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GR9277 #96
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Alternate Solutions |
faham
2006-10-26 22:33:35 | Let:
* N be the number of full wavelengths across the gas-filled cell
* N' be the number of full wavelengths across the cell when it's empty
* lambda_0 is the wavelength of light in vaccuum
* lambda is the wavelength of light in a material of index of refraction n
* d is the length of the cell (5 cm)
* M is the number of fringes seen on the screen
N = d/lambda_0
N' = d/lambda
Now, we know that lambda = lambda_0/n, so
N' = d*n/lambda_0
Be aware that the light goes IN and back OUT of the cell (it's an interferometer!), so
N' - N = M/2
Again, even though you see 40 fringes, it's 20 wavelengths that N gets shifted by.
N' - N = M/2 = d*n/lambda_0 - d/lambda_0
M/2 = d/lambda_0*(n-1)
Solving for n:
n = M*lambda_0/(2*d) + 1
n = (40)*(500nm)/(2*5cm) + 1
n = 1.0002
So the answer is C |  |
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Comments |
Imperate
2008-10-17 07:03:31 | The key here is the optical path difference, given that the mirrors dont move etc everything else is constant except the bit of path going through the gas chamber. The optical path length of this section is 0.1n [metres]. For a maxima the total path difference of the total rays must be a integer multiple of the wavelength  .Therefore one obtains the expression  . Multiply both sides by 10, and one can see that if the optical n changes by 10  a new fringe will appear. Given that 40 fringes went past this implies  .
The refractive index of air is 1 therefore  . Choice C |  | dumbguy
2007-10-10 18:01:07 | What happened to the nanometers? Why did we choose to just leave them out? | | faham
2006-10-26 22:33:35 | Let:
* N be the number of full wavelengths across the gas-filled cell
* N' be the number of full wavelengths across the cell when it's empty
* lambda_0 is the wavelength of light in vaccuum
* lambda is the wavelength of light in a material of index of refraction n
* d is the length of the cell (5 cm)
* M is the number of fringes seen on the screen
N = d/lambda_0
N' = d/lambda
Now, we know that lambda = lambda_0/n, so
N' = d*n/lambda_0
Be aware that the light goes IN and back OUT of the cell (it's an interferometer!), so
N' - N = M/2
Again, even though you see 40 fringes, it's 20 wavelengths that N gets shifted by.
N' - N = M/2 = d*n/lambda_0 - d/lambda_0
M/2 = d/lambda_0*(n-1)
Solving for n:
n = M*lambda_0/(2*d) + 1
n = (40)*(500nm)/(2*5cm) + 1
n = 1.0002
So the answer is C
caffeinated
2008-04-08 12:56:36 |
Why do you use lambda for N' when N' is the empty cell and lambda is lambda in the gas? I don't understand. Why wouldn't you use lambda-0 for N'?
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| | astro_allison
2005-11-26 22:02:48 |  ?
astro_allison
2005-11-26 22:17:25 |
wouldn't  ?
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yosun
2005-11-26 23:06:44 |
astro_allison: thanks for the typo-alerts; both typos have been corrected.
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