Lesson Explainer: Critical Angle for Total Internal Reflection Physics

In this explainer, we will learn how to relate the paths of refracted and internally reflected light rays to the refractive indices of media they travel in.

Light rays often refract, or bend, when they travel between different materials, but there are certain cases in which a light ray will not transmit through a boundary like usual. We will learn how to determine whether or not this phenomenon will occur by calculating the critical angle, which depends on the path of the light ray and the refractive properties of the materials on either side of the boundary.

When we refer to an optical boundary, we are talking about a surface that separates two materials with different refractive indices. A material’s index of refraction, 𝑛, tells how fast or slow light travels through it and, in turn, how much light bends when it enters the material. When a ray of light travels between two materials that have the same index of refraction, there is no change in the ray’s path; thus, there is nothing particularly interesting for us to learn about the system. For this reason, we will only be interested in light rays that move between media with different optical properties. It should be noted that, generally, most rays both reflect and refract when they encounter a boundary, as illustrated below. However, for now, we are focusing only on the part of a ray that transmits and refracts.

Further, we are interested in the angle at which light approaches and leaves a boundary. Recall that we must measure all angles relative to the surface normal, which is just perpendicular to the surface itself. In the diagram below, the surface normal is represented by the dashed line. The relationship between the refractive indices of two materials and the angles of the light rays (relative to the surface normal) is defined by Snell’s law: π‘›πœƒ=π‘›πœƒ.sinsin

It is common to classify the incoming ray and the angle it creates as β€œincident,” so π‘›οŠ§ and πœƒοŠ§ can also be called 𝑛 and πœƒοƒ, where 𝑖 means β€œincident.” Likewise, the material and angle on the other side of the boundary, where the ray is refracted, are often called π‘›οŽ and πœƒοŽ, where π‘Ÿ means β€œrefracted.”

Recall that if a ray passes between two materials with the same optical properties, the light just travels in a straight line. But we are interested in optically different media whose boundaries bend light rays. In this case, when a ray refracts, it can bend either toward or away from the surface normal, depending on the properties of the materials on either side of the boundary. For instance, if π‘›οŠ§ is less than π‘›οŠ¨, Snell’s law dictates that πœƒοŠ§ will be greater than πœƒοŠ¨, so the refracted ray bends toward the normal, which is illustrated below on the left. However, if π‘›οŠ§ is greater than π‘›οŠ¨, πœƒοŠ§ will be less than πœƒοŠ¨, so the refracted ray bends away from the normal, as shown on the right.

In this explainer, we will be focusing on the case where 𝑛>π‘›οŠ§οŠ¨, where the refracted ray bends away from the normal. In this case, the angle of refraction πœƒοŠ¨ is always greater than the angle of incidence πœƒοŠ§. Notice that the larger πœƒοŠ§ becomes, the larger (πœƒοŠ¨) becomes. This is illustrated in the diagram below.

Since all relevant angles are measured between the surface and the surface normal, they can only range in value between zero and 90 degrees. Therefore, if we make an incident angle larger and larger, the refracted angle will eventually reach a limit so that πœƒ=90∘. This is a special case that only happens when the angle of incidence reaches a certain value called the critical angle. When the incident angle is equal to the critical angle, the ray will no longer transmit or pass into the second material; instead, it will just skim along the surface boundary. This phenomenon is illustrated below.

If we make the angle of incidence even larger than the critical angle, we unlock the special case of total internal reflection. During total internal reflection, none of the rays pass through the boundary. The rays completely reflect off the surface, as shown below.

The critical angle defines a limit for an angle of incidence to cause total internal reflection. The measure of the critical angle depends on the properties of the materials on either side of the medium boundary. Let us look at an example to better understand how its value is determined.

Example 1: Deriving the Critical Angle Formula

Which of the following formulas correctly shows the relation between the critical angle for total internal reflection πœƒοŒΌ for a light ray, the refractive index 𝑛 of the substance the light is propagating in, and the refractive index π‘›οŽ of the substance when the light is reflected from its surface?

