Background. A laser can be mode-locked where an acousto-optic modulator is positioned inside the laser cavity. The modulator acts to couple (distribute energy) between cavity modes. The spectrum broadens as a result, to be limited only by (match that of) the laser gain bandwidth. Even more importantly, the cavity modes are locked to a common phase, so delivering a pulsed output. To do this, the acousto-optic frequency is tuned to precisely match the frequency separation of cavity modes (dfm or df as referred to in Th3.5) i.e., fmod = dfm, where dfm = 1/T, where T is the round trip time of the laser. Question. An acousto-optic modulator is inserted into a ring laser cavity, again with a laser wavelength of 580 nm. The modulator frequency is gradually increased until it reaches 1.458 GHz at which repetitive pulsing is seen from the cavity with a pulse-amplitude duration of tp = 10 ps (half-width at e¹ of amplitude). Consider the cavity to be ideally mode-locked. i) ii) iii) iv) Calculate the frequency separation of the cavity modes. Calculate the round-trip optical path-length of the cavity i.e., the path length for a flight field to return to the same plane going in the same direction. Calculate the mode-locked laser bandwidth (half-width at e¹ of amplitude). Give this in units of Hz and also in units of nm. Represent the mode-locked laser-amplitude spectrum with a sketch, labelling the cavity mode separation and the bandwidth with the values you calculated. Background 1. Convolution with a delta function: g(t) (tt) = g(t − to) G(w) (ww₁) = G(w - w₁) eq. 1 eq. 1' i.e., convolution of a function with a delta function reproduces the function positioned where the delta function was. Figure 1. Example || ω ω 0 0 2. Convolution F.T. Theorem: F[g(t) h(t)] = G(w)H(w) F[g(t) h(t)] = (1/2) G(w) ® H(w) where F[g(t)] = G(w) and F[h(t)] = H(w). This also applies for t→ x and w→ K. 3. The Fourier complement of a cosine function is two delta functions: F[cos (wot)] = n(8(w + w₁) + ε(w - w₁)) w 3 Wo eq. 2 eq. 2' It follows that for w₁ = 0: F[1] = 2π8(0) eq. 3 eq. 3'

Background. A laser can be mode-locked where an acousto-optic modulator is positioned inside the laser cavity. The modulator acts to couple (distribute energy) between cavity modes. The spectrum broadens as a result, to be limited only by (match that of) the laser gain bandwidth. Even more importantly, the cavity modes are locked to a common phase, so delivering a pulsed output. To do this, the acousto-optic frequency is tuned to precisely match the frequency separation of cavity modes (dfm or df as referred to in Th3.5) i.e., fmod = dfm, where dfm = 1/T, where T is the round trip time of the laser. Question. An acousto-optic modulator is inserted into a ring laser cavity, again with a laser wavelength of 580 nm. The modulator frequency is gradually increased until it reaches 1.458 GHz at which repetitive pulsing is seen from the cavity with a pulse-amplitude duration of tp = 10 ps (half-width at e¹ of amplitude). Consider the cavity to be ideally mode-locked. i) ii) iii) iv) Calculate the frequency separation of the cavity modes. Calculate the round-trip optical path-length of the cavity i.e., the path length for a flight field to return to the same plane going in the same direction. Calculate the mode-locked laser bandwidth (half-width at e¹ of amplitude). Give this in units of Hz and also in units of nm. Represent the mode-locked laser-amplitude spectrum with a sketch, labelling the cavity mode separation and the bandwidth with the values you calculated. Background 1. Convolution with a delta function: g(t) (tt) = g(t − to) G(w) (ww₁) = G(w - w₁) eq. 1 eq. 1' i.e., convolution of a function with a delta function reproduces the function positioned where the delta function was. Figure 1. Example || ω ω 0 0 2. Convolution F.T. Theorem: F[g(t) h(t)] = G(w)H(w) F[g(t) h(t)] = (1/2) G(w) ® H(w) where F[g(t)] = G(w) and F[h(t)] = H(w). This also applies for t→ x and w→ K. 3. The Fourier complement of a cosine function is two delta functions: F[cos (wot)] = n(8(w + w₁) + ε(w - w₁)) w 3 Wo eq. 2 eq. 2' It follows that for w₁ = 0: F[1] = 2π8(0) eq. 3 eq. 3'