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Calcoolius
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Homework Statement
Finding exact distances / angles for some maximum constructive / destructive interference 'bands' for a two-source interference
- fixed source-source distance, d
- screen parallel with source-source line segment
- screen fixed distance away from sources, L
- wavelength difference for the antinode, λ (let us consider for now only whole wavelength differences i.e. antinodal lines) - the antinode considered lies on the screen (where we see the interference pattern).
- aim is to find the distance between maximum interference and central antinodal line - on the screen.
Homework Equations
Note: please do not try to bring in the good old interference equations (which ARE really approximations, albeit good ones when used for the SUPPOSED small angles and large distance L). It defeats the purpose.
Let the antinode's distances to each source be (x) and (x + λ) ...considering only a single wavelength difference for now...
Let the distance across the screen from the antinode to the central antinodal line be p
So we can form equations using Pythagoras:
x2 = L2 + (0.5d - p)2
(x + λ)2 = L2 + (0.5d + p)2
so that
[L2 + (0.5d - p)2]0.5 = [L2 + (0.5d + p)2]0.5 - λ
The Attempt at a Solution
I manipulated tediously getting:
p2 = λ2(4L2 + d2 - λ2)/(4d2 - 4λ2)
so p is found taking the square-root of this beast(?)
This gives p (for varying L, d and some wavelength difference) with distances to which
- the 'approximations' (i.e. the answers given using the good old equations) are decently close ... though not very satisfyingly much
- If numbers are plugged in, for example, to check, you may confirm the relationship between the wavelength difference between sources, λ, and the distance p from the central antinodal IS NOT REALLY DIRECTLY PROPORTIONAL. This is to mean ... that in the gaps / widths (or however you consider the interference pattern) do not have uniform distances, i.e. not quite evenly spread! How can that be? did I do something wrong?