激光原理第四章ppt課件-文庫吧資料
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【正文】 02 ( )m H qT g l a a???最佳輸出功率 021 ( ) ( 2 ( ) )2m s q H qP A I g l a????深圳大學(xué)電子科學(xué)與技術(shù)學(xué)院 ? Maximizing P with respect to T by setting yields as the condition for the mirror transmission that yields the maximum power output. 0?dTdPalagT mm ?? 2深圳大學(xué)電子科學(xué)與技術(shù)學(xué)院 （ 1） ?q≠?0 （多普勒非均勻加寬） I+和 I兩束光在增益曲線上分別燒兩個(gè)孔，對每一個(gè)孔起飽和作用的分別是 I+或 I，而不是二者的和。 (b) the (useful) loss due to coupling of output power through the partially transmissive reflector. It is obvious that loss (a) should be made as small as possible since it raise the oscillation threshold without contributing to the output power. 深圳大學(xué)電子科學(xué)與技術(shù)學(xué)院 The problem of the coupling loss (b), however, is more subtle. At zero coupling (that is, both mirrors have zero transmission) the threshold will be at its minimum value and the power emitted by the atoms will be maximum. But since none of this power is available as output, this is not a useful state of affairs. If on the other hand, we keep increasing the coupling loss, the increasing threshold pumping will at some point exceed the actual pumping level. When this happens, oscillation will cease and the power output will again be zero. Between these two extremes there exists an optimum value of coupling (that is, mirror transmission) at which the power output is a maximum. 深圳大學(xué)電子科學(xué)與技術(shù)學(xué)院 3）最佳透射率及功率 輸出功率和反射鏡的透射率 T有關(guān)。上式改寫為 aT ???202 ( )1( ) 12HqsqglP AT IaT?????? ?????深圳大學(xué)
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