How bright is a rainbow?
Category: Physics Published: October 12, 2026
By: Christopher S. Baird, author of The Top 50 Science Questions with Surprising Answers and Associate Professor of Physics at West Texas A&M University
The brightness of a rainbow can vary a lot and depends on several things. Keep in mind that the light that you see as a rainbow is sunlight that has refracted and reflected inside spherical raindrops. For this reason, the brightness of a rainbow depends on how bright the sunlight is, which ultimately depends on how hazy it is, the time of day, and your latitude. Additionally, a rainbow will be brighter if there are more raindrops or bigger raindrops. Despite these complications, we can still make some general statements about the brightness of a rainbow. To avoid confusion, let's use the word "rainbow" to mean the thin, banded colorful arc of the primary rainbow, which is what most people mean when they say "rainbow" in everyday life. Also, let's use the word "brightness" to refer to how much energy is being carried by a light wave per second, which scientists call the "optical power" or "radiant flux."
In general, the brightness of a ray of light within a rainbow equals about 4.5% of the brightness of the original ray of sunlight from which it formed. To see where this 4.5% number comes from, let's look at how rainbows are formed and how to calculate the brightness.
Public Domain Image, source: Christopher S. Baird.
A rainbow is formed when rays of sunlight enter a spherical raindrop, refracting and bending in the process, then reflect off the inner back surface of the raindrop, and then refract and bend again when exiting the raindrop. The refraction is the key, because it's what spreads out the white sunlight into its various color components. Because the surface of a raindrop is round, the formation of a rainbow is far more complex than the formation of the rainbow-like pattern created by a triangular glass prism.
It's important to realize that many rays of sunlight enter the raindrop at different heights and take different paths through the raindrop and back out. The height h at which a ray of light enters a raindrop is the vertical distance of the initial ray from a horizontal line that goes through the center of the spherical raindrop. Let's express this entrance height h as a percentage of the radius of the spherical raindrop. Therefore, h = 0% represents an initial ray of light that is headed directly toward the center of the raindrop, h = 50% represents an initial ray that enters the raindrop at a height halfway between the center and the top edge of the raindrop, and h = 100% represents an initial ray that encounters the top edge of the raindrop.
For different entrance heights, the ray of light bends different amounts and therefore ends up traveling in different directions at the end. Let's calculate where each ray of light goes by applying Snell's law of refraction to both refraction events and the law of reflection to the reflection event. Then let's use the Fresnel equations to calculate how much of the light reflects when a ray of light encounters the surface of the raindrop, and how much of it crosses the surface and refracts. In all of the ray images in this article (except the one marked "exaggerated illustration"), the rays of light have been drawn exactly at their physically correct location and traveling in their physically correct direction, as determined by the calculations.
White sunlight is a mixture of all visible colors, but each color experiences a slightly different thing, so let's investigate one color at a time. Let's start by investigating light of the color 675 nm red, which is a middle-of-the-road red color. Red is a good choice to start with because it sits on one end of the visible spectrum. Now let's pick a random entrance height, say h = 67%, and see what happens. The results of the calculations are shown in the image below.
Public Domain Image, source: Christopher S. Baird.
As you can see, the original ray of light encounters the surface of the raindrop and splits, so that 2.6% of the light (in terms of energy flow) reflects off the surface, and the other 97.4% of the light enters the raindrop, refracting and bending in the process. This ray of light that just entered the raindrop then travels until it meets the back, inner surface of the raindrop, where 95% of the original light exits the raindrop and is lost, while the remaining 2.4% of the original light reflects and therefore stays in the raindrop. This ray of light then travels across the raindrop and encounters the surface of the raindrop for a third time, where 0.1% of the original light reflects and stays inside the raindrop, while the remaining 2.3% of the original light exits the raindrop at this point. In this case, the final ray of light that goes on to become part of the rainbow, or part of the faint white area encircled by the rainbow, therefore has a brightness equal to 2.3% of the brightness of the original ray of sunlight from which it came.
