- #1
Daniel Gallimore
- 48
- 17
Just to clarify, this isn't homework. I took an Optics course a few years ago, but in the time between then and now, I've lost all my notes.
Suppose we have an infinitely tall slit of width [itex]a[/itex] and a parallel screen a distance [itex]r_0[/itex] away. Let the [itex]x[/itex] axis be parallel to the width of the slit, with the origin at the center of the slit (such that the slit extends from [itex]x=-a/2[/itex] to [itex]x=a/2[/itex]). The [itex]z[/itex] axis extends toward the screen from the origin. An angle [itex]\theta[/itex] locates a point [itex]P[/itex] on the screen (with respect to the [itex]z[/itex] axis) a distance [itex]x=X[/itex] along the screen. A ray drawn from the origin to [itex]P[/itex] is length [itex]r[/itex].
Given a plane wave incident on the slit, what is the formula for the irradiance of the diffracted wave as it encounters the screen?
I know the answer is similar (if not equal to) [tex]I=I_0\left(\frac{\sin(\beta/2)}{\beta/2}\right)^2[/tex] where [tex]\beta\equiv ka\sin\theta[/tex] I also know that [tex]E=-\frac{ikE_0}{2\pi r_0}\int_{-a/2}^{a/2} e^{ikr} \, dx[/tex] To get from the electric field to the irradiance is not a problem.
Any recommendations?
Suppose we have an infinitely tall slit of width [itex]a[/itex] and a parallel screen a distance [itex]r_0[/itex] away. Let the [itex]x[/itex] axis be parallel to the width of the slit, with the origin at the center of the slit (such that the slit extends from [itex]x=-a/2[/itex] to [itex]x=a/2[/itex]). The [itex]z[/itex] axis extends toward the screen from the origin. An angle [itex]\theta[/itex] locates a point [itex]P[/itex] on the screen (with respect to the [itex]z[/itex] axis) a distance [itex]x=X[/itex] along the screen. A ray drawn from the origin to [itex]P[/itex] is length [itex]r[/itex].
Given a plane wave incident on the slit, what is the formula for the irradiance of the diffracted wave as it encounters the screen?
I know the answer is similar (if not equal to) [tex]I=I_0\left(\frac{\sin(\beta/2)}{\beta/2}\right)^2[/tex] where [tex]\beta\equiv ka\sin\theta[/tex] I also know that [tex]E=-\frac{ikE_0}{2\pi r_0}\int_{-a/2}^{a/2} e^{ikr} \, dx[/tex] To get from the electric field to the irradiance is not a problem.
Any recommendations?