Trig. Project

In my College Trig. class we are having to do a project that involves showing the use of Trigonometry in different aspects of life. I have chossen to do reflection and refraction of light, or something along those lines. But, the only problem is, is that I don't really know of what I could do as a project on the matter. When I say project, I mean something I can physically show to the class how the light reflects off of things, but I'm not sure on how to go about this.

Any help?

-NC

Could you use a mirror and a laser pointer? ^^;

Get a light bulb and place it side by side with a white screen, but slightly behind it so none of the light reflects onto the screen directly from the light-bulb.

Then, stand back, and with a concave mirror (bent inwards like the inside of a spoon), reflect the light from the bulb onto the screen.

Because concave mirrors can focus light, if you hold the mirror at a certain distance away from the screen, you will see an upside-down image on the screen in the shape of the light-bulb filament.
We just did that experiment in my physics class ^^

If you think you could end up doing that, PM me, and I'll explain a little more to you on why this happens.

Or maybe you already know, considering you're in a college level trig class and I'm only in high-school ^^;

yeah, but if this is a trigonometry project you'll have to focus on refraction, Snell's law... (the law of reflection: "the ingoing and reflected angles are equal", is not super trigonometrical...)
so you would need some sort of laser and at least 2 or 3 different transparent (or semi-transparent) materials

Snell's law was something like this: n1 sen(a1) = n2 sen(a2)
n1, n2 are constants of the materials
a1, a2 are the ingoing and refracted angles, measured from the perpendicular...

but I suppose you already know that, if you chose the theme...

So... you can use air and water...
hmmm air and glass (however, a block of glass is difficult to find)
perhaps the three at the same time in a double refraction...
maybe oil might work too...

I believe air's constant is 1, and water is 1.33, only God knows the constant for oil...

but...I need to know, what exactly is your problem with this project? so we all can be more helpful...

Jazz Man: I'm actually in High School, just taking a college level trig. class, but it's like, pretty much basic Trig. I like the idea you gave but Rock's idea is more on the basis of what I'm wanting to do.

Rock: Yea, I've been trying to do some research on Snell's Law but I'm stil not sure exactly on all the information about it. I've seen the formula you gave above and was planning on basing the project around that. Should sen = sin? Just wondering, didn't know if it was a typo or not.

Anyway, since I've finally decided on what I'm going to do the project on, and know a little more about it. If I'm going to use Snels' Law then how exactly coulod I show this. I have some lasers and mirrors, and can probably get some "transperent" materials. What do you have in mind?

-NC

Uh-Oh sorry , in spanish it's seno, not sine, so: "sen"...

do you understand the law?

I'll do the water and air example:

watching it from a lateral perspective you can see the water surface as a line...
the perpendicular is called "normal" (at least in spanish)... if a laser beam enters the water it's refracted by it... I will call "access point" the point where the beam enters through the surface... imagine the perpendicular to the surface that passes through the access point...

then a1 is the angle between the beam and the normal in the point of access (the smallest angle) in the air, while a2 is the same, but underwater...

so.. if you use one of those triangular rulers (sorry, I don't know the english name), the 30-60-90 one, then you can make the beam enter the surface at 60º measured from the same surface... which is 30º from the normal... so a1 = 30, and, as I said before n1 = 1 (air)...

now n2 = 1.33 (water) so the law states that:

1 sin(30º) = (1.33) sin (a2)
which implies this:
a2 = 22.0824º

so the angle predicted by the law is about 22º (68º measured from the surface)...

now you have to use your "triangle ruler" (or any other method to measure angles), a fish tank and a laser... measure the refracted angle with (yes, it happened again) your "round ruler" that measures angles and "really HOPE" it's close to 22º... (if not, then the water constant might be a little different for your water)

for the oil, you can go backwards.. put the laser beam at a determined angle, measure the refracted one, and calculate the oil constant , then use another angle, and predict the refracted one... and finally show the law is true for this example too, doing the same thing you did with water...

one more thing: it will be a lot easier to do this if the laser beam can be seen in the air (not just the red dot in the surface and another one in the tank bottom) but it can be done with just the two dots...
also: be sure the beam draws a parallel line to the borders of the tank when watching it from above.

by the way, please tell me the names of those rulers (I couldn't find them in the dictionary...)


(seriously, we have to put the deadlines in the title... I hope I'm not too late)

Thanks Rock, this will help a lot. I've found some ionformation on other ways to do it as well. So, I'll put together a few ideas and try to turn it into a decent presentation and project.

And I still have plenty of time to do it, so you're not to late.

Thanks again,
-NC