Gaussian beam

Total power

From the definitions (1) and (2): \[ I(r,z) = I_0 \left(\frac{w_0}{w(z)}\right)^2 \exp \left(\frac{-2r^2}{w(z)^2}\right) \tag{1}\] \[w(z) = w_0\sqrt{1 + \left(\frac{z}{z_R}\right)^2} \tag{2}\] we can see that the intensity at the beam waist (\(z=0\)) follows a Gaussian profile: \[ I(r,0) = I_0 \exp \left(\frac{-2r^2}{w_0^2}\right) \tag{3}\] We can find the total power \(P_0\) in the beam by integrating with the 'shell' method (Larson 2014): \[ P_0 = 2\pi \int_0^{\infty} rI(r,0) dr \tag{4}\] which we can write explicitly as: \[ P_0 = 2\pi I_0 \int_0^{\infty} r \exp \left(\frac{-2r^2}{w_0^2}\right) dr \tag{5}\] This could be tricky to solve but fortunately from (3) we see that: \[ \frac{d}{dr} I(r,0) = -\frac{4}{w_0^2} rI(r,0) \tag{6}\] which we can rewrite as: \[ \int rI(r,0) dr = -\frac{1}{4}w_0^2 I(r,0) \tag{7}\] So now we can evaluate (4) to a solvable form: \[ P_0 = 2\pi \left[ -\frac{1}{4}w_0^2 I(r,0) \right]_0^{\infty} \tag{8}\] Tidying up gives the final form: \[ P_0 = \frac{1}{2} \pi I_0 w_0^2 \tag{9}\] or alternativley: \[ I_0 = \frac{2P_0}{\pi w_0^2} \tag{10}\]