---
title: "Soap film thickness from dark bands"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:soap-film-thickness-dark-bands"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/soap-film-thickness-dark-bands
site: BCT Engineering
---

# Soap film thickness from dark bands

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2069 Asar, 5 marks
- Video: [How to Solve Thin Film Interference Problems in Physics](https://www.youtube.com/watch?v=iG18FO71R3w) — Dan the Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/soap-film-thickness-dark-bands
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

White light is incident on a soap film at an angle sin⁻¹(4/5) and the reflected light on examination by a spectrometer shows dark bands. The consecutive dark bands correspond to wavelength 6.1×10⁻⁵ cm and 6.0×10⁻⁵ cm. If μ=1.33 for the film, calculate its thickness.

## Short answer

For consecutive dark bands, 2μt cos r = nλ₁ = (n+1)λ₂. Order n = λ₂/(λ₁−λ₂) = 60. With sin r = sin i/μ, cos r = 0.7989, thickness t = nλ₁/(2μ cos r) ≈ 1.72×10⁻³ cm.

## Answer

For a film viewed by reflected light, a dark band satisfies 2μt cos r = nλ. Two consecutive dark bands of wavelengths λ₁ > λ₂ correspond to orders n and (n+1).

**Given:** sin i = 4/5 = 0.8, μ = 1.33, λ₁ = 6.1×10⁻⁵ cm, λ₂ = 6.0×10⁻⁵ cm.

1. Condition for the two bands: 2μt cos r = nλ₁ = (n+1)λ₂
2. Order:

```text
n = λ₂ / (λ₁ − λ₂) = 6.0×10⁻⁵ / 0.1×10⁻⁵ = 60
```

3. Angle of refraction: sin r = sin i / μ = 0.8 / 1.33 = 0.6015, so cos r = √(1 − 0.3618) = 0.7989
4. Thickness:

```text
t = nλ₁ / (2μ cos r)
  = 60 × 6.1×10⁻⁵ / (2 × 1.33 × 0.7989)
  = 3.66×10⁻³ / 2.125
  = 1.72×10⁻³ cm
```

**Answer: film thickness t ≈ 1.72 × 10⁻³ cm (≈ 17.2 μm = 1.72×10⁻⁵ m).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In reflected light a dark band satisfies:
   A. 2μt cos r = (2n+1)λ/2
   B. 2μt cos r = nλ
   C. 2μt sin r = nλ
   D. μt = nλ
   Answer: B. Including the λ/2 phase change, dark bands occur at 2μt cos r = nλ. (Key point: Dark condition)

2. For consecutive dark bands 6.1×10^-5 cm and 6.0×10^-5 cm, n is:
   A. 6
   B. 30
   C. 60
   D. 610
   Answer: C. nλ1 = (n+1)λ2 gives n = λ2/(λ1 − λ2) = 60. (Key point: Order)

3. How is cos r found?
   A. sin r = sin i / μ, then cos r = sqrt(1 − sin^2 r)
   B. cos r = cos i / μ
   C. cos r = μ sin i
   D. r = i
   Answer: A. Snell's law gives sin r = 0.8/1.33 = 0.6015, so cos r ≈ 0.799. (Key point: Refraction angle)

4. The film thickness is about:
   A. 1.72×10^-3 cm
   B. 1.72×10^-5 cm
   C. 3.66×10^-3 cm
   D. 1.72×10^-1 cm
   Answer: A. t = 60×6.1×10^-5/(2×1.33×0.7989) ≈ 1.72×10^-3 cm. (Key point: Thickness)

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