Problem 5.53 — Path Difference for Maximum in Thin Film

Problem Statement

A soap film ($n = 1.33$, $t = 0.40\;\mu$m) is illuminated at normal incidence. Which visible wavelengths ($400$–$700$ nm) are strongly reflected?

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: Constructive reflection: $2nt = (m+\tfrac{1}{2})\lambda$, so $\lambda = \frac{2nt}{m+1/2} = \frac{2\times1.33\times400}{m+0.5} = \frac{1064}{m+0.5}$ nm.

Step 2 — Apply the relevant physical law or equation:

$m=1$: $\lambda = 1064/1.5 = \mathbf{709}$ nm (just outside visible)
$m=2$: $\lambda = 1064/2.5 = \mathbf{426}$ nm (violet, visible)
$m=3$: $\lambda = 1064/3.5 = \mathbf{304}$ nm (UV, not visible)

Step 3 — Solve algebraically for the unknown: Only $\approx \boxed{426\text{ nm}}$ (violet) is strongly reflected in the visible range.

Worked Calculation

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

$$\boxed{\approx \boxed{426\text{ nm}}}$$

Constructive reflection: $2nt = (m+\tfrac{1}{2})\lambda$, so $\lambda = \frac{2nt}{m+1/2} = \frac{2\times1.33\times400}{m+0.5} = \frac{1064}{m+0.5}$ nm.

$m=1$: $\lambda = 1064/1.5 = \mathbf{709}$ nm (just outside visible)
$m=2$: $\lambda = 1064/2.5 = \mathbf{426}$ nm (violet, visible)
$m=3$: $\lambda = 1064/3.5 = \mathbf{304}$ nm (UV, not visible)

Only $\approx \boxed{426\text{ nm}}$ (violet) is strongly reflected in the visible range.

Answer

$$\boxed{\approx \boxed{426\text{ nm}}}$$

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.