture (a) shows the lunar expedition with the astronauts and their space shuttle. Picture (b) shows rectangular and round shaped bicycle reflectors.

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 [latex]{45^{\circ}}[/latex]. One use of these perfect mirrors is in binoculars, as shown in Figure 6 . Another use is in periscopes found in submarines.

The picture shows binoculars with prisms inside. The light through one of the object lenses enters through the first prism and suffers total internal reflection and then falls on the second prism and gets total internally reflected and emerges out through one of the eyepiece lenses.

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 [latex]{24.4^{\circ}}[/latex], and so when light enters a diamond, it has trouble getting back out. (See Figure 7 .) Although light freely enters the diamond, it can exit only if it makes an angle less than [latex]{24.4^{\circ}}[/latex]. 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 ([latex]{\approx 2.17}[/latex]), 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 Chapter 25.5 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.

$A light ray falls onto one of the faces of a diamond, gets refracted, falls on another face and gets totally internally reflected, and this reflected ray further undergoes multiple reflections when it falls on other faces.$

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.

Section Summary

The incident angle that produces an angle of refraction of [latex]{90^{\circ}}[/latex] is called critical angle.
Total internal reflection is a phenomenon that occurs at the boundary between two mediums, such that if the incident angle in the first medium is greater than the critical angle, then all the light is reflected back into that medium.
Fiber optics involves the transmission of light down fibers of plastic or glass, applying the principle of total internal reflection.
Endoscopes are used to explore the body through various orifices or minor incisions, based on the transmission of light through optical fibers.
Cladding prevents light from being transmitted between fibers in a bundle.
Diamonds sparkle due to total internal reflection coupled with a large index of refraction.

Conceptual Questions

1: A ring with a colorless gemstone is dropped into water. The gemstone becomes invisible when submerged. Can it be a diamond? Explain.

2: A high-quality diamond may be quite clear and colorless, transmitting all visible wavelengths with little absorption. Explain how it can sparkle with flashes of brilliant color when illuminated by white light.

3: Is it possible that total internal reflection plays a role in rainbows? Explain in terms of indices of refraction and angles, perhaps referring to Figure 9 . Some of us have seen the formation of a double rainbow. Is it physically possible to observe a triple rainbow?

A double rainbow with spectacular bands of seven colors.

4: The most common type of mirage is an illusion that light from faraway objects is reflected by a pool of water that is not really there. Mirages are generally observed in deserts, when there is a hot layer of air near the ground. Given that the refractive index of air is lower for air at higher temperatures, explain how mirages can be formed.

Problems & Exercises

1: Verify that the critical angle for light going from water to air is [latex]{48.6^{\circ}}[/latex], as discussed at the end of Example 1 , regarding the critical angle for light traveling in a polystyrene (a type of plastic) pipe surrounded by air.

2: (a) At the end of Example 1 , it was stated that the critical angle for light going from diamond to air is [latex]{24.4^{\circ}}[/latex]. Verify this. (b) What is the critical angle for light going from zircon to air?

3: An optical fiber uses flint glass clad with crown glass. What is the critical angle?

4: At what minimum angle will you get total internal reflection of light traveling in water and reflected from ice?

5: Suppose you are using total internal reflection to make an efficient corner reflector. If there is air outside and the incident angle is [latex]{45.0^{\circ}}[/latex], what must be the minimum index of refraction of the material from which the reflector is made?

6: You can determine the index of refraction of a substance by determining its critical angle. (a) What is the index of refraction of a substance that has a critical angle of [latex]{68.4^{\circ}}[/latex] when submerged in water? What is the substance, based on Chapter 25.3 Table 1 ? (b) What would the critical angle be for this substance in air?

7: A ray of light, emitted beneath the surface of an unknown liquid with air above it, undergoes total internal reflection as shown in Figure 10 . What is the index of refraction for the liquid and its likely identification?

A light ray travels from an object placed in a denser medium n1 at 15.0 centimeter from the boundary and on hitting the boundary gets totally internally reflected with theta c as critical angle. The horizontal distance between the object and the point of incidence is 13.4 centimeters.

8: A light ray entering an optical fiber surrounded by air is first refracted and then reflected as shown in Figure 11 . Show that if the fiber is made from crown glass, any incident ray will be totally internally reflected.

