# The mediums of an equilateral triangle meet at alter

### Total Internal Reflection

The goal of this activity is to fold an equilateral triangle from a square piece of paper. Question 3: One of your variables will be your parameter that you'll change un- til you get the .. Question 1: Find the coordinates of the point P, where the diagonal creases meet. (Assume .. Business cards are a very popular medium in. The three medians of a triangle will always meet at one point in the If the triangle is equilateral, where all sides are equal, all medians will be. A common Physics lab is to sight through the long side of an isosceles triangle at a pin or other object held behind the opposite face. When done so, an unusual.

So let's think about this triangle right over here. Let's think about this triangle, triangle AGB. This is triangle AGB right over there. Those are the same triangles.

And let's compare that to triangle EAG right over here. Let's compare it to this triangle, which is this triangle right over here on the original drawing. Now, they both have the exact same height. If you view EG as their base-- or I the guess their shared base, they don't have the exact same base. The smaller triangle has the base E-- sorry, the smaller orange triangle has GB as it's base. The larger blue triangle has EG as its base, but they definitely both have the same height, or altitude, when you draw it this way.

So their height, in both cases, is this right over here. Now the other thing that we do know is that this blue triangle EAG has twice the area of the orange triangle. How do we know that?

Because it's got two of these triangles in it. So one way to think about it is if this orange triangle has area x, actually let me call it a-- well I already used a, so I'm going to call it area x-- then each of these blue triangles have area x. Or you could say this entire blue region has area 2x.

I'm just applying the formula for area of a triangle. This is our area. Now, let's do the same thing for this orange triangle. Well we can substitute it. If this is equal to x, we can place this entire expression right over here for x. So let's do that.

**Formula of an equilateral triangle**

But instead of x, I'm going to write this here. And now we can just simplify this. Or we could write EG over 2 is equal to-- let me do it in the same color since I've gone this far with the same colors. This is essentially saying that GB is half of EG. So for example, if EG is 2, this is going to be 1. If EG is 4, this is going to be 2. So we've actually proven our result. That is, as the angle of incidence is increased, the brightness of the refracted ray decreases and the brightness of the reflected ray increases.

Finally, we would observe that the angles of the reflection and refraction are not equal. Since the light waves would refract away from the normal a case of the SFA principle of refractionthe angle of refraction would be greater than the angle of incidence. And if this were the case, the angle of refraction would also be greater than the angle of reflection since the angles of reflection and incidence are the same. As the angle of incidence is increased, the angle of refraction would eventually reach a degree angle.

These principles are depicted in the diagram below.

## Proofs concerning equilateral triangles

The maximum possible angle of refraction is degrees. If you think about it a practice that always helpsyou recognize that if the angle of refraction were greater than 90 degrees, then the refracted ray would lie on the incident side of the medium - that's just not possible. So in the case of the laser beam in the water, there is some specific value for the angle of incidence we'll call it the critical angle that yields an angle of refraction of degrees.

Any angle of incidence that is greater than Instead, when the angles of incidence is greater than When this happens, total internal reflection occurs. Two Requirements for Total Internal Reflection Total internal reflection TIR is the phenomenon that involves the reflection of all the incident light off the boundary. TIR only takes place when both of the following two conditions are met: Total internal reflection will not take place unless the incident light is traveling within the more optically dense medium towards the less optically dense medium.

TIR will happen for light traveling from water towards air, but it will not happen for light traveling from air towards water.

TIR occurs because the angle of refraction reaches a degree angle before the angle of incidence reaches a degree angle. The only way for the angle of refraction to be greater than the angle of incidence is for light to bend away from the normal. Since light only bends away from the normal when passing from a more dense medium into a less dense medium, then this would be a necessary condition for total internal reflection.

Total internal reflection only occurs with large angles of incidence. How large is large? As mentioned above, the critical angle for the water-air boundary is So for angles of incidence greater than The actual value of the critical angle is dependent upon the two materials on either side of the boundary.

For the crown glass-air boundary, the critical angle is For the diamond-air boundary, the critical angle is For the diamond-water boundary, the critical angle is The critical angle is different for different media. In the next part of Lesson 3we will investigate how to determine the critical angle for any two materials. For now, let's internalize the idea that TIR can only occur if the angle of incidence is greater than the critical angle for the particular combination of materials.

Light Piping and Optical Fibers Total internal reflection is often demonstrated in a Physics class through a variety of demonstrations. In one such demonstration, a beam of laser light is directed into a coiled plastic thing-a-ma jig.

The plastic served as a light pipe, directing the light through the coils until it finally exits out the opposite end. Once the light entered the plastic, it was in the more dense medium. Every time the light approached the plastic-air boundary, it is approaching at angles greater than the critical angle.

### Proofs concerning equilateral triangles (video) | Khan Academy

The two conditions necessary for TIR are met, and all of the incident light at the plastic-air boundary stays internal to the plastic and undergoes reflection. And with the room lights off, every student becomes quickly aware of the ancient truth that Physics is better than drugs.

This demonstration helps to illustrate the principle by which optical fibers work.

The use of a long strand of plastic or other material such as glass to pipe light from one end of the medium to the other is the basis for modern day use of optical fibers. Optical fibers are used in communication systems and micro-surgeries.