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Optical parametric oscillators have an optical resonant cavity. Why is this cavity so big when the wavelength of light is so small? What will happen if optical cavity length is shortened?

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  • $\begingroup$ I'm not an expert so I don't post an answer, but I think the reason comes from the fact that modulation of optical media is usually extremely weak, so we need a large path length to get an appreciable modulation of the light. In other words, for a given applied voltage, the change of index of refraction of interesting optical media is very tiny. $\endgroup$ – DanielSank Feb 26 '16 at 0:05
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There's no fundamental reason that drives the length. You can make an OPO with a short cavity and the free-spectral range (FSR) will be increased relative to a longer cavity. There are certain situations where you want the pump laser and the frequency of the OPO to be exactly matched, called resonant pumping. In such a case the cavity length of the OPO must be matched to the pump.

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Many applications of OPOs are in nonlinear optics, and so the laser is pulsed, so as to have the largest possible peak intensity for observation of nonlinear effects.

In that case, the cavity length needs to be cf/N, with c the speed of light, f the pulse repetition rate, and N a counting number. $N=1$, and therefore the longest cavity length, is the most straightforward configuration, because the higher the value of N, and the shorter the cavity, the more difficult it is to align the cavity length get two pulses to overlap.

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  • $\begingroup$ Is it high Q you mean to say? If I decrease length Q decreases? $\endgroup$ – user43794 Feb 26 '16 at 8:54
  • $\begingroup$ Q also plays a role (because more round trips and so more loss) but I meant something much more basic: it's more difficult to physically twiddle the knobs that control the cavity length to get the pulses to overlap! Since, for example, if N = 4, then a small adjustment will have a 4x higher effect, and it's harder to hit your target exactly. $\endgroup$ – ptomato Feb 27 '16 at 17:06

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