$The figure shows light traveling from n1 to n2 is incident on a rectangular transparent object at an angle of incidence theta 1. The angle of refraction is theta 2. On refraction, the ray falls onto the long side and gets totally internally reflected with theta 3 as the angle of incidence.$

3: [latex]{66.3^{\circ}}[/latex]

5: > 1.414

7: 1.50, benzene

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Waves travel in all directions in the open ocean, but they always approach the land nearly perpendicular to the shore. Why does this happen?
the angle of incidence
the angle of reflection
the angle of refraction
the angle between the reflected and refracted rays
the angle between the incident and refracted rays
the angle between the incident and reflected rays
Using the protractor in the diagram, measure the angle of incidence.
Using the law of reflection, determine the angle of reflection, then select the lettered ray that best represents the reflected ray.
Using Snell's law, determine the angle of refraction, then select the lettered ray that best represents the transmitted ray.
refraction-drill.pdf The diagrams on the accompanying pdf show a ray of light incident upon an interface. In some cases the ray is traveling through air and entering glass. In other cases it is traveling through glass and entering air. On each diagram, sketch the approximate path of the light in the second medium. Ignore any reflected light. It is not necessary to calculate any angles, but do clearly show the change in direction of the rays, if any, at each surface.
Ice is transparent and so is water, but ice is visible in water. How is this possible?
Acrylic is transparent and so is mineral oil, but acrylic in mineral oil is nearly invisible. Why does this happen?
Find another pair of materials that have the same optical relationship as acrylic and mineral oil.
If you can see the face of a friend who is under water can your friend also see you? Justify your answer.
Is daytime a bit longer or a bit shorter because of atmospheric refraction? Justify your answer.
What optical phenomenon explains this natural phenomenon?
What atmospheric conditions are needed for a Fata Morgana to occur?
Explain why the light takes the path that it does in this kind of mirage.
v = ⅝c
v = ⅔c
v = ¾c
The index of the material is usually chosen so that its index is the square root of the index of the substrate (the material it is deposited on). What would be an appropriate index of refraction for an antireflection coating placed on crown glass (a kind of glass that is still sometimes used for eyeglasses)?
The index of refraction is usually stated for a color of light in the middle of the visible spectrum. What wavelength of light was used to determine the index of refraction you used in part a? What color is this light? Why do you suppose this type of light was chosen?
What is the wavelength of this light in the antireflection coating?
Antireflection coatings of the type described in this problem are usually one-quarter of a wavelength thick. How thick should this coating be?
About how big is an atom (an order of magnitude value will do)? How many atoms thick is the antireflection coating described in this problem (again, an order of magnitude value will do)?

Determine the angular separation between the red and violet components of a beam of white light traveling from diamond to air if the angle of incidence is…

The two basic elements of optical fiber are its core and cladding. The core, or the axial part of the optical fiber, is the light transmission area of the fiber. The cladding is the layer completely surrounding the core. The difference in refractive index between the core and cladding is less than 0.5%. The refractive index of the core is higher than that of the cladding, so that light in the core strikes the interface with the cladding at a bouncing angle and is trapped in the core by total internal reflection Corning Inc., 2008

the index of refraction of the cladding
the critical angle at the core-cladding interface
the critical angle at the cladding-air interface
Determine the angle of incidence.
Determine the angle of reflection.
Determine the angle of refraction.
Determine the index of refraction of the unspecified liquid.
What is the unspecified liquid?

Choose one of the rays at random. On the diagram, trace the transmitted portion of your ray until it emerges from the crystal. Determine the deviation angle between the incident ray and the emergent ray.

Choose one of the rays at random. On the diagram, trace the portion of your ray that is transmitted at the interface, reflects once off the back of the drop, and then emerges. Determine the angle between the incident ray and the emergent ray.

Choose one of the rays at random. On the diagram, trace the portion of your ray that is transmitted at the interface, reflects twice off the back of the drop, and then emerges. Determine the angle between the incident ray and the emergent ray.

What is the critical angle for crown glass?
Which of the rays are refracted at the hemispherical end cap and which experience total internal reflection? (Hint: What is the easiest way to draw a line normal to the circumference of a circle? Second hint: Go find a protractor.)
Draw the refracted rays until they reach the axis. Do they focus?

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Refractive Index: Formula and Solved Problems

In this tutorial, the index of refraction formula is explored with several solved sample problems. All these problems are simple and useful for high schools.

In physics, the ratio of the speed of light in a vacuum to the speed of light in a medium is called the index of refraction. The formula of index of refraction is written as \begin{gather*} n=\frac{\text{Speed of light in a vacuum}}{\text{Speed of light in the medium}}\\\\ n=\frac{c}{v_{medium}}\end{gather*}

Index of Refraction Formula: Definition

When light travels from one medium into another, it is refracted because the speed of light is different in the two media.