All of the light is now accounted for in the image above, as you can verify, because 2.6% plus 95.0% plus 0.1% plus 2.3% equals 100%. However, we only care about the final ray of light on the bottom of the image, because it's the only one that goes on to become part of the rainbow or part of the faint white area encircled by the rainbow.
Hold on! Didn't I say at the beginning that the brightness of a rainbow ray equals about 4.5% of the brightness of the original ray of sunlight from which it formed, not 2.3%? What's going on? The simple answer is that we randomly picked an entrance height of h = 67% the radius of the spherical raindrop, and the corresponding final ray for this entrance height does not actually become part of the rainbow. Rather, it becomes part of the faint white area encircled by the rainbow. A little later, I'll show you the rays of light that enter the raindrop at the right height so that they do indeed become final rays that are part of the rainbow.
For those of you who are curious about the technical details, I calculated how much light reflects off the surface at each surface encounter, and how much crosses the surface, using the Fresnel equations. I did these calculations separately for s-polarized and p-polarized light waves, tracing each polarization all the way through the raindrop and to the observer. I then averaged together the resulting brightness for the s-polarized wave and p-polarized wave for each ray to find the overall brightness of the light that reflects and the overall brightness of the light that crosses the surface. Keep in mind that each original ray of sunlight is unpolarized and therefore acts like an incoherent mixture of equal amounts of an s-polarized wave and a p-polarized wave.
The calculations that led to the brightness values in the image above were done for a certain color of light; namely 675 nm red light. You may wonder, how does a different color entering at the same height compare? Let's now look at what happens to a ray of light with the color of 425 nm violet that enters the raindrop at the same height, h = 67%. Violet is a good choice for the next color to investigate because it sits on the opposite end of the visible spectrum from red. All other visible colors of light lie between red and violet. Therefore, if we understand what happens to red light and violet light, we immediately have a good idea of what is happening to all the other colors.
Public Domain Image, source: Christopher S. Baird.
The image above shows what happens to a ray of 425 nm violet light that enters a raindrop at a height of h = 67%. Compare the image above, which is for violet light, to the previous image, which was for red light. You see that the path through the raindrop taken by the red light is nearly identical to that of the violet light. Although it's hard to tell just by looking at these two images, the two paths are actually not exactly the same, and this is crucial to the formation of the rainbow.
As you can see in the image above, the final ray has a brightness equal to 2.4% of the brightness of the original ray of sunlight from which it came. For comparison, the final ray for red light had a brightness of 2.3%. In terms of the brightness of the final ray, the red light and the violet light are very similar.
Note that the final ray of violet light shown in the image above, which corresponds to an initial ray's entrance height of h = 67%, also does not become part of the rainbow, but instead becomes part of the faint white area encircled by the rainbow. So why did I use an entrance height of 67% when I knew that this light does not become part of the rainbow? I did it to drive home the point that not all rays of sunlight that enter the raindrop end up as part of a rainbow. In fact, most rays of sunlight that enter a raindrop do not become part of a rainbow. Calculations show that only rays of sunlight that enter the raindrop at or near an entrance height of h = 86% end up significantly contributing to the rainbow. (Don't forget that we are using the word "rainbow" to refer only to the thin, banded, colorful arc of the primary rainbow.) Let's redo the calculations that we did above for 675 nm red light and 425 nm violet light, but now for an entrance height h = 86%. The results are shown in the two images below.
Public Domain Image, source: Christopher S. Baird.
Public Domain Image, source: Christopher S. Baird.
As you can see in the two images above, the 675 nm red light that becomes part of the red band of the rainbow has a brightness equal to 4.5% of the brightness of the original ray of sunlight from which it came, and the 425 nm violet light that becomes part of the violet band of the rainbow has a brightness equal to 4.6% of the brightness of the original ray of sunlight from which it came.
In the two images above, the paths taken by the red light and the violet light may appear to be the same, but they are actually slightly different, which is essential to the formation of the rainbow. To demonstrate this, let's superimpose the rays of light in the two images above. While we're at it, let's also calculate and show the rays for 635 nm orange light, 593 nm yellow light, 551 nm green light, 509 nm cyan light, and 467 nm blue light.
