Lesson Explainer: The Dispersive Power of a Prism Physics

In this explainer, we will learn how to calculate the dispersive power of a prism, given the refractive indices of the different colors of light passing through it.

Dispersion is the word we use to describe the process of distributing, or spreading out, something over a large area. In particular, we will discuss the dispersion of light as it travels through a prism. Light is deviated from its original path when it enters the prism and again when it leaves the prism.

A sketch of this kind of dispersion is shown below. We have a beam of light, which contains both red and violet light, entering a prism. Inside the prism, the light separates into its component beams of pure red and pure violet light.

In this explainer, we will explore a quantitative way of explaining how prisms do this separating. We will also see how the index of refraction affects how much each color of light is spread out. For example, violet light refracts at a greater angle than red light, as shown here.

We call any material that affects different wavelengths of light differently β€œdispersive.” A prism is an example of a dispersive material because different wavelengths of light are refracted by different amounts.

Why does a prism cause light of different wavelengths to act differently?

To understand this, we can recall Snell’s law:

Law: Snell’s Law

If a light ray passes through an interface where the index of refraction is different on either side of the interface, let the index of refraction on the first side (the side the light ray hits the interface from) of the interface be π‘›οŠ§ and the index of refraction on the other side (the side the light ray passes into) of the interface be π‘›οŠ¨. The angle of incidence of the light is πœƒοŠ§ and the angle of refraction is πœƒοŠ¨. Recall that the angle of incidence is the angle that a beam makes with the normal of the interface as it enters one medium from another, while the angle of refraction is the angle that the beam makes with the normal of the interface as it travels through the second medium. Then, these angles obey the relationship π‘›πœƒ=π‘›πœƒ.sinsin

Note that the angle of incidence and the angle of refraction are measured from the line that is perpendicular to the interface and not with the interface itself.

How does Snell’s law tell us that the angle of refraction is different for different wavelengths of light in a prism?

Imagine a light ray, made up of all wavelengths of light, entering a prism, as shown in the following diagram.

The light ray enters from the left of the prism. Since the ray contains all wavelengths of light, we see that all the colors of light will enter the prism with the same angle of incidence.

However, if the index of refraction on the outside of the prism π‘›οŠ§ is different to the index of refraction on the inside of the prism, then we see an interesting effect.

From Snell’s law, we have the relationship π‘›πœƒ=π‘›πœƒοŠ§οŠ§οŠ¨οŠ¨sinsin.

Recall that the refractive index of a material is a measure of how fast light travels through that material. If the refractive index of a material is equal to 1, then light travels through the material at the same (very high) speed as it travels through a vacuum. As the refractive index increases from 1, this tells us that light travels slower through the material.

Let’s assume that the refractive index on the outside of the prism is the same for all wavelengths of light. An example of a material in which this is true is air. In the language of Snell’s law, this means π‘›οŠ§ is the same for all wavelengths of light.

However, we know that prisms refract different wavelengths of light by different amounts. For example, red light is refracted less than violet light. This means that the angle of deviation for red light, which we will call 𝛼r, is smaller than the angle of deviation for violet light, which we will call 𝛼v.

The reason that this happens is that the prism has a different index of refraction for different wavelengths of light. We can see this by comparing the cases of red light and violet light. The following diagram shows how different colors of light are refracted by different amounts. We have labeled the refractive index outside the prism π‘›οŠ§ and the angle of incidence πœƒοŠ§. We have also labeled the angles of refraction for red and violet light as πœƒr and πœƒv respectively.

Before each of the red and violet wavelengths enters the prism, they both have the same refractive index and the same angle of incidence. This means that the left-hand side of Snell’s law, π‘›πœƒοŠ§οŠ§sin, is the same for both wavelengths.

For the right-hand side of Snell’s law, we know that the angle of refraction is different for the two wavelengths. In order to satisfy π‘›πœƒ=π‘›πœƒοŠ§οŠ§οŠ¨sinsinr and π‘›πœƒ=π‘›πœƒοŠ§οŠ§οŠ¨sinsinv, π‘›οŠ¨ must not be the same for the two different wavelengths!