The index of refraction shows a kind of strength of a substance in refracting the incident light.

In the following sample problems, we see that the higher the index of refraction, the slower the speed of light in a material.

On the other side, the lower the speed of light in a substance, the smaller the angle of refraction.

Let's solve a few problems on the index of refraction.

Problem (1): The speed of a ray of light in ice is $2.29\times 10^{8}\,{\rm m/s}$. Find the index of refraction of ice.

Solution : The index of refraction of a transparent material, $n$, is defined as the speed of light in the vacuum divided by the speed of light in that material. \[n=\frac{c}{v}\] Note that $v$ is the speed of light in that medium. Substituting the numerical values into the index of refraction formula, we get \[n_{ice}=\frac{c}{v_{ice}}=\frac{3\times 10^{8}}{2.29\times 10^{8}}=1.31\] The index of refraction is a dimensionless quantity in physics.

Problem (2): The index of refraction of a medium is measured to be 1.2. What is the speed of light through that medium?

Solution : The speed of light moving through a dense medium is found using the formula \[n=\frac{c}{v}\] Rearranging this and solving for $v$, we can find the speed of light in a medium. \[v=\frac{c}{n}=\frac{3\times 10^8}{1.2}=2\times 10^8\quad {\rm m/s}\] Pay attention that the speed of light in vacuum is the maximum speed in nature and consequently, $v$ is always less than $c$.

Problem (3): The speed of light in a particular transparent medium is $82\%$ of its value in water. What is the index of refraction of this medium? (The index of refraction of water is 1.33)

Solution : The speed of light in a transparent material is determined by the formula $v=\frac{c}{n}$. Here, we are told that the speed of light in an unknown material $v_u$ is $82\%$ of its value in water so $v_u=0.82v_w$. Applying the above formula, we will have \begin{align*} \frac{c}{n_u}&=0.82\,\frac{c}{n_w}\\\\ \Rightarrow \quad n_w&=0.82 n_u\end{align*} So, the refractive index of the unknown medium is found to be \[n_u=\frac{n_w}{0.82}=\frac{1.33}{0.82}=1.62\]

Problem (4): A beam of light of wavelength $589\,{\rm nm}$ in the vacuum passes through a piece of Silicon of the index of refraction $n=3.5$. (a) How fast does light travel in Silicon? (b) How does the wavelength of light change in Silicon? (c) What is the frequency of light in Silicon?

Solution : When light traveling in one medium enters the second medium, its velocity is changed. This change in velocity is the basic principle of refraction in physics.

(a) The ratio of the speed of light in the vacuum to the speed in a transparent medium is defined as the index of refraction of that medium i.e. $n=\frac{c}{v}$. Solving this equation for $v$, we have \[v=\frac{c}{n}=\frac{3\times 10^8}{3.5}=8.57\times 10^7\,{\rm m/s}\] (b) In a similar way, when light moves from one medium into another, its wavelength also changes. In particular, when a beam of light of wavelength $\lambda_0$ in the vacuum passes through a dense medium having a refractive index of $n$, its wavelength reduces by the amount of \[\lambda_n=\frac{\lambda_0}{n}\] where $\lambda_n$ is the wavelength in the medium having index of refraction $n$. With these explanations, we have \[\lambda_n=\frac{589\,{\rm nm}}{3.5}=168.3\quad {\rm nm}\] (c) As light travels from one medium to the second one, its frequency does not change. The only things that change when a wave moves from one medium to another are wavelength and speed in the medium.

So, the frequency of this ray of light in Silicon is the same as in the vacuum. To find the frequency, we use the wave speed formula $v=f\lambda$. In the vacuum $v=c$ so $c=\lambda f$. Thus, the frequency in of light in Silicon is found to be \[f=\frac{c}{\lambda}=\frac{3\times 10^8\,{\rm m/s}}{589\times 10^{-9}}\,{\rm m}=5.09\times 10^{14}\,{\rm Hz}\]

Problem (5): The index of refraction of diamond $n=2.41$. What is the speed of light through the diamond?

Solution : According to the definition of the index of refraction, $n=\frac{c}{v}$, the speed of a light ray in a medium, say diamond, is measured to be \[v=\frac{c}{n}=\frac{3\times 10^8}{2.41}=1.25\times 10^8\quad {\rm m/s}\] As expected because the speed of light in a medium is always less than the speed in the vacuum $c$.

