Calculating Interference Pattern Intensity for Double-Slit Experiment

[Honda47]

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...

 

[maverick280857]

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:

[tex]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't need to compute $\theta$ if you observe that<br /> <br /> $d\sin\theta$ = Path difference = $\Delta x$<br /> <br /> Cheers<br /> Vivek[/tex]

 

[OlderDan]

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?

 

[maverick280857]

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.