This means we have an index of refraction for red light, 𝑛r, and a different index of refraction for violet light, 𝑛v. Snell’s law for the two wavelengths should therefore read π‘›πœƒ=π‘›πœƒοŠ§οŠ§sinsinrr and π‘›πœƒ=π‘›πœƒοŠ§οŠ§sinsinvv.

In fact, since each different wavelength of light has a different angle of refraction in a prism, we know that each wavelength has a different index of refraction.

We can use this property as a way of defining a dispersive material.

Definition: Dispersive Material

A dispersive material is a material whose index of refraction varies with wavelength.

We will now explore a way of quantifying how much dispersion happens in a given prism.

Let’s return to our prism, but just consider the paths of the red and violet light. We choose to consider these two wavelengths of light because red is the longest wavelength of visible light and violet is the shortest wavelength of visible light. This means we see the two most extreme cases, and we study the largest range of wavelengths that we can.

In the diagram below, we can see that both of these wavelengths of light have deviated from the path they are following due to the prism. If they had not changed path, they would have continued to travel along the black arrow shown here.

We can call the angle that the red beam deviates from the black path 𝛼min, since it is the smallest angle that any visible light will be deviated from. This label is the Greek letter β€œalpha.”

Similarly, we can call the angle of deviation for violet light 𝛼max, since it is the largest angle of deviation that any visible light will experience.

We can quantify the amount of dispersion that a particular prism causes using these two angles of deviation: 𝛼min and 𝛼max.

We do this by defining a quantity called the β€œdispersive power” of the prism, which we denote by πœ”ο΅, where πœ” is the Greek letter β€œomega.” This dispersive power is given by the equation πœ”=π›Όβˆ’π›Όο€»ο‡.maxminmaxmin

This equation looks a bit complicated, but if we break it down, it is fairly simple to see what is happening.

On the top of the fraction, we have the difference between the maximum angle of deviation and the minimum angle of deviation, which is π›Όβˆ’π›Όmaxmin. This tells us the range of angles of deviation for the light going into the prism.

On the bottom of the fraction, we have the average value of 𝛼max and 𝛼min, i.e., the average angular deviation caused by the prism. This is what we get when we calculate the quantity 𝛼+𝛼2maxmin. This also corresponds to the angle of deviation of the β€œaverage” or β€œmean” wavelength of light to enter the prism. In terms of colors of light, if it is white light that enters the prism, this will be a color in between yellow and green.

To get the dispersive power of a prism, we need to find these two things and then work out the fraction given above. Let’s also notice that the dispersive power of a prism does not have any units. This means we call it a β€œdimensionless number.” Each angle 𝛼min and 𝛼max has units of degrees, but in the equation for dispersive power, all of these units cancel out and leave us with no units.

Definition: Dispersive Power

The dispersive power of a prism, πœ”ο΅, is a measure of the difference in refraction of the lights of the highest wavelength and the lowest wavelength that enter the prism. It is given by πœ”=π›Όβˆ’π›Όο€»ο‡ο΅ο΅οŠ°ο΅οŠ¨maxminmaxmin where 𝛼max is the angle of deviation of the shortest wavelength of light to enter the prism and 𝛼min is the angle of deviation of the longest wavelength of light to enter the prism.

Note that if 𝛼max and 𝛼min are the same, then πœ”=0. This means that the largest and smallest angles of deviation are the same and the prism is not dispersive at all. So, the smallest value of πœ”ο΅ is therefore zero (for nondispersive materials), and it increases to larger values for prisms that disperse light in a large amount. However, typical values for most glass prisms, or similar, are between 0.01 and 0.1.

It is important to remember that, even though the name β€œdispersive power” uses the word β€œpower,” this does not have the usual meaning it has in physics of an amount of energy over an amount of time. This is just a name we use that happens to reuse the word β€œpower.” Dispersive power is just a quantitative description of how much a prism spreads out light.