Problem (6): A ray of red light takes $2.63\,{\rm ns}$ to travel $50\,{\rm cm}$ through an optically dense medium. What is the index of refraction of this medium?

Solution : This problem involves a kinematics problem. The known information is the time taken $t$ and distance traveled $d$ in the medium. Recall that the ratio of distance traveled to the time taken is the definition of speed in physics, $v=\frac{d}{t}$.

This light ray is moving in a dense medium so its speed can be related to the index of refraction of that medium using equation $n=\frac{c}{v}$. Combining these two above equations, and solving for the unknown $n$, we get \begin{align*} v&=\frac{d}{t}=\frac{c}{n}\\\\ \Rightarrow \quad n&=\frac{ct}{d}\\\\ &=\frac{(3\times 10^8)(2.63\times 10^{-9})}{0.5}\\\\&=1.578\end{align*}

Problem (7): The index of refraction for crown glass is $1.52$. A beam of light of wavelength $6500\text{\AA}$ in a vacuum passes through it. What is the wavelength of this light in the crown glass?

Solution : When a light of wavelength $\lambda_0$ in a vacuum moves through a dense medium having a refractive index of $n$, its wavelength in that medium is found to be $\lambda_n=\frac{\lambda_0}{n}$. Using this formula, we have \[\lambda_n=\frac{6500}{1.52}=4276.3 \text{\AA} \] As expected, the wavelength in a dense medium is reduced.

Problem (8): A ray of light having wavelength $\lambda$ in a vacuum enters three different transparent mediums. From largest to smallest, rank the indices of refraction of the media.

$Index of refraction formula for wavelength$

Solution : The extension of the wavelength of light in each medium is shown. As you can see, the wavelength in $b$ is greater than $a$ than $c$. \[\lambda_b>\lambda_a>\lambda_c\] According to the refractive index formula for wavelength, $\lambda_n=\frac{\lambda}{n}$, the wavelength in any transparent medium is inversely proportional with the index of refraction of that medium. So, the larger the wavelength, the greater the index of refraction. Hence, the order of the indices of refraction for the given mediums is as follows \[n_c>n_a>n_b\]

Problem (9): Two data centers, 300-m away, are connected with each other through optical fibers having a refractive index of 1.55. How long does it take the light to travel through this optical fiber?

Solution : The distance traveled $d=300\,{\rm m}$ and the index of refraction of the optical fiber $n=1.55$ are given. The refractive index formula,$n=\frac{c}{v}$, tells us about the velocity of a light ray in a medium like optical fiber. Having speed and distance traveled, we can use the definition of speed, $v=\frac{d}{t}$, to find the time elapsed.

Combining these two equations, gives us \[n=\frac{c}{d/t}=\frac{ct}{d}\] Solving this for $t$ and substituting the numerical values we will have \begin{align*} t&=\frac{nd}{c}\\\\&=\frac{(1.55)(300)}{3\times 10^{8}}\\\\&=1.55\times 10^{-6}\quad{\rm s}\end{align*} So, it takes about $1.5\,{\rm \mu s}$ to light travels this distance between the data centers.

Index of Refraction Formula in Snell's law

Another method to find the index of refraction of a material is using Snell's law of refraction. According to this rule, when a ray of light, strikes at an angle of $\theta_1$ with normal upon a boundary of another medium, it is refracted and enters into the second medium at an angle of $\theta_2$ with normal.

The first angle is called the angle of incidence and the second one is the angle of refraction. Snell's law relates these two angles with each other using the following formula \[n_1 \sin\theta_1=n_2 \sin\theta_2\] where $n_1$ and $n_2$ are the indices of refractive of the corresponding media.

Suppose a light ray is in the air and is to enter an unknown material. Knowing the angles of incidence and refraction, we can use this equation to find the index of refraction of that material.

In the following, we use Snell's law to find the index of refraction using some worked sample problems.

Problem (10): A ray of light traveling in the air, strikes the boundary of an unknown transparent material at an angle of incidence $54^\circ$. The angle of refraction is $23^\circ$. What is the index of refraction of that material?

Solution : When a light ray travels through a medium and strikes the boundary of another medium, it is refracted and bends toward or away from the normal. Snell's law of refraction governs this behavior. According to this rule, $n_1 \sin\theta_1=n_2\sin \theta_2$, where $\theta_1$ and $\theta_2$ are the angles of incidence and refraction with normal.