  1. πœƒ=π‘›π‘›οŒΌοŽοƒ
  2. πœƒ=π‘›π‘›οŒΌοŽοƒsin
  3. sinπœƒ=π‘›π‘›οŒΌοŽοƒ
  4. sinπœƒ=π‘›π‘›οŒΌοƒοŽ
  5. sinsinπœƒ=π‘›π‘›οŒΌοŽοƒ


We want to find an equation to relate the critical angle and indices of refraction of two different media. To begin, let us look at a diagram that shows a surface separating two optically different media.

In this case, the medium on the left side of the boundary is where the ray will be incident from, and it has an index of refraction 𝑛. The medium on the right side of the boundary has an index of refraction π‘›οŽ. We want to find a relationship between these indices of refraction and a ray that is incident at the critical angle.

When a light ray is incident at the critical angle πœƒοŒΌ, as measured relative to the surface normal, the ray does not transmit and refract in the other medium, but it does not reflect back into the incident medium either. Instead, it skims along the boundary, as shown in the diagram above. Although the ray is not technically refracting, we can still measure its angle of refraction to be 90∘ here, which is the maximum possible value.

In order to devise a mathematical relation, let us recall Snell’s law, the formula that tells how light refracts between two different materials: π‘›πœƒ=π‘›πœƒ.οƒοƒοŽοŽsinsin

We can plug in some values to begin to solve for a relationship. Because the ray is incident at the critical angle, let us substitute πœƒοŒΌ in for πœƒοƒ, and further, we can substitute 90∘ for the angle of refraction: π‘›πœƒ=𝑛90.οƒοŒΌοŽβˆ˜sinsin

We know that sin90=1∘, so the equation simplifies to π‘›πœƒ=𝑛.οƒοŒΌοŽsin

Let us rearrange the equation so that 𝑛 and π‘›οŽ are on the same side. To do this, we divide both sides of the equation by 𝑛: sinπœƒ=𝑛𝑛.οŒΌοŽοƒ

This equation correctly relates the critical angle to the refractive indices of the materials on either side of the surface boundary, so C is the correct answer.

This example unpacked the mathematical relationship that we can use to determine the critical angle for total internal reflection, given the indices of refraction of the media on either side of the boundary. If we want to go a step further, we can solve for the critical angle by taking the inverse sine, or arcsin, of both sides of the equation, which will undo the sine operation on πœƒοŒΌ. Thus, we have πœƒ=𝑛𝑛.οŒΌοŽοƒarcsin

Let us formally define this relationship.

Definition: The Critical Angle for Total Internal Reflection

The critical angle, πœƒοŒΌ, for light rays traveling in a medium of refractive index π‘›οŠ§ at a boundary with a medium of refractive index π‘›οŠ¨ can be calculated using sinπœƒ=𝑛𝑛, where 𝑛<π‘›οŠ¨οŠ§.

Notice that, in this definition, we used the notation of π‘›οŠ§ and π‘›οŠ¨, as opposed to 𝑛 and π‘›οŽ, which we used in the first example. It is generally accepted that β€œ1” represents quantities associated with the incident ray and β€œ2” represents quantities associated with the refracted ray. Let us work through a couple of examples to practice applying the formula.

Example 2: Calculating the Critical Angle given 𝑛 Values

What is the critical angle for a light ray traveling in water with a refractive index of 1.33 that is incident on the surface of water above which air is with a refractive index of 1.00? Answer to the nearest degree.


To begin, let us recall the equation to determine the critical angle: sinπœƒ=𝑛𝑛.

Here, we know that a ray of light is traveling through water, approaching the boundary of the air above. Water is the incident material, so we will use 𝑛=1.33 and 𝑛=1.00. First, we will rewrite the equation to solve for πœƒοŒΌ, and since we have values for two of the three variables in the equation, we can substitute them to solve for the critical angle: sinarcsinarcsinπœƒ=π‘›π‘›πœƒ=𝑛𝑛=1.001.33=48.8.∘

Rounding to the nearest degree, we have determined that the critical angle between water and air exists in water and occurs at 49∘.

