Problem Statement
A thin glass film ($n = 1.5$) in air appears dark in reflected light for $\lambda = 640$ nm. Find the minimum nonzero thickness.
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: Dark in reflection (destructive): two phase reversals cancel, condition is $2nt = m\lambda$.
Step 2 — Apply the relevant physical law or equation: Minimum nonzero ($m = 1$):
Step 3 — Solve algebraically for the unknown: $$t = \frac{\lambda}{2n} = \frac{640}{2\times1.5} = \frac{640}{3} \approx \boxed{213\text{ nm}}$$
Worked Calculation
$$t = \frac{\lambda}{2n} = \frac{640}{2\times1.5} = \frac{640}{3} \approx \boxed{213\text{ nm}}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{t = \frac{\lambda}{2n} = \frac{640}{2\times1.5} = \frac{640}{3} \approx \boxed{213\text{ nm}}}$$
Dark in reflection (destructive): two phase reversals cancel, condition is $2nt = m\lambda$.
Minimum nonzero ($m = 1$):
$$t = \frac{\lambda}{2n} = \frac{640}{2\times1.5} = \frac{640}{3} \approx \boxed{213\text{ nm}}$$
Answer
$$t = \frac{\lambda}{2n} = \frac{640}{2\times1.5} = \frac{640}{3} \approx \boxed{213\text{ nm}}$$
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.
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