Substituting the numerical values into Snell's law formula and solving for the refractive index of unknown material, we have \begin{align*} n_1 \sin\theta_1&=n_2\sin \theta_2\\\\ (1)\sin 53^\circ&=n_2\sin 23^\circ\\\\ 0.8&=(0.39)n_2\\\\ \Rightarrow n_2&= 2.05 \end{align*} To practice more problems on Snell's law, refer to the following page: Snell's law Problems with answers

Problem (11): An unknown material is surrounded by water. A ray of the yellow light of wavelength $589\,{\rm nm}$ strikes the boundary between them at $45^\circ$ and is refracted at $32^\circ$. What is the index of refraction of the material? (The refractive index of water is 1.33).

$Snell's law to find the refractive index of formula$

Solution : The angle of incidence $\theta_1=45^\circ$ and of refraction $\theta_2=32^\circ$ are given. Using Snell's law, we can a formula for index of refraction of that medium \begin{align*} n_1 \sin\theta_1&=n_2\sin \theta_2\\\\ (1.33)\sin 45^\circ&=n_2\sin 32^\circ\\\\ (1.33)(0.707)&=(0.530)n_2\\\\ \Rightarrow n_2&= 1.774 \end{align*} As you have seen, by measuring the incident and refracted angles of a light ray strike the boundary between two different media and using Snell's law, one can find a useful equation for the index of refraction of an unknown substance.

Problem (12): What is the critical angle for total internal reflection for a beam of light leaving the crown glass and entering the air? (The refractive index of crown glass is 1.52)

Solution : Critical angle is the incident angle at which the angle of refraction is $90^\circ$. So, putting this into Snell's law we will have \begin{align*}n_1 \sin\theta_1&=n_2\sin\theta_2\\\\ (1.52)\sin\theta_1&=(1)\sin 90^\circ\\\\ \Rightarrow \quad \sin\theta_1&=\frac{1}{1.52}\\\\&=0.658\end{align*} Taking the inverse of sine of both sides, the angle whose sine is 0.658 is obtained \[\theta_1=\sin^{-1}(0.658)=41.1^\circ\] So, the critical angle for crown glass is about $41.1^\circ$.

On the other side, total internal reflection occurs when the two conditions below are satisfied: (i) The incident light is in a medium that has a higher refractive index. (ii) The angle of incidence is greater than the critical angle.

So, for incident angles $\alpha>41.1^\circ$, all incoming light rays striking the glass-air boundary are reflected back into the original medium.

For more problems on total internal reflection, visit the following page Total internal reflection problems with solution

Problem (13): A ray of light originally is in an unknown substance, at an angle of $68.4^\circ$ strike the glass having the refractive index of $n_g=1.333$. The refracted ray exit along the boundary. What is the index of refraction of the unknown substance?

$Critical angle to find the refractive index of formula$

Solution : Because the refracted ray is parallel to the boundary so its angle with the normal (perpendicular) to the boundary is $\theta_2=90^\circ$. Plugging the numerical value into Snell's law of refraction and solving for $n$, we can find a formula for refractive index of the initial substance \begin{align*}n_1 \sin\theta_1&=n_2\sin\theta_2\\\\ n_1\sin 68.4^\circ&=(1.333)\sin 90^\circ\\\\ \Rightarrow \quad n_1&=\frac{1.333\times \sin 90^\circ}{\sin 68.4^\circ}\\\\&=\frac{1.333\times 1}{0.930}\\\\&=1.43\end{align*} Thus, the index of refraction of the unknown substance is $1.43$.

This particular incident angle at which the refracted ray is $90^\circ$ is called the critical angle. In this problem, if the incident angle becomes greater than this critical one, $100\%$ of the light ray is reflected back (total internal reflection) into the original substance without any part of it entering the glass.

In fact, in this situation, the boundary acts as a perfect mirror.

In this article, the formula for the refractive index,$n$, of a material is discussed using a problem-solving strategy.

We learned that:

According to the index of refraction formula, $n=\frac{c}{v}$, the index of refraction of any substance is always greater than or equal to one. This is because the speed of light in a medium is always less than $c$.

As light travels from one medium into another, its frequency does not change but its wavelength changes as $\lambda_n=\frac{\lambda}{n}$, where $\lambda$ is the wavelength in the vacuum.

Using the refractive index formula, we can write Snell's law in an alternative form below \[n_1 \lambda_1=n_2\lambda_2\]

If light moves from one medium into another where its speed is greater, then the refracted ray bends away from the normal. This is the case for light traveling from water to air.