The larger the value of πœ”ο΅, the larger the range of dispersion that light experiences through the prism. The smaller the value of πœ”ο΅, the smaller the range of dispersion that light experiences through the prism.

This is shown in the diagram below. The prism on the left has a small dispersive power and hence the different colors of light do not spread out very much. The prism on the right has a large dispersive power and hence spreads out the different colors to a greater extent.

This means the prisms must be made of different materials because they have different indices of refraction. This is shown here by the different colors of the two prisms.

Prisms with more dispersive power spread out light more, while prisms with less dispersive power spread out light less.

Example 1: Calculating Dispersive Power from Angles of Deviation

White light dispersed by a prism has wavelengths ranging from 400 nm to 700 nm. The 400 nm wavelength light has a minimum angle of deviation of 22.9∘, the 700 nm wavelength light has a minimum angle of deviation of 22.1∘, and the 550 nm wavelength light has a minimum angle of deviation of 22.5∘. What is the dispersive power of the prism? Give your answer to three decimal places.

Answer

Here, we have a prism dispersing white light, and we are told the minimum angles of deviation experienced by some of the wavelengths in the white light.

We are told that the longest wavelength of light that enters the prism is 700 nm, and this wavelength will experience an angle of deviation of at least 22.1∘. This wavelength corresponds to red light and is the longest wavelength here, so we know it will experience the smallest angle of deviation, and we can say that 𝛼=22.1min∘.

The shortest wavelength of light that enters the prism is 400 nm, and this wavelength will experience an angle of deviation of at least 22.9∘. This wavelength corresponds to violet light and is the shortest wavelength here, so we can say that 𝛼=22.9max∘.

We can answer this question using just this information. We are asked to calculate the dispersive power of this prism, so let’s recall that the equation for dispersive power is πœ”=π›Όβˆ’π›Όο€»ο‡.maxminmaxmin

We now need to substitute 𝛼min and 𝛼max with the values we know. If we do this, we find πœ”=22.9βˆ’22.1.ο΅βˆ˜βˆ˜οŠ¨οŠ¨οŽ–οŠ―οŠ°οŠ¨οŠ¨οŽ–οŠ§οŠ¨βˆ˜βˆ˜

Let’s first work out the top of the fraction, which is the difference between the maximum and minimum angles of deviation. This gives us 22.9βˆ’22.1=0.8.∘∘∘

The bottom of the fraction tells us the mean, or average, angle of deviation that a wavelength of light will experience in the prism. We find this to be 22.9+22.12=452=22.5.∘∘∘∘

We can now combine these two results, and we find the dispersive power to be πœ”=0.822.5=0.03555….∘∘

Note that, in the final line, the degrees have cancelled out, and we are left with a unitless number. We are asked to give our answer to three decimal places, so our final answer is πœ”=0.036.

Let’s note that we did not use the information in the question that 550 nm wavelength light has a minimum angle of deviation of 22.5∘. While we did not need this information to solve the question, we could have used it to take a shortcut.

Since we know that the shortest wavelength is 400 nm and the longest wavelength is 700 nm, 550 nm is the middle wavelength. We are also told the deviation that this middle wavelength experiences is 22.5∘. We can see that this is the exact result we found for the average deviation, which we worked out for the bottom of the fraction in the equation for the dispersive power.

We could have input this result directly from the question by knowing that the deviation for the middle wavelength will be the average of the deviations from the longest and shortest wavelengths. This means we would not have had to calculate the average ourselves and we would have saved ourselves a little bit of time.

Either way, the result we find is exactly the same.

Recall that we have also said that we can define a dispersive material as a material whose refractive index varies with the wavelength of light. We will now see that we can rewrite the equation for the dispersive power of a prism in terms of the refractive indices of the prism.

We can do this because there is a certain refractive index that corresponds to the wavelength of light that is deviated by the angle 𝛼max and a certain different refractive index that corresponds to the wavelength of light that is deviated by the angle 𝛼min.

