Scattering from a single slit

Consider a beam of amplitude A₀ normally incident on a single slit of width w.

The scattering density is described as follows:

ρ(x) = 1 when -w/2 ≤ x ≤ w/2
ρ(x) = 0 otherwise.

The equation for A(S) can be solved by setting the integration limits appropriately:

$${\rm{A(}}{\bf{S}}{\rm{)}} = {{\rm{A}}_0}\int\limits_{ - \infty }^\infty {\rho ({\rm{x}})\exp \left( {2\pi {\rm{i}}{{{\bf{x}} \cdot {\bf{S}}} \over \lambda }} \right){\rm{d}}x} = {{\rm{A}}_0}\int\limits_{ - w/2}^{w/2} {\exp \left( {2\pi {\rm{i}}{{{\bf{x}} \cdot {\bf{S}}} \over \lambda }} \right){\rm{d}}x} $$

We must evaluate x•S before proceeding.

So the scattered amplitude varies as a function of S (and hence as a function of 2θ) as follows:

$${\rm{A(}}{\bf{S}}{\rm{) = }}{{\rm{A}}_0}\int\limits_{ - w/2}^{w/2} {\exp \left( {2\pi {\rm{i}}{{{\bf{x}} \cdot {\bf{S}}} \over \lambda }} \right){\rm{d}}x} = {{\rm{A}}_0}\int\limits_{ - w/2}^{w/2} {\exp \left( {2\pi {\rm{i}}{{{\bf{x}}\sin 2\theta } \over \lambda }} \right){\rm{d}}x} $$ $${\rm{ = }}{{\rm{A}}_0}\left[ {{{\exp \left( {{{2\pi {\rm{i}}{\bf{x}}\sin 2\theta } \over \lambda }} \right)} \over {{{2\pi {\rm{i}}\sin 2\theta } \over \lambda }}}} \right]_{ - w/2}^{w/2}$$ $${\rm{ = }}{{\rm{A}}_0}\left( {{{\sin \left( {{{\pi w\sin 2\theta } \over \lambda }} \right)} \over {\left( {{{\pi \sin 2\theta } \over \lambda }} \right)}}} \right)$$

The intensity of the diffraction pattern is proportional to the square of the amplitude:

$${\rm{/(S) = (A(S)}}{{\rm{)}}^2}{\rm{ = }}{{\rm{A}}_0}^2\left( {{{{{\sin }^2}\left( {{{\pi w\sin 2\theta } \over \lambda }} \right)} \over {{{\left( {{{\pi \sin 2\theta } \over \lambda }} \right)}^2}}}} \right)$$

The form of this relationship is a sinc² function. The variation of I(S) with the variable (sin 2θ / λ) is shown below. The central maximum is much more intense than the diffracted maxima - this central maximum corresponds to the undiffracted beam.