QB_PH_SEC6_008: Lasers and Quantum Optics

A He-Ne laser cavity has two plane mirrors separated by a distance of $L = 30\text{ cm}$. What is the frequency separation (in MHz) between two adjacent longitudinal cavity modes? (Take the speed of light $c = 3.0 \times 10^8\text{ m/s}$)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Boundary Condition**: A standing optical wave in a Fabry-Pérot cavity of length $L$ requires: $L = q \frac{\lambda_q}{2} = q \frac{c}{2\nu_q} \implies \nu_q = q \frac{c}{2L}$ where $q$ is an integer mode index. 2. **Adjacent Mode Separation**: $\Delta \nu = \nu_{q+1} - \nu_q = \frac{c}{2L}$ 3. **Numerical Computation**: - $L = 30\text{ cm} = 0.30\text{ m}$ - $c = 3.0 \times 10^8\text{ m/s}$ $\Delta \nu = \frac{3.0 \times 10^8\text{ m/s}}{2 \times 0.30\text{ m}} = \frac{3.0 \times 10^8}{0.60} = 5.0 \times 10^8\text{ Hz} = 500\text{ MHz}$ 4. **Official Integer Answer**: 500.