Optics B/C

Test your knowledge of various Science Olympiad events.
kenniky
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Re: Optics B/C

Postby kenniky » February 23rd, 2017, 5:51 pm

Didn't even know Fabry-Perot was a thing until now
From what I've managed to find on the subject, it looks like it should be wavelength squared over two times the length of the cavity, and we can probably safely ignore the slight difference between refractive index of air and vacuum. Using this, I got a result of 0.0039... nanometers, I think? Not really sure if I did this right.
that's not what I'm getting... check again?
Well if that didn't work I'm not really sure how to do it... how did you do the calculation?
Your formula is right but for some reason I'm getting a different result than you did, check your calculation again I think
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Re: Optics B/C

Postby Avogadro » February 23rd, 2017, 6:18 pm

that's not what I'm getting... check again?
Well if that didn't work I'm not really sure how to do it... how did you do the calculation?
Your formula is right but for some reason I'm getting a different result than you did, check your calculation again I think
Got 0.00196 nm this time. Sounding better?
Lower Merion 2017
Subtitled: Revenge of the Non-Harriton

Placement Record:

Code: Islip | Conestoga | Tiger | Regionals | States
Out of: 61 | 42 | 36 | 37 | 36

Chemistry Lab: 9 | - | - | 4 | 4
Astronomy: 14 | - | 5 | 10 | 3
Material Science: 12 | 19 | 9 | 5 | 9
Optics: 14 | 7 | 3 | 4 | 2

kenniky
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Re: Optics B/C

Postby kenniky » February 24th, 2017, 7:23 am

Well if that didn't work I'm not really sure how to do it... how did you do the calculation?
Your formula is right but for some reason I'm getting a different result than you did, check your calculation again I think
answer
Got 0.00196 nm this time. Sounding better?
Yup, your turn
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Re: Optics B/C

Postby Avogadro » February 24th, 2017, 12:45 pm

Oh good, I was getting worried there.

White light is passed first through a filter absorbing wavelengths of light around 700 nm, and then through another filter absorbing wavelengths of light around 450 nm. It then hits a (yellow) lemon. Assuming no other light is present, what color does the lemon appear?
Lower Merion 2017
Subtitled: Revenge of the Non-Harriton

Placement Record:

Code: Islip | Conestoga | Tiger | Regionals | States
Out of: 61 | 42 | 36 | 37 | 36

Chemistry Lab: 9 | - | - | 4 | 4
Astronomy: 14 | - | 5 | 10 | 3
Material Science: 12 | 19 | 9 | 5 | 9
Optics: 14 | 7 | 3 | 4 | 2

UTF-8 U+6211 U+662F
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Re: Optics B/C

Postby UTF-8 U+6211 U+662F » February 26th, 2017, 11:34 am

Answer
If 700 nm is red and 450 nm is blue, then green?

Avogadro
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Re: Optics B/C

Postby Avogadro » February 26th, 2017, 3:20 pm

Answer
If 700 nm is red and 450 nm is blue, then green?
Correct! Your turn/
Lower Merion 2017
Subtitled: Revenge of the Non-Harriton

Placement Record:

Code: Islip | Conestoga | Tiger | Regionals | States
Out of: 61 | 42 | 36 | 37 | 36

Chemistry Lab: 9 | - | - | 4 | 4
Astronomy: 14 | - | 5 | 10 | 3
Material Science: 12 | 19 | 9 | 5 | 9
Optics: 14 | 7 | 3 | 4 | 2

UTF-8 U+6211 U+662F
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Re: Optics B/C

Postby UTF-8 U+6211 U+662F » February 27th, 2017, 4:29 pm

A laser is pointed at mirror A with an angle of incidence of 50 degrees. Mirror A is connected to Mirror B at a 30 degree angle. What is the angle of reflection off of mirror B?

(The mirrors look something like this:

Mirror A
<
Mirror B

)

Avogadro
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Re: Optics B/C

Postby Avogadro » March 2nd, 2017, 6:23 pm

A laser is pointed at mirror A with an angle of incidence of 50 degrees. Mirror A is connected to Mirror B at a 30 degree angle. What is the angle of reflection off of mirror B?

(The mirrors look something like this:

Mirror A
<
Mirror B

)
Attempt
20°?
Lower Merion 2017
Subtitled: Revenge of the Non-Harriton

Placement Record:

Code: Islip | Conestoga | Tiger | Regionals | States
Out of: 61 | 42 | 36 | 37 | 36

Chemistry Lab: 9 | - | - | 4 | 4
Astronomy: 14 | - | 5 | 10 | 3
Material Science: 12 | 19 | 9 | 5 | 9
Optics: 14 | 7 | 3 | 4 | 2

UTF-8 U+6211 U+662F
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Re: Optics B/C

Postby UTF-8 U+6211 U+662F » March 3rd, 2017, 6:10 pm

That's what I got, yep!

Avogadro
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Re: Optics B/C

Postby Avogadro » March 4th, 2017, 6:47 am

Cool.