The image above shows the exact paths taken by red, orange, yellow, green, cyan, and blue rays of light that enter a raindrop at a height of exactly h = 86% and therefore become main contributors to the rainbow that you see. Click on the image above to enlarge it. As you can see, the colors present in each ray of white sunlight spread out into a series of rainbow colors.
Hold on! Aren't the colors in the image above in the wrong order? Isn't red actually on the top of the rainbow? The image above is indeed correct, as it should be, because it was directly generated by calculations that correctly use the laws of physics. You need to realize that to see rays of light that exit the raindrops at different angles, you have to look in different directions at different raindrops.
The exaggerated illustration above demonstrates why you see the order of the colors in a rainbow as you do. Sunlight travels from the left to the right of the illustration, where it then encounters many different raindrops. Each raindrop splits the white light into various colors headed down at different directions, with the red light headed down at the steepest angle, the orange light headed down at the next steepest angle, and so forth, as we already demonstrated. However, your eye is only lined up to receive red light from the top raindrop in this illustration, orange light from the middle raindrop, and yellow light from the bottom raindrop. Therefore, what your eye sees is red on the top (i.e. at the outer edge of the rainbow), orange farther down, yellow even farther down, and so forth. Note that in the illustration above, the angle differences between the various colors in a rainbow have been greatly exaggerated in order to clearly illustrate the point.
But how do we know that the light that enters the raindrop at a height of h = 86% is the main contributor to the rainbow? Let's again investigate what happens to 675 nm light, but now let's calculate the paths and brightness levels for many rays that enter the raindrop at different heights.
As you can see in the image above, many rays of light enter the raindrop at different entrance heights, and they become final rays traveling down in different directions. This image only shows the rays that exit the raindrop on the third surface encounter, because these are the only ones that become part of the primary rainbow or part of the faint white area surrounded by the rainbow. Each number in the image above specifies the brightness of that final ray as a percent of the brightness of the original ray of sunlight from which it came. As you can see in the image above, the final rays of light that travel down at a steeper angle (the ones near the bottom of the image) have higher brightness values and therefore contribute more to what you see when you look toward a rainbow. To make this more obvious, let's shade the red color of each ray in the image above according to its physical brightness in the real world.
As you can see in the image above, the final rays are all rather dim compared to the initial rays. What's more important to notice, though, is that the final rays with the steepest angle are brighter and more bunched together than the other rays. (Note that these brightness values are for individual rays and therefore do not include wave interference effects.) Let's do some calculations and re-create the same type of image as above, but now with many initial rays, to get a better idea of what is going on.
As you can see in the image above, the final rays with the steepest angles, which correspond to the largest viewing angles, are bunched together the most. This bunching makes the light that you see when you look in that direction brighter than the other final rays.
Another important thing to notice in the image above is that there is a limit to the steepness of the final rays, such that there are no final rays beyond this limit. The final ray with the steepest angle, and therefore the highest viewing angle as seen by the person, forms a distinct edge of the group of all final rays of 675 nm red light. Furthermore, this edge is where the final rays of red light are the brightest and the most closely bunched together. This is why you see a distinct, red outer edge to the rainbow. To be clear, the "viewing angle" is the angle between the visual center of the entire rainbow and where the observer sees a particular bit of light. It equals the inner angle between a final ray of light and the direction that the original sunlight is traveling. I didn't label the brightness of each final ray in the image above because there was no room in the image. However, we can still show this information by shading each red ray in the image above according to its physical brightness in the real world.
The image above demonstrates that both individual ray brightness and ray bunching contribute to the lower edge of the group of final rays being far brighter than any other part of the group of final rays. Looking at the data that produced the image above, I found that the final rays at the lower edge of the group, which are the brightest ones and therefore the ones that dominantly contribute to the red band of a rainbow, all corresponded to initial rays that entered the raindrop at or near a height of h = 86%.
