Lasers and Laser Light — Question 3

Question 3

A Gaussian laser beam has a finite Rayleigh range and divergence

For an ideal Gaussian beam in vacuum use $z_R=\pi w_0^2/\lambda$, $w(z)=w_0\sqrt{1+(z/z_R)^2}$, and far-field half-angle $\theta=\lambda/(\pi w_0)$. The intensity is proportional to $e^{-2r^2/w^2}$. Let $w_0=.400\,\mathrm{mm}$ and $\lambda=800\,\mathrm{nm}$.

1. Find $z_R$ and $\theta$.

2. Find the radius at $z=2z_R$ and the on-axis intensity relative to its waist value for fixed power.

Question 3 - Solution

A Gaussian laser beam has a finite Rayleigh range and divergence

(a) Method and calculation

$$\boxed{z_R=\pi(.400\times10^{-3})^2/(800\times10^{-9})=.6283\,\mathrm m}.$$$$\boxed{\theta=800\times10^{-9}/[\pi(.400\times10^{-3})]=.6366\,\mathrm{mrad}}.$$

(b) Interpretation and further calculation

$$\boxed{w(2z_R)=\sqrt5w_0=.8944\,\mathrm{mm}}.$$$$I(0,z)\propto w(z)^{-2}\Rightarrow\boxed{I(0,2z_R)/I(0,0)=1/5}.$$

Physical check

The radius is the intensity 1/e-squared radius, and the divergence is a half-angle. Consistent definitions avoid factors of two in beam-size calculations.