A beam of light enters a prism with an index of refraction of 1.35. If each of the angles of the prism is 60°, what maximum angle must the angle of incidence be so that no light comes out the other side?
Lower Merion 2017
Subtitled: Revenge of the Non-Harriton

Placement Record:

Code: Islip | Conestoga | Tiger | Regionals | States
Out of: 61 | 42 | 36 | 37 | 36

Chemistry Lab: 9 | - | - | 4 | 4
Astronomy: 14 | - | 5 | 10 | 3
Material Science: 12 | 19 | 9 | 5 | 9
Optics: 14 | 7 | 3 | 4 | 2

UTF-8 U+6211 U+662F
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Re: Optics B/C

Postby UTF-8 U+6211 U+662F » March 5th, 2017, 10:03 am

Cool.

A beam of light enters a prism with an index of refraction of 1.35. If each of the angles of the prism is 60°, what maximum angle must the angle of incidence be so that no light comes out the other side?
Answer?
Assuming a surrounding index of refraction of 1, arcsin(1/1.35) = 47.8 deg, which requires an angle of refraction of 77.8 deg, so 1*sin(theta) = 1.35*sin(77.8 deg), theta = arcsin(1.35*sin(77.8 deg)), and theta = undefined. Therefore, light always comes out the other side?

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Re: Optics B/C

Postby Avogadro » March 7th, 2017, 7:58 am

Cool.

A beam of light enters a prism with an index of refraction of 1.35. If each of the angles of the prism is 60°, what maximum angle must the angle of incidence be so that no light comes out the other side?
Answer?
Assuming a surrounding index of refraction of 1, arcsin(1/1.35) = 47.8 deg, which requires an angle of refraction of 77.8 deg, so 1*sin(theta) = 1.35*sin(77.8 deg), theta = arcsin(1.35*sin(77.8 deg)), and theta = undefined. Therefore, light always comes out the other side?
Yeah, looks like I accidentally made a trick question lol. Good practice anyway! :P
Lower Merion 2017
Subtitled: Revenge of the Non-Harriton

Placement Record:

Code: Islip | Conestoga | Tiger | Regionals | States
Out of: 61 | 42 | 36 | 37 | 36

Chemistry Lab: 9 | - | - | 4 | 4
Astronomy: 14 | - | 5 | 10 | 3
Material Science: 12 | 19 | 9 | 5 | 9
Optics: 14 | 7 | 3 | 4 | 2

Tom_MS
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Re: Optics B/C

Postby Tom_MS » March 8th, 2017, 11:37 am

Cool.

A beam of light enters a prism with an index of refraction of 1.35. If each of the angles of the prism is 60°, what maximum angle must the angle of incidence be so that no light comes out the other side?
Answer?
Assuming a surrounding index of refraction of 1, arcsin(1/1.35) = 47.8 deg, which requires an angle of refraction of 77.8 deg, so 1*sin(theta) = 1.35*sin(77.8 deg), theta = arcsin(1.35*sin(77.8 deg)), and theta = undefined. Therefore, light always comes out the other side?
Yeah, looks like I accidentally made a trick question lol. Good practice anyway! :P
I'm not getting that result.
It looks like you just used snell's law and the critical angle and didn't use the 60 degree apex angle. Using geometry (which is kind of complicated to explain here), I'm getting the formula of theta=arcsin( n*sin( 60 - arcsin(1/n))). This gives be 16.58 degrees. If this angle was decreased (I think) any more, no light would escape.

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Re: Optics B/C

Postby UTF-8 U+6211 U+662F » March 8th, 2017, 12:54 pm

Answer?
Assuming a surrounding index of refraction of 1, arcsin(1/1.35) = 47.8 deg, which requires an angle of refraction of 77.8 deg, so 1*sin(theta) = 1.35*sin(77.8 deg), theta = arcsin(1.35*sin(77.8 deg)), and theta = undefined. Therefore, light always comes out the other side?
Yeah, looks like I accidentally made a trick question lol. Good practice anyway! :P
I'm not getting that result.
It looks like you just used snell's law and the critical angle and didn't use the 60 degree apex angle. Using geometry (which is kind of complicated to explain here), I'm getting the formula of theta=arcsin( n*sin( 60 - arcsin(1/n))). This gives be 16.58 degrees. If this angle was decreased (I think) any more, no light would escape.
The 60 degree apex angle was used to determine
the angle of 77.8 deg
Anyways new question:
A Science Olympiad Optics supervisor holds up two white flashlights. The first has a magenta filter, and the second has a cyan filter. What color is the overlap of the two flashlights?

jonboyage
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Re: Optics B/C

Postby jonboyage » March 8th, 2017, 2:12 pm

Yeah, looks like I accidentally made a trick question lol. Good practice anyway! :P
I'm not getting that result.
It looks like you just used snell's law and the critical angle and didn't use the 60 degree apex angle. Using geometry (which is kind of complicated to explain here), I'm getting the formula of theta=arcsin( n*sin( 60 - arcsin(1/n))). This gives be 16.58 degrees. If this angle was decreased (I think) any more, no light would escape.
The 60 degree apex angle was used to determine
the angle of 77.8 deg
Anyways new question:
A Science Olympiad Optics supervisor holds up two white flashlights. The first has a magenta filter, and the second has a cyan filter. What color is the overlap of the two flashlights?
I have to agree with Tom here
I'm getting the same exact answer of 16.5836 degrees and I made sure it was right by doing it the long way and going through each step that leads to the condensed formula Tom described
Answer
I'm not 100% sure about this but is it light blue?
I was in a bin

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