Let's do the same type of calculation, but now for 425 nm violet light, in order to see if different colors experience wildly different things. The image below shows the results.
As you can see in the image above (zoom in and look carefully!), the final rays at the steepest angle are the brightest and the most closely bunched together. It may not be obvious from the two images above, but red and violet actually have their bundle of brightest final rays at slightly different viewing angles. This is why the red band and violet band of a rainbow are visually separated from each other, at different positions. The same is true of the other colors. Each color has its little bundle of brightest final rays at slightly different viewing angles from the other colors, which is why a rainbow has the colors visually separated out to slightly different locations. To make this more obvious, let's plot the brightness of each final ray as a function of its viewing angle, for seven colors spread across the visible spectrum. To do this, I mathematically shot hundreds of rays for each color into a raindrop at different entrance heights ranging from h = 80% to 90%, and then calculated the brightness and viewing angle of each final ray. The results are shown below.
As you can see in the image above, each color has its maximum viewing angle at a different viewing angle from the other colors. Furthermore, the plot above shows that the order of the colors in terms of maximum viewing angle is the exact same as the order of the colors that you see in a rainbow, as it should be, because this is what determines the order of the colors in a rainbow. Don't forget that the maximum viewing angle of a particular color is where that color is the brightest. Red has the largest maximum viewing angle than the other colors, so you should see it as the outermost color of a rainbow. And indeed, you do! As the image above makes clear, to see the middle of the red band of a rainbow, you should look at a viewing angle of about 42.4°.
Note that there appear to be gaps between the different colored curves in the image above. However, that's just because I only did the calculations for seven specific colors. If you did the same type of calculation for all colors, and plotted all the data, the gaps would go away. The graph would instead show a continuous range of rainbow colors, just like we see in a real rainbow.
The image above shows that, for each color, as you start increasing the initial ray's entrance height from 80%, the associated final ray's viewing angle and brightness continuously increase, until you reach an entrance height of 86%. After that point, as you continue to increase the entrance height, the associated final ray's viewing angle decreases, but its brightness continues to increase.
When you look at a rainbow, the large area that is visually enclosed by the rainbow has a faint white appearance because all colors travel out equally at those viewing angles, and white is what you get when you mix equal amounts of all colors. The visually colorful bands of the rainbow arise because the different colors have their bunch of brightest final rays, which happens at the maximum viewing angle for the color, at slightly different viewing angles.
From the image above, we can immediately see some interesting things. The spot in a rainbow that dominantly has the color of 675 nm red sits at a viewing angle of about 42.4° and the spot in a rainbow that dominantly has a color of 425 nm violet sits at a viewing angle of about 41.3°. The visible spectrum actually extends a bit above 700 nm and a bit below 400 nm, meaning that there are some longer-wavelength reds than the 675 nm red shown here and some shorter-wavelength violets than the 425 nm violet shown here. As a result, the entire rainbow actually extends in viewing angle from about 40.6° to about 42.5°. Thus, a rainbow's colors visually extend across about 2° of viewing angle. That's about the same visual width as the width of your pinky finger held outstretched at arm's length, times two.
The image above also makes clear that each color has its final-ray maximum viewing angle for almost the exact same initial ray entrance height as the other colors, which turns out to be h = 86% when you look at the data. Additionally, the image above makes clear that each color has its brightest rays (the ones at the maximum viewing angle) almost exactly at the same brightness level, which is 4.5%.
The last important thing to notice in the image above is that for one particular color and one particular viewing angle, there are actually two final rays that end up at that viewing angle. These two final rays came from two initial rays that entered the raindrop at different heights, but did so just right so that they ended up with the same final-ray viewing angle. For instance, for 425 nm violet light and a viewing angle of 40.8°, the above image tells us that there is ray of light with a brightness of 3.5% that came from an initial ray with an entrance height of about h = 81%, and there is a ray of light with a brightness of 6.1% that came from an initial ray with an entrance height of h = 90%. Of course, you don't actually see these rays as separate rays when you look at a rainbow, because they are coming at you from the same raindrop, at the same angle, and they are the same color. Therefore, what you actually see is the combination of these two rays. For this reason, let's take the data shown in the image above and average together the brightness of the two final rays of the same color at the same viewing angle. And let's do it for all viewing angles and for our seven chosen colors. The results are shown in the image below.
