Light: interference - two slit

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SUMMARY

The discussion focuses on calculating the position of minimum light intensity due to destructive interference from two isotropic point sources, S1 and S2, separated by 2.7 micrometers and emitting light at a wavelength of 900 nm. The relevant equation for determining the position of interference minima is Y = (λL (m+½)) / d, where Y is the position of the minima, L is the distance from the sources to the detector, d is the separation of the sources, and m is the mode. The maximum value of Xp for destructive interference is determined to be 7.88 x 10^-6 m.

PREREQUISITES
  • Understanding of wave interference principles
  • Familiarity with the two-slit interference equation
  • Knowledge of phase difference calculations
  • Basic geometry of light propagation in a two-dimensional plane
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  • Learn about the significance of path length differences in wave interference
  • Investigate the impact of varying wavelengths on interference patterns
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Students studying optics, physics educators, and anyone interested in understanding wave interference phenomena in light. This discussion is particularly beneficial for those tackling problems related to interference patterns in experimental setups.

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Homework Statement



Two isotropic point sources of light (s1 and s2) are separated by 2.7micrometers along a y-axis and emit in phase at wavelength 900 nm and at the same amplitude. A point detector is located at point P at coordinate Xp on the x-axis (s1 is also on the x axis, at (0,0).) What is the greatest value of Xp at which the detected light is minimum due to destructive interference?

I don't know how to make a diagram for this on the computer :blushing: but say there is an x,y plane: S1 is on the origin, P is at distance Xp to the right of S1 (on the X axis) and S2 is at 2.7um from S2 (lower, on the Y axis.)


Homework Equations


Y = (λL (m+½)) / d
Where Y is the position of the interference minima
L is the distance Xp
D is the slit space (2.7 um)
and λ is the wavelength.
m= mode.

1.[itex]d\sin\theta=m\lambda[/itex] for constructive interference
2.[itex]d\sin\theta=(m+\frac{1}{2})\lambda[/itex] for destructive interference
3.[itex]\phi=\frac{2\pi}{\lambda}(r_2-r_1)[/itex] phase difference related to path difference


The Attempt at a Solution



I figure since the center of interference is always between the two slits (or in this case point sources), y = 1/2 of 2.7um, i.e. 1.35 micrometers, and thus

Y = (λL (m+½)) / d
1.35 x 10 ^- 6 = [ (900 * 10^-9) * L (m+½) / 2.7 * 10 ^ 0.6 ]
or
4.05 x 10 ^ -6 = L (m+½).

And I'm stuck there, I don't know if I'm to suppose that m is 0 seeing as how that would give the biggest L.

The answer is 7.88 x 10^-6 m, but I can't get it...
 
Last edited:
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Rather than try to force fit the formula for two-slit interference to your problem, why not attack the problem directly? Imagine your detector on the x-axis, starting at the origin and moving outward. Now consider the path length difference between the light reaching the detector from both sources. As you move the dectector further along the x-axis, does the path length difference--and thus the phase difference--increase or decrease? What's the minimum path length difference that will lead to destructive interference? Write an expression for the path difference as a function of x, and use the insights from answering the above questions to solve for the maximum value of x.
 
Thank you, kind sir!
 
Last edited:

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