Optics B/C

Avogadro · Post by **Avogadro** » Fri Feb 10, 2017 4:49 pm

kenniky wrote:
Avogadro wrote:Here's your question:

The absorption spectra for chlorophyll a and chlorophyll b are shown above. How does this help to explain why leaves appear green?
ANSWER (Click to Show Hidden Text)
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.

kenniky · Post by **kenniky** » Fri Feb 10, 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

Avogadro · Post by **Avogadro** » Mon Feb 13, 2017 6:47 pm

kenniky wrote: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 (Click to Show Hidden Text)

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.

kenniky · Post by **kenniky** » Mon Feb 13, 2017 7:00 pm

Avogadro wrote:
kenniky wrote: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 (Click to Show Hidden Text)
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?

Avogadro · Post by **Avogadro** » Thu Feb 23, 2017 5:24 pm

kenniky wrote:
Avogadro wrote:
kenniky wrote: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 (Click to Show Hidden Text)
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?

kenniky · Post by **kenniky** » Thu Feb 23, 2017 5:51 pm

Avogadro wrote:
kenniky wrote:
Avogadro wrote:
Didn't even know Fabry-Perot was a thing until now (Click to Show Hidden Text)
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

Avogadro · Post by **Avogadro** » Thu Feb 23, 2017 6:18 pm

kenniky wrote:
Avogadro wrote:
kenniky wrote: 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?

kenniky · Post by **kenniky** » Fri Feb 24, 2017 7:23 am

Avogadro wrote:
kenniky wrote:
Avogadro wrote: 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 (Click to Show Hidden Text)
Got 0.00196 nm this time. Sounding better?

Yup, your turn

Avogadro · Post by **Avogadro** » Fri Feb 24, 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?

Sun Feb 26, 2017 11:34 am

Answer (Click to Show Hidden Text)

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

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