The image above shows that what you actually see when you look across the different colors of a rainbow is light with brightness levels that have very little variation. Note that the image above and the previous image can be a bit misleading because they are just showing the brightness levels of individual rays, whereas what you actually see depends on the brightness of individual rays and how closely they are bunched together. The two images above could mislead you into thinking, for instance, that light of the color 675 nm is brightest in the rainbow at a viewing angle of 41.9°, or even lower, which is incorrect.
The brightness that you actually see when you look at a rainbow depends both on the brightness of the individual rays and on how closely the rays are bunched together. To get a better idea of this, let's create and plot data that includes both effects. I did this by mathematically shooting 25,000 initial rays of light of the same color into a raindrop, evenly spread across all possible entrance heights, then I calculated the brightness and viewing angle of each associated final ray, and then I added up the brightness of all of the final rays that end up in the same little viewing angle bin, repeated for all viewing angle bins. (This is similar to creating a histogram, but I'm not adding the number of rays that end up in the same viewing angle bin; I'm adding their brightness levels. Therefore, I'm actually numerically integrating the brightness across each viewing angle bin.) The results are shown below.
As you can see in the image above, when you include both ray brightness and ray bunching effects, the result becomes very pronounced. Each color is definitely brightest at the final-ray maximum viewing angle for that color. Let's take the image above and zoom in on the peaks, to see them more clearly.
Now you can clearly see in the image above that each color is definitely the brightest at its maximum viewing angle. Furthermore, each color has a different maximum viewing angle, and therefore each color is brightest at a different viewing angle. This is why we see a rainbow as a series of elementary colors separated out into different bands at different locations.
Note that because I started with a ray-specific definition of brightness, but then I added together the brightness of the rays that end up in the same viewing angle bin, to take into account ray bunching effects, the resulting units of brightness are not particularly meaningful. For this reason, I simply labeled the brightness axis units in the graph above as "arbitrary units." But that's okay, because all that we care about in the graph above is the shape of the curves.
Also note that the results in the graph above do not include wave interference effects. It is straightforward to include wave interference effects in the calculations that produced the graph above. However, I intentionally omitted wave interference effects because they depend strongly on the size of the raindrop, and I wanted to demonstrate general effects that are the same for almost all raindrops of all sizes; and because wave interference effects are usually not significant. For most of the rainbows seen in everyday life, the variations that arise from wave interference effects are not noticeable. If you are curious about what wave interference effects can give rise to in rainbows, look up "supernumerary rainbows."
Now that the image above has demonstrated that each color is almost completely only seen in the rainbow at its maximum viewing angle, let's go back to the ray-specific image that had meaningful brightness units and erase all data points for each color except at and near the maximum viewing angle of each color. The result is shown below.
The image above is now the clearest representation of where each color is actually seen in a rainbow and how bright each color is actually seen to be. Again, there are gaps between the colors in the graph above just because I only did the calculations for seven specific colors. However, the results for these seven colors give you a good idea of what it would be for all other colors. If you repeated the calculations for all colors, and plotted all the data, there would be no gaps in the image above. Rather, it would be a continuous series of elementary colors, exactly as you see in a real rainbow.
Amazingly, the results in the image above show that all points in a rainbow are almost exactly the same brightness, with a ray brightness value equal to 4.5% of the brightness of the original ray from which the final ray came from.
Another interesting aspect of rainbows is their polarization. Although humans cannot see the polarization of light with the naked eyes, we can indirectly explore polarization effects using polarization filters. The polarization of a light wave specifies the direction that its electric field points. Any polarization can be represented as a combination of two fundamental polarizations: an s-polarized wave, in which the electric field points perpendicular to the plane containing the rays, and a p-polarized wave, in which the electric field points parallel to the plane containing the rays. The Fresnel equations enable us to calculate how much light is transmitted and how much light is reflected when a ray of light encounters a surface, for each polarization.
