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we can solve for I in terms of X.
 
we can solve for I in terms of X.
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<math> I(x)
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<math> I(x)= I (\frac{\sin(\frac{KR}{2}\sin^{-1}(\frac{R}{L}))}{\frac{KR}{2}\sin^{-1}(\frac{R}{L})})^2</math>
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Since
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<math>\sin(\lambda) = 0</math>
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our zeroes are at
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<math> \frac{KR}{2}\sin^{-1}(\frac{R}{L}) = 0 </math>
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since this makes the sine function in the numerator zero.
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This means that
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<math> L = \frac{R}{2\tan(\sin^{-1}(\frac{\lambda}{R}))}</math>
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Unfortunately, this gives us absurd values for the necessary length of the optical path; for R to be plausibly small, L quickly approaches similar orders of magnitude. This approach is therefore invalid.
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If we use the double-slit approach, treating the diamond as the spacing between the slits, the slits become arbitrarily large. Under this approximation,
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== Color of the Laser ==
 
== Color of the Laser ==
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* Compensate for interference
 
* Compensate for interference
 
* Compensate for other sources of error
 
* Compensate for other sources of error
* Calculate expected error</math>
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* Calculate expected error
135

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