Diffraction and the Wave Theory of Light — Question 5

Question 5

Fresnel similarity requires distance to scale with aperture size squared

For one-dimensional paraxial propagation use $U(x,L)=C(L)\int_{-a/2}^{a/2}A(u)\exp[i\pi(x-u)^2/(\lambda L)]du$. Compare geometrically similar normalized aperture profiles after scaling all transverse coordinates by s, at fixed wavelength.

1. Find how L must scale to leave the phase kernel unchanged.

2. If the slit width grows from $0.20$ to $0.60\,\mathrm{mm}$, find the new propagation distance and feature coordinate corresponding to $L=0.40\,\mathrm m$ and $x=0.80\,\mathrm{mm}$.

Question 5 - Solution

Fresnel similarity requires distance to scale with aperture size squared

(a) Method and calculation

$$u^\prime=su,\quad x^\prime=sx:\quad\frac{(x^\prime-u^\prime)^2}{\lambda L^\prime}=\frac{s^2(x-u)^2}{\lambda L^\prime}.$$$$\boxed{L^\prime=s^2L}\Longrightarrow\text{same normalized Fresnel pattern}.$$

(b) Interpretation and further calculation

$$s=0.60/0.20=3\Longrightarrow\boxed{L^\prime=3.60\,\mathrm m}.$$$$\boxed{x^\prime=sx=2.40\,\mathrm{mm}}.$$

Physical check

This is a near-field similarity transformation. Scaling distance linearly with aperture width would not preserve the quadratic phase.