By comparing the brightness of the two polarizations for the same ray of light, we can calculate the degree of polarization of that ray and the type of polarization. Specifically, the "degree of polarization" is defined as the optical power of the s-polarized wave minus the optical power of the p-polarized wave, divided by the sum of the two. Therefore, a degree of polarization of 0% corresponds to equal amounts of s-polarized light and p-polarized light, and thus designates unpolarized light. In contrast, a degree of polarization of +100% indicates a ray of light that is completely polarized in the direction perpendicular to the plane containing the rays, while a degree of polarization of -100% indicates a ray of light that is completely polarized in the direction parallel to the plane containing the rays.
Public Domain Image, source: Christopher S. Baird.
The image above shows the calculated degree of polarization of a ray of 675 nm red light that starts out unpolarized (sunlight is naturally unpolarized) and enters the raindrop at a height of h = 86%, thus eventually becoming the middle of the red band of a rainbow. The electric field vectors of p-polarized light are represented in the image above as the black arrows, while the electric field vectors of s-polarized light are represented in the image above by the gray arrows pointing out of the screen directly toward the person reading this article, and they also point into the screen.
As you can see in the image above, upon entering the raindrop, this ray of light becomes slightly polarized. It's 5.6% polarized in the p-polarization direction. When it then reflects off the inner back surface of the raindrop it becomes almost completely polarized. It's 93.3% polarized in the s-polarization direction. When it leaves the raindrop, it becomes slightly less polarized. It's now 92.5% polarized in the s-polarization direction. However, 92.5% polarized is still almost completely polarized. This final ray, which contributes to the middle of the red band of a rainbow, is therefore 92.5% polarized in the direction perpendicular to the plane containing the rays. From the viewpoint of the person looking at the rainbow, the light is therefore almost completely polarized in the direction tangential to the rainbow's arc. This is true for all colors seen in the rainbow. Of course, humans can't directly see polarization.
Public Domain Image, source: Christopher S. Baird.
The image above shows the same type of polarization calculations, but now for a ray of 425 nm violet light that becomes part of the middle of the violet band of the rainbow. The degree of polarization numbers are nearly the same as for red.
We can now repeat these types of calculations for many rays entering at different heights, for our chosen seven colors. The results are shown below.
Public Domain Image, source: Christopher S. Baird.
For each color, when you start increasing the initial ray's entrance height from h = 80%, the viewing angle of the associated final ray increases toward a maximum viewing angle that corresponds to h = 86% (as we already demonstrated), and the polarization of the light steadily decreases. As h increases from 86% to 90%, the viewing angle then decreases while the degree of polarization of the light continues to decrease. As we already discussed, for a certain color and a certain viewing angle, two rays of light end up at that viewing angle and they combine with each other. The observed ray polarization at a particular viewing angle for a particular color is therefore the average of the polarization of these two final rays. However, we have to calculate a weighted average, using the brightness of each polarization as the weighting factors, because a brighter ray of light contributes more of its polarization to the overall polarization. The image below shows the results when you take the data shown in the image above and calculate the weighted average of the degree of polarization of the two final rays of the same color at the same viewing angle, for all viewing angles.
Public Domain Image, source: Christopher S. Baird.
As you can see in the image above, the different colors exhibit similar trends, and there seems to be a broad range in degree of polarization between 85% and 95%. However, keep in mind that the final rays of a certain color are by far the brightest at the maximum viewing angle of that color and those rays are what mostly constitute the rainbow. Therefore, to get a more accurate picture, we take the data shown in the image above and erase all the data points except for those at and near the maximum viewing angle, for each color. The results are shown below.
The image above makes clear that the different color bands in a rainbow have about the same degree of polarization. This means that when you are looking at the different bands of color of a rainbow, you are looking at light that is about 92% to 94% polarized in the direction tangential to the rainbow. Of course, as a human, you can't directly see the polarization. But it's still there.