My Math Forum geometrical contradiction in rainbow formation

 Physics Physics Forum

 April 11th, 2016, 09:55 PM #1 Member     Joined: May 2014 From: India Posts: 87 Thanks: 5 Math Focus: Abstract maths! geometrical contradiction in rainbow formation If the angle of incidence($i$) is such that the angle of refraction($c$) is greater than the critical angle for the water droplet, then the next angle of incidence will also be $c$ as radii of circle are equal. As the diagram shows, the light ray shall suffer total internal reflection. This process shall continue infinitely because every time the angle of incidence is $c$. The light ray gets trapped and shall never escape. However, according to multiple books and multiple websites, the sunlight enters a spherical water droplet, (gets dispersed obviously), suffers total internal reflection, COMES OUT!, and reaches the eye giving the perception of rainbow. Either I am wrong or the explanation is wrong. The phenomenon of rainbow is surely not wrong, I've seen one myself! If you know where the problem lies, please reply. I wait for an answer...
April 12th, 2016, 02:14 AM   #2
Senior Member

Joined: Apr 2014
From: Glasgow

Posts: 2,150
Thanks: 730

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
 Originally Posted by Rishabh If the angle of incidence($i$) is such that the angle of refraction($c$) is greater than the critical angle for the water droplet, then the next angle of incidence will also be $c$ as radii of circle are equal. As the diagram shows, the light ray shall suffer total internal reflection. This process shall continue infinitely because every time the angle of incidence is $c$. The light ray gets trapped and shall never escape. However, according to multiple books and multiple websites, the sunlight enters a spherical water droplet, (gets dispersed obviously), suffers total internal reflection, COMES OUT!, and reaches the eye giving the perception of rainbow. Either I am wrong or the explanation is wrong. The phenomenon of rainbow is surely not wrong, I've seen one myself! If you know where the problem lies, please reply. I wait for an answer...
I don't know for sure, but it seems to me that given some random ray and collection of (moving) water droplets, it is unlikely that:

1. the ray will hit at the exact point on the droplet and angle needed to cause repeated total internal reflections; and
2. all of the water droplets are exactly spherical and chances are they change as a function of time as they float about in the air... there will be little asymmetries in the shape of the droplet.

So, given some sample of water droplets and a bunch of rays, the vast majority of droplets are going to have their light escape at a different outgoing angle (scattering). However, I don't see it being impossible for many total internal reflections to happen before the ray eventually escapes.

 April 12th, 2016, 02:40 AM #3 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timey-wimey stuff. I have encountered a few Physicists that say my idea doesn't work and it has to be a reflection. On the other hand I've never seen a rainbow when looking at the sun. That means the light is backscattered somehow. But I think it's more likely to be a diffraction phenomenon rather than reflection. At least, that's what makes sense to me. -Dan
 April 17th, 2016, 06:31 AM #4 Member   Joined: Mar 2016 From: US Posts: 30 Thanks: 1 Hello, How are a DVD's grouping of same colors similar to a rainbow phenomenon of grouping colors, if spherical water drops are not involved? Also when I look in a mirror in a mirror, it appears to get darker and change color the farther back the reflections get. If the angle of incidence is not changing from the bathroom light, then is the angle of refraction changing? In regards to reflection? Can different colors on the DVD reflect on each other (maybe after bouncing off the silver surface, running into the original ray, then bouncing off again as a different color)? Therefore changing the angle of refraction, by reflecting off the original ray? Last edited by GreenBeast; April 17th, 2016 at 06:40 AM.
 April 17th, 2016, 07:18 AM #5 Member   Joined: Mar 2016 From: US Posts: 30 Thanks: 1 Hello, How are a DVD's grouping of same colors similar to a rainbow phenomenon of grouping colors, if spherical water drops are not involved? Also when I look in a mirror in a mirror, it appears to get darker and change color the farther back the reflections get. If the angle of incidence is not changing from the bathroom light, then is the angle of refraction changing? In regards to reflection? Can different colors on the DVD reflect on each other (maybe after bouncing off the silver surface, running into the originating ray, and so the originating ray maybe refracts the reflected part inside of the virtual reflection, and the originating ray is slightly refracted also in the virtual reflection, and so both are exiting as a different color)? So flat surfaces reflecting off each other, spherical water drops, and triangular prisms, can all cause color changes with refraction and reflection. I am still mystified by the rainbow. It is almost like a DVD's circular plane meets the spherical drops and the drops translate the colors usually running towards the DVD edge, into the colors organized in a semi circle. Anyone else have any ideas?
May 13th, 2016, 10:03 AM   #6
Senior Member

