Problem Statement
A wedge-shaped air film between two flat glass plates has an angle $\alpha = 20 $ (arc seconds). For $\lambda = 550$ nm at normal incidence, find the spacing between adjacent dark fringes.
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: Fringe spacing in a wedge (dark fringes in reflected light, air wedge):
Step 2 — Apply the relevant physical law or equation: $$\Delta x = \frac{\lambda}{2\alpha}$$
Step 3 — Solve algebraically for the unknown: Convert: $20” = 20/(3600)° = 9.696\times10^{-5}$ rad.
Step 4 — Substitute numerical values with units: $$\Delta x = \frac{550\times10^{-9}}{2\times9.696\times10^{-5}} = \frac{550\times10^{-9}}{1.939\times10^{-4}} \approx 2.84\times10^{-3}\text{ m} \approx \boxed{2.8\text{ mm}}$$
Worked Calculation
$$\Delta x = \frac{\lambda}{2\alpha}$$
$$\Delta x = \frac{550\times10^{-9}}{2\times9.696\times10^{-5}} = \frac{550\times10^{-9}}{1.939\times10^{-4}} \approx 2.84\times10^{-3}\text{ m} \approx \boxed{2.8\text{ mm}}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
Fringe spacing in a wedge (dark fringes in reflected light, air wedge):
$$\Delta x = \frac{\lambda}{2\alpha}$$
Convert: $20” = 20/(3600)° = 9.696\times10^{-5}$ rad.
$$\Delta x = \frac{550\times10^{-9}}{2\times9.696\times10^{-5}} = \frac{550\times10^{-9}}{1.939\times10^{-4}} \approx 2.84\times10^{-3}\text{ m} \approx \boxed{2.8\text{ mm}}$$
Answer
$$\Delta x = \frac{550\times10^{-9}}{2\times9.696\times10^{-5}} = \frac{550\times10^{-9}}{1.939\times10^{-4}} \approx 2.84\times10^{-3}\text{ m} \approx \boxed{2.8\text{ mm}}$$
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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