MHB Solving for Slit Width-Wavelength Ratio at $\pm90°$ Diffraction Pattern

AI Thread Summary
The discussion centers on determining the ratio of slit width to wavelength for the first minima of a single-slit diffraction pattern at ±90°. The formula used is sin(θ) = nλ/a, with n set to 1 for the first minima. There is a debate about whether to use -90° for sin(θ), which would yield a negative ratio, but it is clarified that the sign can be associated with n, where positive n indicates minima to the right and negative n to the left of the central maximum. Ultimately, it is suggested to take the absolute value of the ratio, as negative values lack meaning in this context. The conversation concludes with an affirmation of understanding regarding the interpretation of the angles.
MermaidWonders
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For what ratio of slit width to wavelength will the first minima of a single-slit diffraction pattern occur at $\pm 90°$?

The thing is, when I did it, I used the formula $sin\theta = \frac{n\lambda}{a}$, and used the fact that $m = 1$ and $\pm 90°$ to solve for $\frac{a}{\lambda}$. However, I don't know if we're supposed to plug in $-90°$ for $sin\theta$, because that would mean that our ratio of $\frac{n\lambda}{a}$ would be $-1$ as opposed to just $1$ (when $sin90°$ was plugged in)...
 
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MermaidWonders said:
For what ratio of slit width to wavelength will the first minima of a single-slit diffraction pattern occur at $\pm 90°$?

The thing is, when I did it, I used the formula $sin\theta = \frac{n\lambda}{a}$, and used the fact that $m = 1$ and $\pm 90°$ to solve for $\frac{a}{\lambda}$. However, I don't know if we're supposed to plug in $-90°$ for $sin\theta$, because that would mean that our ratio of $\frac{n\lambda}{a}$ would be $-1$ as opposed to just $1$ (when $sin90°$ was plugged in)...
More or less we can take the sign of [math]sin( \theta )[/math] to be "attached" to the n. A positive n describes the nth minima to the right of the central maximum and a negative n describes the nth minima to the left of the central maximum.

-Dan
 
topsquark said:
More or less we can take the sign of [math]sin( \theta )[/math] to be "attached" to the n. A positive n describes the nth minima to the right of the central maximum and a negative n describes the nth minima to the left of the central maximum.

-Dan

OK, makes sense. So should I take the absolute value, since a negative ratio wouldn't be very meaningful in this context?
 
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MermaidWonders said:
OK, makes sense. So should I take the absolute value, since a negative ratio wouldn't be very meaningful in this context?
Yup.

Double check the angle on the |n| = 1 interpretation I gave you. I might have screwed up left and right. But in the long run, no, it doesn't really matter.

-Dan
 
Yeah, OK, makes sense. Thanks so much!
 
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