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- Thread starter CINA
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- #1

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Thanks

- #2

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If the screen is not infinite away from the slits but still far away, and assuming that the point on the screen that you want to calculate the intensity is above the top slit, then you'd still drop a perpendicular and calculate the excess path which would be d*sin(theta), but also in addition there'll be a term [d*cos(theta)]^2/(2r1), for a total difference in length of path:

[tex]dsin(\theta)+\frac{(dcos(\theta))^2}{2r_1} [/tex]

where d is the distance between slits, theta is the angle to the screen from the top slit, and r_1 is the distance to the point on the screen from the slits.

At least I think this is right. My geometry is not so good, as are my skills at keeping track what order approximations I'm using (I also have trouble with significant figures).

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