Optics B/C

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

Postby kenniky » February 2nd, 2017, 4:45 pm

Are you sure?
Perfectly unpolarized light does lose half of its intensity when going through the first polarizer. According to Wikipedia, "A beam of unpolarized light can be thought of as containing a uniform mixture of linear polarizations at all possible angles. Since the average value of cos^2 is 1/2, the transmission coefficient becomes I/I_o=1/2"
Next question
An upright image is reduced to 1/4 of the object’s height when the object is placed 26.9 cm from the lens. What is the focal length of the lens
Huh, I took a test that said differently... weird

Either way...
answer
the distance of the image is 26.9/4 = 6.725 cm Upright so it's virtual so distance is negative 1/f = 1/29.6 + 1/-6.725 = 1/29.6 - 4/29.6 = -3/29.6 So f = 29.6/-3 = - 9.86666... cm
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jonboyage
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Re: Optics B/C

Postby jonboyage » February 2nd, 2017, 7:20 pm

Are you sure?
Perfectly unpolarized light does lose half of its intensity when going through the first polarizer. According to Wikipedia, "A beam of unpolarized light can be thought of as containing a uniform mixture of linear polarizations at all possible angles. Since the average value of cos^2 is 1/2, the transmission coefficient becomes I/I_o=1/2"
Next question
An upright image is reduced to 1/4 of the object’s height when the object is placed 26.9 cm from the lens. What is the focal length of the lens
Huh, I took a test that said differently... weird

Either way...
answer
the distance of the image is 26.9/4 = 6.725 cm Upright so it's virtual so distance is negative 1/f = 1/29.6 + 1/-6.725 = 1/29.6 - 4/29.6 = -3/29.6 So f = 29.6/-3 = - 9.86666... cm
Whoops!
You accidentally turned 26.9 into 29.6 halfway through. The answer should be -8.966...cm. You did it the right way, though.
Your turn.
I was in a bin

Rustin '19
UPenn '23

kenniky
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Re: Optics B/C

Postby kenniky » February 5th, 2017, 12:54 pm

A telescope has a focal length of 1.25 m and an eyepiece diameter of 5.00 mm, what is the magnification?
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Avogadro
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Re: Optics B/C

Postby Avogadro » February 8th, 2017, 12:28 pm

A telescope has a focal length of 1.25 m and an eyepiece diameter of 5.00 mm, what is the magnification?
Seems pretty simple
250 times?
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

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Re: Optics B/C

Postby CVMSAvalacheStudent » February 8th, 2017, 4:47 pm

[quote="kenniky"]A telescope has a focal length of 1.25 m and an eyepiece diameter of 5.00 mm, what is the magnification?[/quote]
[hide]Why/idk[/hide]
"Great things in business are never done by one person. They're done by a team of people."-Steve Jobs

kenniky
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Re: Optics B/C

Postby kenniky » February 8th, 2017, 8:02 pm

A telescope has a focal length of 1.25 m and an eyepiece diameter of 5.00 mm, what is the magnification?
Seems pretty simple
250 times?
Correct, your turn
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Tom_MS
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Re: Optics B/C

Postby Tom_MS » February 9th, 2017, 8:56 am

A telescope has a focal length of 1.25 m and an eyepiece diameter of 5.00 mm, what is the magnification?
Seems pretty simple
250 times?
I'm used to the formula for magnification being the focal length of the objective divided by the focal length of the eyepiece. If you treat the eyepiece diameter as the focal length of the eyepiece, you get this result; however, the focal length should be independent of the diameter for the eyepiece. I looked at what this question may have originated from on wikipedia: https://en.wikipedia.org/wiki/Optical_t ... nification and I believe that the section where it has "(or diameter)" for the magnification formula means that it is the ratio of both diameters, not the focal length of one and the diameter of the other. I found the referenced site for that sentence on wikipedia, and I found this page:http://www.nexstarsite.com/_RAC/form.html. This page leads to this conclusion that I just explained. If you work out the math yourself, you will also see what I mean.

I saw this question on the MIT test, and I really don't think it makes sense.

kenniky
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Re: Optics B/C

Postby kenniky » February 9th, 2017, 9:00 am

I'm used to the formula for magnification being the focal length of the objective divided by the focal length of the eyepiece. If you treat the eyepiece diameter as the focal length of the eyepiece, you get this result; however, the focal length should be independent of the diameter for the eyepiece. I looked at what this question may have originated from on wikipedia: https://en.wikipedia.org/wiki/Optical_t ... nification and I believe that the section where it has "(or diameter)" for the magnification formula means that it is the ratio of both diameters, not the focal length of one and the diameter of the other. I found the referenced site for that sentence on wikipedia, and I found this page:http://www.nexstarsite.com/_RAC/form.html. This page leads to this conclusion that I just explained. If you work out the math yourself, you will also see what I mean.

I saw this question on the MIT test, and I really don't think it makes sense.
Interesting... I will admit my question was blatantly stolen from the MIT Test with different numbers lol

This will be helpful to know at least
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Avogadro
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Re: Optics B/C

Postby Avogadro » February 10th, 2017, 2:24 pm

Here's your question:
Image
The absorption spectra for chlorophyll a and chlorophyll b are shown above. How does this help to explain why leaves appear green?
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 10th, 2017, 3:45 pm

Here's your question:
Image
The absorption spectra for chlorophyll a and chlorophyll b are shown above. How does this help to explain why leaves appear green?
ANSWER
Almost none of the yellow and green light is absorbed. It is instead reflected and picked up by our eyes, so the leaves appear green.
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Avogadro
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Re: Optics B/C

Postby Avogadro » February 10th, 2017, 4:49 pm

Here's your question:
Image
The absorption spectra for chlorophyll a and chlorophyll b are shown above. How does this help to explain why leaves appear green?
ANSWER
Almost none of the yellow and green light is absorbed. It is instead reflected and picked up by our eyes, so the leaves appear 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

kenniky
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Re: Optics B/C

Postby kenniky » February 10th, 2017, 7:23 pm

What is the free-spectral range of a Fabry-Perot cavity in air, of length 31 cm, at a wavelength of 1105 nm?
also shamelessly stolen from the MIT test, mostly because I don't think their process is right
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Re: Optics B/C

Postby Avogadro » February 13th, 2017, 6:47 pm

What is the free-spectral range of a Fabry-Perot cavity in air, of length 31 cm, at a wavelength of 1105 nm?
also shamelessly stolen from the MIT test, mostly because I don't think their process is right
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.
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 13th, 2017, 7:00 pm

What is the free-spectral range of a Fabry-Perot cavity in air, of length 31 cm, at a wavelength of 1105 nm?
also shamelessly stolen from the MIT test, mostly because I don't think their process is right
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?
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Avogadro
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Re: Optics B/C

Postby Avogadro » February 23rd, 2017, 5:24 pm

What is the free-spectral range of a Fabry-Perot cavity in air, of length 31 cm, at a wavelength of 1105 nm?
also shamelessly stolen from the MIT test, mostly because I don't think their process is right
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?
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


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