Optics B/C

[foreverphysics]

...You just wouldn't get a score, in that case. And the laser shoot is worth 40 points, so you should probably at least make an attempt. Especially seeing as you're going to be competing against the powerhouse teams of AL.

[cngu23]

...You just wouldn't get a score, in that case. And the laser shoot is worth 40 points, so you should probably at least make an attempt. Especially seeing as you're going to be competing against the powerhouse teams of AL.

You get 5 points for each mirror the laser hits, so without preparation, I think 10, or maybe even 20 points is definitely possible, by placing the mirrors close to each other.

[AlphaTauri]

...You just wouldn't get a score, in that case. And the laser shoot is worth 40 points, so you should probably at least make an attempt. Especially seeing as you're going to be competing against the powerhouse teams of AL.

You get 5 points for each mirror the laser hits, so without preparation, I think 10, or maybe even 20 points is definitely possible, by placing the mirrors close to each other.

From a layman's perspective, 4 mirrors shouldn't be tough at all, assuming you can get them to go very precisely at 45* angles and line them up with each other, like corners of a rectangle...the C-shape will bounce it nicely around an obstacle, too. That being said, I really ought to go yell at our Optics people...I don't think they did so hot last time with the laser shoot...

[sarahrutledge]

will the target be exact opposite from the laser or will it be somewhere along the wall opposite the wall with the laser?

[Schrodingerscat]

will the target be exact opposite from the laser or will it be somewhere along the wall opposite the wall with the laser?

The target can be anywhere on the opposite wall (checked to confirm B and C have exact same rule on this).

[DivineBbbbbeast]

will the target be exact opposite from the laser or will it be somewhere along the wall opposite the wall with the laser?

My best suggestion for you is to place the mirrors all close to each other, and hope you hit all 5. Then at least you'll get 20/40 instead of a 0/40.

[Osennecho]

Yeah, I have experienced the exact same thing, particularly when the laser beam travels large distances after being reflected at an angle, increasing the result of error, that it strikes the very edge in a way that it hits both the front and side of the mirror and reflects separately.

New FAQ: http://soinc.org/node/925

-----That is precisely how it has happened for us. Much less common this year I would imagine due to the increase in mirrors, but it does occur often enough if using templates where a slight change can be a large thing. 45 degree angles essentially eliminate this worry, but last year with only 2 mirrors that was never the case.

------Thank you for the new FAQ. Printed and will be added to our binder as of tomorrow

[Infinity Flat]

Can someone explain why when you wear sunglasses inside, you perceive white on the chalkboard as green? And how much information should I include about separate parts of the eye?
And also, this problem? (From the Lake Erie/Niagra Regionals C)

54) Doppler shift for the light of wavelength 6000 Å emitted from the sun is 0.04 Å. If the radius of the sun is 7 x 108 m, then the time period of rotation of the sun will be
a. 30 days
b. 365 days
c. 24 hours
d. 25 days

As for the first question, I would think it's because the cones in your eyes are most sensitive to green light.

For the second question, I don't really want to do the calculation, but here are the steps:
Use the radius to find the circumference of the sun.
Use the doppler shift to find the rotational speed of the sun.
Divide the circumference by the rotational sped to get the period.

[cngu23]

Can someone explain why when you wear sunglasses inside, you perceive white on the chalkboard as green? And how much information should I include about separate parts of the eye?
And also, this problem? (From the Lake Erie/Niagra Regionals C)

54) Doppler shift for the light of wavelength 6000 Å emitted from the sun is 0.04 Å. If the radius of the sun is 7 x 108 m, then the time period of rotation of the sun will be
a. 30 days
b. 365 days
c. 24 hours
d. 25 days

As for the first question, I would think it's because the cones in your eyes are most sensitive to green light.

For the second question, I don't really want to do the calculation, but here are the steps:
Use the radius to find the circumference of the sun.
Use the doppler shift to find the rotational speed of the sun.
Divide the circumference by the rotational sped to get the period.

So for finding the doppler shift to get the rotational speed, is that just the formula for doppler effect?
Is the speed of light the speed to use for the equation?
And since you are observing the sun, the speed of observer should be zero?

[Infinity Flat]

Can someone explain why when you wear sunglasses inside, you perceive white on the chalkboard as green? And how much information should I include about separate parts of the eye?
And also, this problem? (From the Lake Erie/Niagra Regionals C)

54) Doppler shift for the light of wavelength 6000 Å emitted from the sun is 0.04 Å. If the radius of the sun is 7 x 108 m, then the time period of rotation of the sun will be
a. 30 days
b. 365 days
c. 24 hours
d. 25 days

As for the first question, I would think it's because the cones in your eyes are most sensitive to green light.

For the second question, I don't really want to do the calculation, but here are the steps:
Use the radius to find the circumference of the sun.
Use the doppler shift to find the rotational speed of the sun.
Divide the circumference by the rotational sped to get the period.

So for finding the doppler shift to get the rotational speed, is that just the formula for doppler effect?
Is the speed of light the speed to use for the equation?
And since you are observing the sun, the speed of observer should be zero?

Because the rotational speed is low enough, you can just use the regular Doppler shift formula. (not the relativistic one)
The speed of the wave is the speed of light.
Yes, the calculated velocity will entirely represent the rotational speed of the sun.