Joined: Apr 2015
From: Planet Earth

Posts: 141
Thanks: 25

Quote:
 Originally Posted by Rishabh If the angle of incidence($i$) is such that the angle of refraction($c$) is greater than the critical angle for the water droplet, then the next angle of incidence will also be $c$ as radii of circle are equal.
The angle angle of incidence ($\displaystyle i$) can't be such.

Changing your notation to what I'm more used to, if $\displaystyle A$ is the angle light makes with the surface normal on the outside of the spherical raindrop, and $B$ is angle it makes on the inside, then $\displaystyle sin(A) = n*sin(B)$, where $\displaystyle n$ the index of refraction of water (about 4/3). And this formula holds whether the light is entering, or exiting, the drop.

The critical angle $\displaystyle C$ is the interior angle $\displaystyle B$ that corresponds to $\displaystyle A=90°$. Total Internal Reflection occurs if $\displaystyle B>C$, since there is no solution to the formula. But as you noticed, $\displaystyle B$ can't be greater than $\displaystyle C$.

I don't know what explanation you refer to, but I'd say 90% of the explanations you find on the internet are wrong in some way. The errors range from the blatantly false - like Total Internal Reflection is responsible for the colored bands, or the secondary's colors are backwards because the second reflection inverts the image - to the simply inaccurate - like the raindrops act like prisms.

When most people say that "dispersion" is synonymous with "separation of colors," they are really thinking "isolation of colors." Newton cut a vertical slit in a window shade and intercepted the beam of sunlight with a triangular prism held vertically. The resulting spectrum was a rectangle to the right or left of the beam, with red isolated on one side, violet on the other, and ROYGBIV in between.

If he had held it horizontally, he would have gotten a line instead of the rectangle, and above or below instead of to the side. It would be red on one end, violet on the other, but with composite colors in between as the spectra were projected on top of each other. Colors are still "separating" due to dispersion, but recombining from different original positions. Rainbows are actually closer to this experiment, than the one with the vertical prism.

Using my notation, light gets deflected thru the angle $\displaystyle A-B$ when it enters the drop, and when it exits. Each time it reflects internally, it deflects $\displaystyle 180°-2B$. So light that exits after one reflection deflects a total of $\displaystyle 180°+2A-4B$. But this is from its original direction; it's easier subtract this from 180° to get the angle from the sun, to the drop, then to you: $\displaystyle D=4B-2A$. My first attachment shows a plot of this for three colors.

So each color gets deflected thru all angles from 0° to about 40° to 42°. Something like the beam of a wide flashlight. But each color is brightest, by far, in the outer rim of this beam (see second attachment). So, while the red band is pure red, the violet one is composed of all colors in ROYGBIV; but violet is brightest. So it isn't a true spectrum at all, but it is closely related. And inside the violet band, the rainbow still exists as a white disk that extends to the horizon. (What is usually noticed is that the sky is darker outside.)
Attached Images
 RainbowDeflection.jpg (58.7 KB, 2 views) Rainbow Power.jpg (38.9 KB, 1 views)

Last edited by JeffJo; May 13th, 2016 at 10:06 AM.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post BenFRayfield Physics 3 January 14th, 2015 07:46 AM H_e_c_t_o_r Linear Algebra 5 November 21st, 2013 09:37 AM Chikis Physics 0 September 7th, 2012 06:15 AM H_e_c_t_o_r Abstract Algebra 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top