Ring lasers can emit bandwidths of less than 1 kHz in the visible region of the spectrum (mid-1014 Hz). This constitutes the nearest we have achieved to a delta function light-spectrum. This results from there being no standing waves for this type of cavity. 1. A ring laser with a low transmission (highly reflective) output coupler gives an output field of Ein cos (wit), a pure cosine wave of constant amplitude, centred in the visible. An acousto-optic modulator is placed in the output laser beam, external to the laser. This device provides a sinusoidal modulation in transmission amplitude, h(t) = (1 - A) + Acos (@modt) where A << 1, at a frequency Wmod = 2πfmod (typically a few GHz). This modulates the laser beam amplitude in time and as a consequence it modifies the spectrum too, creating new frequencies as so-called side bands. i) ii) iii) iv) v) Illustrate h(t) with a sketched graph, clearly showing the modulation and labelling the magnitudes of features. Determine and simplify an expression for F[h(t)] = H(w). Include the amplitude scaling of features. Determine and simplify an expression for F[Ein(t)] = Ein(w). Include the amplitude scaling of features. Using the convolution theorem, give an expression for the transmitted field, Eout(w) written as a convolution of two functions. Use the symbol for your answer. Illustrate Ein (w) and H (w) with two aligned sketched graphs (one above the other), clearly labelling the positions and magnitudes of all features. Aligned below these two graphs, sketch Eout(w), clearly labelling the positions and magnitudes. You don't need to write integrals to do this, rather you may pull on point 1 of the background and your figures to determine and explain Eout(w) visually. For A = 0.05, fmod = 2.000 GHz and for a laser wavelength of 580 nm (precisely), calculate the intensity of the main spectral peak at 580 nm relative to that of the side-bands and calculate the frequency separation of either side-band from the main peak. 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'

Ring lasers can emit bandwidths of less than 1 kHz in the visible region of the spectrum (mid-1014 Hz). This constitutes the nearest we have achieved to a delta function light-spectrum. This results from there being no standing waves for this type of cavity. 1. A ring laser with a low transmission (highly reflective) output coupler gives an output field of Ein cos (wit), a pure cosine wave of constant amplitude, centred in the visible. An acousto-optic modulator is placed in the output laser beam, external to the laser. This device provides a sinusoidal modulation in transmission amplitude, h(t) = (1 - A) + Acos (@modt) where A << 1, at a frequency Wmod = 2πfmod (typically a few GHz). This modulates the laser beam amplitude in time and as a consequence it modifies the spectrum too, creating new frequencies as so-called side bands. i) ii) iii) iv) v) Illustrate h(t) with a sketched graph, clearly showing the modulation and labelling the magnitudes of features. Determine and simplify an expression for F[h(t)] = H(w). Include the amplitude scaling of features. Determine and simplify an expression for F[Ein(t)] = Ein(w). Include the amplitude scaling of features. Using the convolution theorem, give an expression for the transmitted field, Eout(w) written as a convolution of two functions. Use the symbol for your answer. Illustrate Ein (w) and H (w) with two aligned sketched graphs (one above the other), clearly labelling the positions and magnitudes of all features. Aligned below these two graphs, sketch Eout(w), clearly labelling the positions and magnitudes. You don't need to write integrals to do this, rather you may pull on point 1 of the background and your figures to determine and explain Eout(w) visually. For A = 0.05, fmod = 2.000 GHz and for a laser wavelength of 580 nm (precisely), calculate the intensity of the main spectral peak at 580 nm relative to that of the side-bands and calculate the frequency separation of either side-band from the main peak. 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'