 Solution

The incident intensity radius is $w_L=2.5\mathrm{mm}$, half the quoted diameter. For an ideal collimated Gaussian beam focused by a thin lens, the diffraction-limited waist isThe incident Rayleigh range $\pi w_L^2/{\lambda }_0\simeq 24.5\mathrm{m}$ is much larger than $f$, justifying this approximation. The pulse energy is $U=P_{\mathrm{av}}/R=0.25\mu \mathrm{J}$. Taking its duration as an effective energy-to-peak-power width gives $P_{\mathrm{pk}}\simeq U/\tau =7.14\times 10^5\mathrm{W}$. Since $I(r)=I_0{e}^{-2r^2/w_0^2}$ has integrated power $I_0\pi w_0^2/2$,If $350\mathrm{fs}$ denotes the temporal intensity full width at half maximum of a Gaussian pulse, the time integral instead gives $P_{\mathrm{pk}}=2\sqrt{\ln 2}U/(\sqrt{\pi }\tau )$, and $I_{\mathrm{pk}}=1.03\times 10^{13}\mathrm{W}\mathrm{c}{\mathrm{m}}^{-2}$. The pulse shape is unspecified, so the estimate's order-one convention is explicit. Aberration, clipping and propagation through glass were excluded from this calculation as stipulated.