Recall that the definition for the critical angle includes the condition that π‘›οŠ¨ must be less than π‘›οŠ§. To explore this concept, let us revisit the example above and attempt to solve it a different way. We can begin with the formula solved for the critical angle: πœƒ=𝑛𝑛.arcsin

At this point, we just need to plug in values for π‘›οŠ§ and π‘›οŠ¨. Recall that we are examining a ray traveling through water, approaching the surface boundary of air. It might seem that either medium can be represented in the equation with either 𝑛 value, but this is untrue as it does matter which material is represented by which variable. To understand why, we can try to perform the calculation with air, rather than water, as the incident material. In this case, we will use 𝑛=1.00 (air) and 𝑛=1.33 (water): πœƒ=1.331.00.arcsin

If we attempt to enter this into a calculator, the result will either be an β€œerror” message or an imaginary number, and neither result is useful here. The reason this expression cannot be meaningfully computed is that the fraction 1.331.00 is greater than one, and there is no angle whose sine value is greater than one. For this reason, 𝑛 must be greater than π‘›οŽ.

In order for the critical angle to exist, we must be dealing with a light ray that is incident in a medium of a higher index of refraction, approaching the surface of a medium with a lower index of refraction. Knowing this, we can conclude that the critical angle must exist in the medium with the higher 𝑛 value, which is the water in this case.

Moving on, let us work through another example using the critical angle equation.

Example 3: Calculating the Critical Angle given 𝑛 Values

What is the critical angle for a light ray traveling in water with a refractive index of 1.33 that is incident on the surface of water above which there is ice with a refractive index of 1.31? Answer to the nearest degree.


Here, we are given the refractive indices for two materials, and we will use 𝑛=1.33 (water) and 𝑛=1.31 (ice). Because the critical angle only exists in the incident material with a higher refractive index, πœƒοŒΌ exists in the water. We can find its value using the critical angle formula, solved for πœƒοŒΌ: sinarcsinarcsinπœƒ=π‘›π‘›πœƒ=𝑛𝑛=1.311.33=80.1.∘

Rounding to the nearest degree, we have found that the critical angle exists in water at 80∘.

Now that we have worked through the critical angle equation a couple of times, let us focus on an example that explores several angles in a system with three optically different materials.

Example 4: Relating Angles in Several Optically Different Materials

The diagram shows a light ray that is transmitted from substance I to substance II at angle πœƒοŠ§ to the boundary between the substances. The ray is totally internally reflected back into substance II at the boundary to substance III. For any angle of πœƒ greater than πœƒοŠ§, the light ray is transmitted to substance II. Find the angle πœƒοŠ§ to the nearest degree.


The ray experiences total internal reflection at the boundary between substances II and III, so it must be incident on the boundary at an angle no smaller than the critical angle. We also know that the ray transmits through this boundary when it is incident on the first boundary between substances I and II at an angle larger than πœƒοŠ§, so let us think about what this implies. If the angle at πœƒοŠ§ is made larger, the angle of incidence at the boundary of substances II and III is made smaller. We know that if this angle is made any smaller, the light will not experience total internal reflection; thus, πœƒοŠ§ is set at a special value so that the angle of incidence between substances II and III is approximate to the critical angle. Since we know the refractive indices of the materials on either side of the boundary, we can calculate the critical angle to learn more about this setup. To calculate the critical angle here, we can rearrange the critical angle formula (from Snell’s law) to solve for πœƒοŒΌ and plug in the values 𝑛=1.00 and 𝑛=1.50: πœƒ=𝑛𝑛=1.001.50=41.8.οŒΌοŽοƒβˆ˜arcsinarcsin

Knowing this, we can use the properties of a right triangle to determine the angle of refraction as the ray passes from substance I to II. Below is a diagram that helps relate these angles to each other using a right triangle, shown in blue. Notice that the vertical leg of the blue triangle is parallel to the surface normal between substances II and III. Because of the alternate interior angles theorem, we know that the top internal angle of the right triangle, which is the angle of refraction for the ray passing from substance I to II, is congruent, or equal, to πœƒοŒΌ. This is shown in the diagram below.