The reason that violet light experiences the greatest angle of deviation 𝛼max is because violet light experiences the greatest difference in the refractive index as it enters the prism. This means that the index of refraction for violet light is the largest of all the possible values for the index of refraction that visible light can experience in the prism. To denote this, we call the index of refraction for violet light 𝑛max. This means that the largest angle of deviation 𝛼max corresponds to the largest refractive index.

Similarly, red light experiences the smallest angle of deviation 𝛼min because red light experiences the smallest change in the refractive index as it enters the prism. This means that the index of refraction for red light is the smallest of all the possible values for the index of refraction that visible light can experience in the prism. We will call the index of refraction for red light 𝑛min, so the smallest angle of deviation corresponds to the smallest refractive index. The following diagram is an illustration of this, where different colors of light are shown to have different refractive indices.

In order to rewrite our equation for the dispersive power in terms of the refractive index instead of in terms of the angle of deviation, it is almost as simple as replacing 𝛼max with 𝑛max and replacing 𝛼min with 𝑛min. However, there is one small extra step we must remember, too.

In terms of the minimum and maximum refractive indices, the dispersive power of a prism can be written as πœ”=π‘›βˆ’π‘›ο€»ο‡βˆ’1.maxminmaxmin

We can see that this equation is very similar to the previous expression we had for the dispersive power. The biggest difference is that we need to include a βˆ’1 in the denominator. This is a convention that is used to account for the fact that prisms are usually surrounded by air, which has a refractive index very close to 1.

This means that, whether we know the maximum and minimum angles of deviation of a prism or we know the maximum and minimum indices of refraction of a prism, we can work out the dispersive power of that prism. This tells us how much that prisms spreads out light as it passes through the prism.

Example 2: Identifying the Relationship between Wavelength and Angle of Deviation

The diagram represents red, green, and blue light being deviated through an angle 𝛼 by a prism. Which of the following correctly describes how πœ†, the wavelength of the light, relates to the angle 𝛼?

  1. As πœ† increases, 𝛼 decreases.
  2. As πœ† increases, 𝛼 increases.
  3. πœ† and 𝛼 are unrelated.

Answer

We know that red light has the longest wavelength of any color of visible light. Out of the colors shown on the diagram, the color that has the shortest wavelength is blue.

This means that πœ† is larger for red light, and πœ† is smaller for blue light. So, as we go from red to green to blue light, πœ† decreases.

We can also see that red light is deviated the least out of all of the colors in the diagram, while green light is deviated a bit more and blue light is deviated the most. This means that 𝛼, the angle of deviation, is smallest for red light. This angle 𝛼 is larger for green light, and 𝛼 increases even more for blue light.

So we have seen that, as we go from red to green to blue light, the wavelength πœ† decreases, but the angle of deviation 𝛼 gets larger. This means that, if we go the other way, from blue light to green light to red light, the wavelength πœ† gets larger but the angle of deviation 𝛼 gets smaller.

We can summarize what we have said as the following statement, which we give as our answer: As πœ† increases, 𝛼 decreases.

Let’s note here that we also know that the index of refraction 𝑛 increases (or decreases) in the same way the 𝛼 does. So, we also know that as πœ† increases, 𝑛 decreases.

Example 3: Calculating Dispersive Power from the Index of Refraction

White light is dispersed by a prism. For the shortest wavelength light in the white light, the prism has a refractive index of 1.48, and for the longest wavelength light, the prism has a refractive index of 1.44. What is the dispersive power of the prism? Give your answer to three decimal places.

Answer

For this question, we are asked to find the dispersive power of a prism, given the refractive index of the longest and shortest wavelengths of light.

We are told that the longest wavelength of light has a refractive index of 1.44 in the prism. We are also told that the shortest wavelength of light has a refractive index of 1.48 in the prism.

