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My textbook only considers two-dimensional scattering, so I will stick to that. When explaining Bragg's law, it states that the incidence angle and the scattering angle must be equal so that all trajectories from different atoms (from the same row) to a point $P$ in a screen are the same. Then, it says that between different rows separated by a distance $d$, the equation $2d\sin\theta=m\lambda$ must also be satisficed for constructive interference to occur.

What I do not understand is why do the incidence and scattering angles must be equal? Isn't it possible for $\theta_i$ and $\theta_s$ to be different, and account for the difference in trajectories so that it differs in an integer number of wavelenghts? It will end up as a system of equations somehow a bit more complicated than the expression from above, but I find it completely possible.

TheScientist
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If you have a uniformly spaced linear array of $M$ identical omni-directional emitters such that they have equal amplitudes say $1$ but the phases are in an arithmetic series, that is the $m^{th}$ has phase $m\alpha$ then the sum of all at a large distance such that $R>>Md$ and in direction $\theta$, will be ($\kappa = 2\pi d/\lambda$ and $d$ is the spacing): $$A(\theta) \propto e^{-\mathfrak j \kappa R} \sum_{m=0}^{M-1} e^{\mathfrak j m(\kappa d\rm{ cos}(\theta) - \alpha)}.$$ This is a geometric series that sums to $$A(\theta) \propto e^{-\mathfrak j \kappa R} \frac {e^{\mathfrak j M(\kappa d\rm{ cos}(\theta) - \alpha)}-1}{e^{\mathfrak j (\kappa d\rm{ cos}(\theta) - \alpha)}-1}.$$ This obviously has the peak in the direction $\hat \theta $ where $\kappa d\rm{cos}(\hat \theta) = \alpha$ in which case they all add up coherently resulting in $|A(\hat \theta )| \propto M$ Now from symmetry consideration this would be the same angle from the other side exciting the array with a plane wave resulting the same phase steps exciting the individual emitters in an arithmetic series.

hyportnex
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