Intensity in Single-Slit Diffraction, Quantitatively — Question 1

Question 1

Integrating the aperture field yields the normalized sinc-squared law

A slit of width a has constant field amplitude $A_0$. In the far field $F(q)=\int_{-a/2}^{a/2}A_0e^{-iqu}du$, where $q=(2\pi/\lambda)\sin\theta$. Define $\beta=qa/2$.

1. Evaluate the integral and normalize the field to its axial value.

2. Derive the intensity formula and evaluate its finite limit on axis.

Question 1 - Solution

Integrating the aperture field yields the normalized sinc-squared law

(a) Method and calculation

$$F(q)=A_0\left[\frac{e^{-iqu}}{-iq}\right]_{-a/2}^{a/2}=\frac{2A_0\sin(qa/2)}q.$$$$\boxed{F(q)=A_0a\frac{\sin\beta}{\beta},\quad F(0)=A_0a}.$$

(b) Interpretation and further calculation

$$\boxed{I(\theta)=I_c\left(\frac{\sin\beta}{\beta}\right)^2,\quad\beta=\frac{\pi a\sin\theta}{\lambda}}.$$$$\lim_{\beta\to0}\sin\beta/\beta=1\Longrightarrow\boxed{I(0)=I_c}.$$

Physical check

The apparent zero-over-zero form is removable. Intensity normalization must use the total axial amplitude, not the amplitude of one aperture element.