Problem 5.215 — Optical Pumping and Laser Threshold

Problem Statement

Solve the optics problem: A 4-level laser medium has stimulated emission cross section $\sigma = 3\times10^{-23}$ m² and upper state lifetime $\tau = 1\;\mu$s. The cavity has length $L = 0.3$ m, output coupler $R = 0.95$, and internal loss $\alpha_i = 0.1$ m$^{-1}$. Find the threshold population inversion density. Threshold

Given Information

  • Refractive index $n$ or focal length $f$ as given
  • Object distance $u$ (negative for real objects in Cartesian convention)
  • Radius of curvature $R$ or lens/mirror parameters as given

Physical Concepts & Formulas

Geometric optics is governed by Snell’s Law ($n_1 \sin\theta_1 = n_2 \sin\theta_2$) at each interface and the mirror/lens formulas in the paraxial limit. The Cartesian sign convention assigns the incident direction as positive: distances measured opposite to light are negative. For mirrors: $1/v + 1/u = 2/R = 1/f$. For thin lenses: $1/v – 1/u = 1/f$. Magnification $m = -v/u$ for mirrors and $m = v/u$ for lenses. A real image has $v > 0$ for lenses; a virtual image has $v < 0$.

  • $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ — mirror formula
  • $\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f}$ — thin lens formula (Cartesian)
  • $n_1 \sin\theta_1 = n_2 \sin\theta_2$ — Snell’s Law
  • $m = -v/u$ (mirror) or $m = v/u$ (lens) — linear magnification

Step-by-Step Solution

Step 1 — Apply correct sign convention: Real object: $u < 0$. Concave mirror/converging lens: $f > 0$.

Step 2 — Use the appropriate formula:

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\quad\text{(mirror)} \quad\text{or}\quad \frac{1}{v} = \frac{1}{f} + \frac{1}{u}\quad\text{(lens with Cartesian)}$$

Step 3 — Solve for image distance $v$ and compute magnification.

Worked Calculation

Substituting all values with units:

Substitute given values of $u$, $f$ into the formula and solve for $v$.

Answer

$$\boxed{\frac{1}{v} – \frac{1}{u} = \frac{1}{f}}$$

Physical Interpretation

A positive $v$ means the image forms on the real side (same side as outgoing light for lenses). Magnification $|m| > 1$ means the image is enlarged; $|m| < 1$ means diminished. A negative $m$ indicates an inverted image.


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