For white light, we know that red light has the longest wavelength and experiences the smallest refractive index, and violet light has the shortest wavelength and experiences the largest refractive index. This means that, here, we can identify our maximum and minimum refractive indices in the same way.

The longest wavelength of light will experience the smallest refractive index, so here we can set 𝑛=1.44min. We also know that the shortest wavelength of light will experience the largest refractive index, so we can set 𝑛=1.48max.

We can then recall that the dispersive power of a prism, in terms of refractive index, is given by πœ”=π‘›βˆ’π‘›ο€»ο‡βˆ’1.maxminmaxmin

Let’s first calculate the top part of the fraction here, which we find to be π‘›βˆ’π‘›=1.48βˆ’1.44=0.04.maxmin

We can then calculate the bottom part of the fraction, which here is given by 𝑛+𝑛2βˆ’1=1.48+1.442βˆ’1=2.922βˆ’1=1.46βˆ’1=0.46.maxmin

This means that, by combining these two results, we can calculate the dispersive power of this prism to be πœ”=0.040.46=0.0869565….

Since the question asks us to give our answer to three decimal places, our final answer is that the dispersive power of the prism is πœ”=0.087.

Let’s now give symbols to name the average angle of deviation of a prism and the average refractive index of a prism: 𝛼=𝛼+𝛼2,𝑛=𝑛+𝑛2.avgmaxminavgmaxmin

Recall that the β€œapex angle” of a prism is the angle at the top, or tip, of the prism. In the following diagram, we see two prisms with two different apex angles, which are labeled 𝐴 and 𝐴. Both of these prisms have the same refractive indices.

The left-hand prism has a small apex angle 𝐴, while the right-hand prism has a large apex angle 𝐴. The larger the apex angle of a prism, the larger the angle of deviation is for each wavelength of light. Another way to say this is that if the apex angle increases, then the average angle of deviation, 𝛼avg, increases too.

In the following diagram, we now show two prisms with the same apex angle, labeled 𝐴, but different indices of refraction. This is demonstrated by the different colors of the two prisms.

We can see that if the refractive index of the prism increases, then the angle of deviation of a light beam increases too. If the prism is made of a dispersive material and if we then increase the average refractive index, 𝑛avg, then the average angle of deviation, 𝛼avg, increases.

From now on, let’s consider β€œthin prisms,” which are prisms whose apex angle 𝐴 is small. In this context, an apex angle 𝐴 of approximately 10∘ or less is considered small.

Let’s consider the quantity π›Όπ‘›βˆ’1avgavg for a thin prism. If we want to increase this quantity, what would we need to change in the prism to make this quantity get larger?

We know that if we increase 𝑛avg, then this increases 𝛼avg. However, we are considering the fraction π›Όπ‘›βˆ’1avgavg, not just 𝛼avg. This means that if we increase 𝑛avg, this does make 𝛼avg get bigger, but the overall fraction does not get bigger because increasing the denominator of a fraction makes the whole fraction smaller. This means that increasing 𝑛avg cannot be the answer.

Instead, the answer is the apex angle, 𝐴. If the increase 𝐴, then the quantity π›Όπ‘›βˆ’1avgavg will increase too. In fact, for a thin prism, we can actually set these two things equal to each other, so we have 𝐴=π›Όπ‘›βˆ’1.avgavg

Let’s check if the units of this equation are consistent. On the left-hand side, 𝐴 is an angle, so it has the unit of degrees. On the right-hand side, we have a fraction. The numerator of this fraction is the average angle of deviation, which is an angle, so this also has the unit of degrees. The denominator of the right-hand side is the average refractive index, which is a dimensionless, or unitless, number. This means that, overall, the right-hand side of the equation also has the unit of degrees, so our equation has consistent units.

We have also seen two different expressions for the dispersive power of a prism, one in terms of angles of deflection and one in terms of the refractive index. Since these expressions are equivalent, we can combine them to find an expression that relates angles of deviation to the refractive index.

