Diffraction and the Wave Theory of Light — Question 8

Question 8

Gaussian illumination has reciprocal transverse and angular widths

An aperture-plane field is $A(u)=A_0e^{-u^2/(4\sigma^2)}$. Use $F(q)=\int A(u)e^{-iqu}du=2\sqrt\pi\sigma A_0e^{-\sigma^2q^2}$. Define widths as standard deviations of the aperture intensity and the far-field intensity in q.

1. Find both intensity variances and their width product.

2. Using $q\simeq k\theta$, calculate the angular standard deviation for $\sigma=0.10\,\mathrm{mm}$ and $\lambda=500\,\mathrm{nm}$.

Question 8 - Solution

Gaussian illumination has reciprocal transverse and angular widths

(a) Method and calculation

$$|A|^2\propto e^{-u^2/(2\sigma^2)}\Longrightarrow\boxed{\sigma_u=\sigma}.$$$$|F|^2\propto e^{-2\sigma^2q^2}\Longrightarrow\boxed{\sigma_q=1/(2\sigma),\quad\sigma_u\sigma_q=1/2}.$$

(b) Interpretation and further calculation

$$\sigma_\theta=\sigma_q/k=\lambda/(4\pi\sigma).$$$$\boxed{\sigma_\theta=3.97887\times10^{-4}\,\mathrm{rad}}.$$

Physical check

The width convention is explicit: a field-envelope width and an intensity standard deviation differ by numerical factors.