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

[MermaidWonders]

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)...

 

[topsquark]

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

 

[MermaidWonders]

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?

 

[topsquark]

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

 

[MermaidWonders]

Yeah, OK, makes sense. Thanks so much!