How Is the Third Dark Fringe Calculated in Diffraction Problems?

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SUMMARY

The calculation of the third dark fringe in diffraction problems involves using the formula sin(theta) = m(wavelength/width), where the wavelength is 668 nm and the slit width is 6.73 x 10^-6 m. The setup requires determining the angle theta using the tangent function, leading to the equation tan(theta) = x/1.85 m. The user calculated the distance x to be 0.55 m, while the book states the correct distance is 0.576 m, indicating a discrepancy that needs resolution.

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Homework Statement



Light that has a wavelength of 668 nm passes through a slit 6.73 x 10^-6 m wide and falls on a screen that is 1.85 m away. What is the distance on the screen from the center of the central bright fringe in the third dark fringe on either side?

Homework Equations



I will discuss this in the attempt at the solution; however, my attempt will based on sin (theta) = m(wavelength/ width)

The Attempt at a Solution



I have drawn a diagram of what the problem basically states. Anyways...I first tried to find the angle between the length (1.85 m) and the area where the fringe resides (I called it X). So...my setup for the angle looks like this tan (theta) = x/1.85.

Then, knowing that sin(theta) is similar (~) to tan (theta) I proceeded to find the answer by setting up x/1.85 = 3 (668x10^-9 m/ 6.73 x 10^-6 m) and found that x = 0.55 m.

^I used this equation: sin (theta) = m(wavelength/ width). Just in case there is any confusion!

I am confused with the answer discrepensy from my book and my approach...the book has the solution 0.576 m for the bright fringe. I am not sure whether my approach is wrong OR that the discrepensy is fine...Please let me know - even if there is a mathematical error (which I don't think is the case).

I appreciate all the help I can Get!
 
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