Question

(b) The gain coefficient is given by $\alpha = \frac{\alpha_0}{1 + I/I_s}$ where $\alpha_0$ is the small-signal gain and $I_s$ is the saturation intensity. Show that the expression for the amplification of the beam can now be solved to yield: $\alpha_0 z = ln\left[\frac{I(z)}{I(0)}\right] + \frac{I(z) - I(0)}{I_s}$

          (b) The gain coefficient is given by
$\alpha = \frac{\alpha_0}{1 + I/I_s}$
where $\alpha_0$ is the small-signal gain and $I_s$ is the saturation intensity. Show that the expression for the
amplification of the beam can now be solved to yield:
$\alpha_0 z = ln\left[\frac{I(z)}{I(0)}\right] + \frac{I(z) - I(0)}{I_s}$

(b) The gain coefficient is given by
α = (α0)/(1 + I/Is)
where α0 is the small-signal gain and Is is the saturation intensity. Show that the expression for the
amplification of the beam can now be solved to yield:
α0 z = ln[(I(z))/(I(0))] + (I(z) - I(0))/(Is)

Added by Julia W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

11/21/2023

Step 1

Step 1: Start with the expression for the gain coefficient α = (α₀) / (1 + (I) / (Iₛ)) Show more…

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(b) The gain coefficient is given by α = (α₀) / (1 + (I) / (Iₛ)) where α₀ is the small-signal gain and Iₛ is the saturation intensity. Show that the expression for the amplification of the beam can now be solved to yield: α₀z = ln[(I(z)) / (I(0))] + (I(z) - I(0)) / (Iₛ)