Thus, at this boundary, πœƒ=41.8∘, and we also know the indices of refraction for the materials on either side of the boundary, so we can use Snell’s law to solve for the angle of incidence there: π‘›πœƒ=π‘›πœƒπœƒ=ο€½π‘›π‘›πœƒο‰=ο€Ό1.501.3341.8=48.7.οƒοƒοŽοŽοƒοŽοƒοŽβˆ˜βˆ˜sinsinarcsinsinarcsinsin

Because πœƒοŠ§ is complementary to πœƒοƒ, we can find the value of πœƒοŠ§ using 90βˆ’48.7=41.3∘∘∘.

Rounding to the nearest degree, we have found that πœƒ=41∘.

Beyond using 𝑛 values to determine the critical angle, it is important to be able to use the critical angle equation to solve for different values, such as the refractive index of a given material. We will practice this in the following example.

Example 5: Calculating a Refractive Index Using the Critical Angle

Light rays travel through a layer of kerosene floating on the surface of water that has a refractive index of 1.33. Light rays that are incident on the interface of kerosene and water at angles of 16.9∘ from the surface or less are totally internally reflected. What is the refractive index of the kerosene? Give your answer to two decimal places.


We have been given the value of an angle, but it is not equal to the critical angle because it is measured relative to the surface rather than the surface normal. However, the critical angle can be easily computed from the angle given because they are complementary: 90βˆ’16.9=73.1.∘∘∘

Thus, we know that πœƒ=73.1∘. Now that we have a value for the critical angle, and since we know that 𝑛=1.33water, we have two out of the three variables that appear in our equation for the critical angle: sinπœƒ=𝑛𝑛.οŒΌοŽοƒ

We can rearrange this equation to solve for one of the 𝑛 values. Because the light ray is traveling from the kerosene to water, we know that 𝑛=𝑛water, and we need to solve for 𝑛, the refractive index of kerosene. To do this, we can multiply both sides of the equation by π‘›πœƒοƒοŒΌsin: 𝑛=π‘›πœƒ.οƒοŽοŒΌsin

We are now ready to plug in πœƒ=73.1∘ and 𝑛=1.33: 𝑛=π‘›πœƒ=1.3373.1=1.39.οƒοŽοŒΌβˆ˜sinsin

Therefore, kerosene has a refractive index of 1.39.

Total internal reflection can be used to explain some everyday phenomena that we may already be familiar with. For instance, fiber optics is one of the most widespread applications of the concept. Optical fibers are long thin strands of material layered around a center core. The innermost layer has a higher index of refraction than the surrounding layer. Light is sent into the strand at one end, and the light experiences total internal reflection many times as it travels down the fiber. The light bounces off the boundary surrounding the core until it emerges at the other end. A diagram of this process is shown below. Fiber optics has many practical uses in fields such as medicine and engineeringβ€”for example, systems of long of optical fibers have been laid on sea floors to carry communication signals across large distances. Optical fibers are even used in toys and decor due to their interesting appearance as the light escapes at the ends of the fibers, as shown below.

Bunch of Fiber Optic dynamic flying from deep

We can consider another application of total internal reflection that causes an optical illusion. Mirage is a phenomenon that happens on hot days in which the image of an object appears reflected from the ground, thus giving the impression that a body of water may be reflecting that image. This happens because as the sun heats a surface, such as sand or pavement, the air near the surface is also warmed. This hot air has a lower density and a lower index of refraction than the cooler air above, and thus, there exists a critical angle between the layers of air. When rays of light are incident upon this boundary at certain angles, the rays experience total internal reflection, which redirects the light back upward and creates an inverted image, as illustrated below. This gives the illusion that the light is bouncing off a reflective surface, such as water.

Let us finish by summarizing a few important concepts.

Key Points

  • When a ray of light is incident on a boundary and it gets refracted at 90∘, such that it skims along the medium surface, it has an angle of incidence called the critical angle, πœƒοŒΌ.
  • The critical angle πœƒοŒΌ can be calculated with the formula sinπœƒ=π‘›π‘›οŒΌοŽοƒ.
  • In order for the critical angle to exist, 𝑛 must be greater than π‘›οŽ.
  • If a light ray is incident at an angle greater than πœƒοŒΌ, the light is totally internally reflected and does not escape into the second medium.

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