Calculating Interference Pattern Intensity for Double-Slit Experiment

AI Thread Summary
The discussion focuses on calculating the intensity of an interference pattern in a double-slit experiment using the correct equations. The primary equation mentioned is I=Imaxcos^2(πdsinθ/λ), but users are encouraged to simplify the process by avoiding the direct calculation of θ. Instead, they can use the path difference Δx, defined as Δx = d sinθ, and relate it to the y-coordinate of the fringe. Additionally, attention is drawn to the importance of unit conversions among mm, cm, and nm to ensure accurate calculations. The conversation emphasizes clarity in using the right formulas and understanding the relationships involved in the experiment.
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Two slits are separated by 0.180 mm. An interference pattern is formed on a screen 80.0 cm away by 656.3-nm light. Calculate the fraction of the maximum intensity 0.600 cm above the central maximum

I was using the equation I=Imaxcos^2(pie(d)sintheta/wavelength and I don't seem to be getting anywhere is there another equation because this one doesn't seem right and it's the only one in the chapter...
 
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You are indeed using the right equation. But you are probably getting stuck trying to find \theta. The intensity as a function of the phase \Delta\phi is:

I = I_{max}\cos^2\frac{\Delta\phi}{2}[/itex]<br /> <br /> Now, \Delta\phi = \frac{2\pi}{\lambda}\Delta x<br /> <br /> This should see you through...you don&#039;t need to compute \theta if you observe that<br /> <br /> d\sin\theta = Path difference = \Delta x<br /> <br /> Cheers<br /> Vivek
 
Looks like the correct equation except that you do not have a closing parenthesis. Make sure you have the units right. You have mm, cm, and nm in the problem. Are you making the necessary conversions?
 
OOps..I didn't see you have the y-coordinate of the fringe as well...in that case, for small theta,

\sin\theta = \frac{y}{D} and that should do it.
 
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