That is, we have the following: πœ”=π›Όβˆ’π›Όο€»ο‡=π‘›βˆ’π‘›ο€»ο‡βˆ’1.maxminmaxminmaxminmaxmin

We can rewrite this with our new notation for the average angle of deviation and the average refractive index. Doing this gives us πœ”=π›Όβˆ’π›Όπ›Ό=π‘›βˆ’π‘›π‘›βˆ’1.maxminavgmaxminavg

Let’s now take just the second equality here, π›Όβˆ’π›Όπ›Ό=π‘›βˆ’π‘›π‘›βˆ’1,maxminavgmaxminavg and multiply both sides of the equation by 𝛼avg. This gives us π›Όβˆ’π›Ό=π›ΌΓ—π‘›βˆ’π‘›π‘›βˆ’1.maxminavgmaxminavg

We can rewrite the right-hand side of this in the following way: π›Όβˆ’π›Ό=π›Όπ‘›βˆ’1(π‘›βˆ’π‘›)=𝐴(π‘›βˆ’π‘›).maxminavgavgmaxminmaxmin

In the second line here, we have used our equation for the apex angle 𝐴 in terms of 𝛼avg and 𝑛avg, which is only true for thin prisms, so here we are assuming that 𝐴 is small.

Let’s think about how we might use this new equation for thin prisms: π›Όβˆ’π›Ό=𝐴(π‘›βˆ’π‘›)maxminmaxmin.

Suppose we have a prism that has an apex angle of 8∘, which is a small enough angle for us to consider this a thin prism. This prism has different refractive indices for different wavelengths of light that are transmitted though the prism. This refractive index for the longest wavelength of light is 1.512, and the refractive index for the shortest wavelength of light is 1.535. We call the difference between the deviation angle for the most deviated and least deviated rays passing through the prism its β€œangular dispersion.” We can calculate this as follows.

The following diagram shows the refraction of light in this thin prism.

Angular dispersion is the difference between the maximum angle of deflection and the minimum angle of deflection for the prism. We call the angular dispersion Δ𝛼, where Ξ” is the Greek capital letter delta. Note that Δ𝛼 should be read as one quantity, and it is not a multiplication of Ξ” times 𝛼. This means that Δ𝛼=π›Όβˆ’π›Όmaxmin.

Let’s use our equation relating refractive index to angle of dispersion, which is π›Όβˆ’π›Ό=𝐴(π‘›βˆ’π‘›)maxminmaxmin. We can replace the left-hand side of this equation with the angular dispersion, so we have Δ𝛼=𝐴(π‘›βˆ’π‘›).maxmin

We can now input the values we are told in the question, which are 𝐴=8∘, 𝑛=1.535max, and 𝑛=1.512min. This gives us Δ𝛼=8Γ—(1.535βˆ’1.512)=8Γ—(0.023)=0.184.∘∘∘

This means that our answer for the angular dispersion of the prism is Δ𝛼=0.184∘.

Let’s now summarize what has been learned in this explainer.

Key Points

  • Prisms disperse white light, spreading it out according to the different wavelengths of different colors of light.
  • This dispersion is caused because the refractive index of the material that makes up the prism varies with wavelength.
  • The amount that a prism disperses light can be quantified by a quantity called the β€œdispersive power” of the prism, denoted as πœ”ο΅.
  • If 𝛼max and 𝛼min are the maximum and minimum angles of deviation that different wavelengths of light experience when they pass through the prism, then πœ”=π›Όβˆ’π›Όο€»ο‡.maxminmaxmin
  • If 𝑛max and 𝑛min are the maximum and minimum refractive indices of different wavelengths of light in a prism, then πœ”=π‘›βˆ’π‘›ο€»ο‡βˆ’1.maxminmaxmin
  • For thin prisms, which have a small apex angle 𝐴, we can relate the angular dispersion of a prism to the apex angle of the prism and the refractive indices of the prism using the equation π›Όβˆ’π›Ό=𝐴(π‘›βˆ’π‘›).maxminmaxmin

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