Author : Ali Nemati Date Published : 9/15/2021

Physics Problems with Solutions

Refraction of light rays, examples and solutions, refraction of light rays at interfaces.

Light rays travel in different mediums at different speeds. In vacuum, for example, light travels at the speed of 3×10 8 m/s. This is the highest speed possible in physics. One of the most important parameters that measures optical properties of a medium is the index of refraction . It is defined as the ratio of the speed of light in vacuum (3×10 8 m/s) and the speed of light v in the medium in question. The refractive index measure the optical density of a given material. A material with higher refractive is more optically dense than a material with lower refractive index.

What is the index of refraction in a medium where the speed of light is 1.5×10 8 m/s?

n = c / v = 3×10 8 m/s / 1.5×10 8 m/s = 2

What is the speed of light in water whose refractive index is 1.33?

1.33 = 3×10 8 m/s / v Solve for v v = 3×10 8 m/s / 1.33 ≈ 2.26 × 10 8 m/s

Snell's Laws of refraction

Reflection of Light Rays on a Reflecting Surface

n 1 sin (i) = n 2 sin (t)

Refraction explains physical phenomena such the appearance of a pen bent inside a glass of water and the separation of white light into different colors such as the rainbow as shown below.

$example 3 - Refraction of light$

Use Snell's law to write: 1.52 sin 25° = 1.32 sin t t = arcsin((1.52/1.32) sin 25) ≈ 29.1 °

$example 4 - Refraction of light$

Assume refractive index of air is equal to 1. Use Snell's law to write: 1 sin (i) = 1.32 sin 30° i = arcsin (1.32 sin 30°) ≈ 41.3 °

$example 5 - Refraction of light$

a) Snell's at point A n 1 sin i = n 2 sin j Snell's law at point B n 2 sin k = n 1 sin i' The normals NA and NB to the interfaces at A and B are parallel to each other and therefore angles j and k are equal. Hence n 1 sin i = n 2 sin j = n 2 sin k = n 1 sin i' Therefor i'= i The incident and emerging have equal angles with the normals therefore the emerging and incident rays are parallel. b) Find d Consider the right triangle ABO. sin α = d / AB Consider triangle ABO' cos j = w / AB or AB = w / cos j α = i - j sin (i - j) = d / AB Substitute AB by w / cos j in the above equation sin(i - j) = d cos j / w d = sin(i - j) W / cos j Use Snell's law n 1 sin i = n 2 sin j j = arcsin [ (n 1 /n 2 ) sin i ] Hence d = w sin(i - j) / cos j where j = arcsin [ (n 1 /n 2 ) sin i ] c) Calculate d for n 1 = 1, n 2 = 1.55, w = 3 cm and i = 32°. j = arcsin [ (n 1 /n 2 ) sin i ] = arcsin [ (1/1.55) sin 32°] ≈ 20 ° d = 3 sin(32 - 20) / cos 20 ≈ 0.7 cm

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Physics - Solved Example Problems for Reflection of Sound Waves | 11th Physics : UNIT 11 : Waves

Chapter: 11th physics : unit 11 : waves, solved example problems for reflection of sound waves.

EXAMPLE 11.10

Suppose a man stands at a distance from a cliff and claps his hands. He receives an echo from the cliff after 4 second. Calculate the distance between the man and the cliff. Assume the speed of sound to be 343 m s-1.

The time taken by the sound to come back as echo is 2 t = 4 ⇒ t = 2 s

∴ The distance is d = vt =(343 m s -1 )(2 s) = 686 m.

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problem-solving. Problem categorization based upon similarity of solution is a strategy to help them reflect about the deep features of the problems related to the physics principles involved. We find that there is a large overlap between the introductory and graduate students in their ability to categorize.
Solved Example Problems for Reflection of Sound Waves
Physics : Waves - Solved Example Problems for Reflection of Sound Waves. EXAMPLE 11.10. Suppose a man stands at a distance from a cliff and claps his hands. He receives an echo from the cliff after 4 second. Calculate the distance between the man and the cliff. Assume the speed of sound to be 343 m s-1.
Refraction and Lenses Problem-Sets
Problem Set RL4 - Snell's Law 1. Use Snell's Law to determine and angle of incidence or an angle of refraction for light refracting at a boundary. Includes 5 problems. Problem Set RL5 - Snell's Law 2. Use Snell's Law to determine an index of refraction value of an unknown material from knowledge of the angles of incidence and refraction.