Problem 5.49 — Fabry-Perot Etalon Free Spectral Range

Problem Statement

A Fabry-Perot etalon has plate separation $d = 5.0$ mm. Find the free spectral range near $\lambda = 500$ nm.

Given Information

• See problem statement for all given quantities.

Physical Concepts & Formulas

Projectile motion decomposes into independent horizontal and vertical components. Horizontal: constant velocity (no air resistance). Vertical: constant downward acceleration $g$. The trajectory is a parabola. Maximum range occurs at $45°$ launch angle; max height at $90°$.

• $x = v_0\cos\theta \cdot t$, $y = v_0\sin\theta \cdot t – \frac{1}{2}gt^2$
• $R = v_0^2\sin 2\theta/g$ — horizontal range
• $H = v_0^2\sin^2\theta/(2g)$ — maximum height
• $T = 2v_0\sin\theta/g$ — total flight time

Step-by-Step Solution

Step 1 — Identify given quantities and set up the problem: $$\delta\lambda_{FSR} = \frac{\lambda^2}{2d} = \frac{(500\times10^{-9})^2}{2\times5.0\times10^{-3}} = \frac{2.5\times10^{-13}}{10^{-2}} = 2.5\times10^{-11}\text{ m} = \boxed{0.025\text{ nm}}$$

Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Worked Calculation

$$\delta\lambda_{FSR} = \frac{\lambda^2}{2d} = \frac{(500\times10^{-9})^2}{2\times5.0\times10^{-3}} = \frac{2.5\times10^{-13}}{10^{-2}} = 2.5\times10^{-11}\text{ m} = \boxed{0.025\text{ nm}}$$

$$\text{Numerical result} = \text{given expression substituted with values}$$

$$\boxed{\delta\lambda_{FSR} = \frac{\lambda^2}{2d} = \frac{(500\times10^{-9})^2}{2\times5.0\times10^{-3}} = \frac{2.5\times10^{-13}}{10^{-2}} = 2.5\times10^{-11}\text{ m} = \boxed{0.025\text{ nm}}}$$

$$\delta\lambda_{FSR} = \frac{\lambda^2}{2d} = \frac{(500\times10^{-9})^2}{2\times5.0\times10^{-3}} = \frac{2.5\times10^{-13}}{10^{-2}} = 2.5\times10^{-11}\text{ m} = \boxed{0.025\text{ nm}}$$

Answer

$$\delta\lambda_{FSR} = \frac{\lambda^2}{2d} = \frac{(500\times10^{-9})^2}{2\times5.0\times10^{-3}} = \frac{2.5\times10^{-13}}{10^{-2}} = 2.5\times10^{-11}\text{ m} = \boxed{0.025\text{ nm}}$$

Physical Interpretation

The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.