Diffraction — Question 2

Question 2

The quadratic path correction tests a far-field approximation

A slit extends from $u=-a/2$ to $u=a/2$ and is observed on axis at distance L. The exact path is $r(u)=\sqrt{L^2+u^2}$. Let $a=1.0\,\mathrm{mm}$, $\lambda=500\,\mathrm{nm}$. Require the largest neglected quadratic phase to be at most $\pi/10$.

1. Expand r through second order and express the maximum quadratic phase in terms of a, L, and wavelength.

2. Find the minimum L satisfying the stated phase tolerance.

Question 2 - Solution

The quadratic path correction tests a far-field approximation

(a) Method and calculation

$$r(u)=L\sqrt{1+u^2/L^2}\simeq L+\frac{u^2}{2L}.$$$$\boxed{\Delta\phi_{\max}=\frac{2\pi}{\lambda}\frac{(a/2)^2}{2L}=\frac{\pi a^2}{4\lambda L}}.$$

(b) Interpretation and further calculation

$$\frac{\pi a^2}{4\lambda L}\leq\frac{\pi}{10}\Longrightarrow L\geq\frac{5a^2}{2\lambda}.$$$$\boxed{L_{\min}=5.0\,\mathrm m};\qquad a/L_{\min}=2.0\times10^{-4}.$$

Physical check

The small geometrical ratio a/L alone does not guarantee a negligible phase error, because optical wavelengths are very short.