Problem 5.36 — Phase Difference in Double-Slit Experiment

Problem Statement

In Young’s experiment, what is the phase difference between two interfering waves at a point on the screen that is 2 mm from the central maximum? Given: $d = 0.4$ mm, $L = 1.0$ m, $\lambda = 600$ nm.

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: Path difference at distance $y$ from centre:

Step 2 — Apply the relevant physical law or equation: $$\Delta = \frac{yd}{L} = \frac{2\times10^{-3}\times0.4\times10^{-3}}{1.0} = 8\times10^{-7}\text{ m} = 800\text{ nm}$$

Step 3 — Solve algebraically for the unknown: Phase difference:

Step 4 — Substitute numerical values with units: $$\phi = \frac{2\pi}{\lambda}\Delta = \frac{2\pi}{600}\times800 = \frac{8\pi}{3} \approx \boxed{8.38\text{ rad}}$$

Worked Calculation

$$\Delta = \frac{yd}{L} = \frac{2\times10^{-3}\times0.4\times10^{-3}}{1.0} = 8\times10^{-7}\text{ m} = 800\text{ nm}$$

$$\phi = \frac{2\pi}{\lambda}\Delta = \frac{2\pi}{600}\times800 = \frac{8\pi}{3} \approx \boxed{8.38\text{ rad}}$$

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

Path difference at distance $y$ from centre:

$$\Delta = \frac{yd}{L} = \frac{2\times10^{-3}\times0.4\times10^{-3}}{1.0} = 8\times10^{-7}\text{ m} = 800\text{ nm}$$

Phase difference:

$$\phi = \frac{2\pi}{\lambda}\Delta = \frac{2\pi}{600}\times800 = \frac{8\pi}{3} \approx \boxed{8.38\text{ rad}}$$

Answer

$$\phi = \frac{2\pi}{\lambda}\Delta = \frac{2\pi}{600}\times800 = \frac{8\pi}{3} \approx \boxed{8.38\text{ rad}}$$

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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