Problem Statement
A plane wave ($\lambda = 500$ nm) falls on a circular aperture. The aperture is adjusted to expose exactly $m = 3$ Fresnel half-period zones to a point $P$ at distance $b = 1.0$ m. Find the aperture radius.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Radius of the $m$th Fresnel zone for a plane wave:
Step 2 — Apply the relevant physical law or equation: $$r_m = \sqrt{m\lambda b} = \sqrt{3\times500\times10^{-9}\times1.0} = \sqrt{1.5\times10^{-6}} = 1.22\times10^{-3}\text{ m}$$
$$\boxed{r_3 \approx 1.22\text{ mm}}$$
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$r_m = \sqrt{m\lambda b} = \sqrt{3\times500\times10^{-9}\times1.0} = \sqrt{1.5\times10^{-6}} = 1.22\times10^{-3}\text{ m}$$
$$\boxed{r_3 \approx 1.22\text{ mm}}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
Radius of the $m$th Fresnel zone for a plane wave:
$$r_m = \sqrt{m\lambda b} = \sqrt{3\times500\times10^{-9}\times1.0} = \sqrt{1.5\times10^{-6}} = 1.22\times10^{-3}\text{ m}$$
$$\boxed{r_3 \approx 1.22\text{ mm}}$$
Answer
$$\boxed{r_3 \approx 1.22\text{ mm}}$$
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.
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