---
title: "Thin film interference by reflected light"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:thin-film-reflected-interference-theory"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/thin-film-reflected-interference-theory
site: BCT Engineering
---

# Thin film interference by reflected light

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2073 Chaitra, 5 marks; 2071 Shrawan, 5 marks (asked 2 times)
- Video: [Thin Film Interference | Soap Bubble & Oil Film | Optics | Physics | BSc | BTech | UPSC | GATE | JEE](https://www.youtube.com/watch?v=jqEitP6tyjk) — Gautam Varde
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/thin-film-reflected-interference-theory
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

Write down the conditions for interference of light. Give the necessary theory for interference in a thin film due to reflected light (derive the relations for interference by reflected light).

## Short answer

Coherent monochromatic sources with same polarization are needed. For a thin film of refractive index μ and thickness t, reflected light path difference is 2μt cos r, with an extra λ/2 from reflection at the top surface. Effective path difference Δ = 2μt cos r − λ/2. Bright: 2μt cos r = (2n+1)λ/2; dark: 2μt cos r = nλ.

## Answer

#### Conditions for interference
1. The two sources must be **coherent**: a constant phase difference and the same frequency.
2. They should emit light of the **same wavelength** (monochromatic) and the same state of polarization.
3. The amplitudes should be nearly equal for good contrast.
4. The sources must be **narrow**, and the path difference small (within the coherence length).
5. The screen should be sufficiently far from the sources.

#### Thin film: reflected light
```text
   air:       \ 1        / 1'
               \ i     /
   -----------A-\\---C---------  top surface
   film (μ)    r \ /  r      t
   -------------B-------------  bottom surface
```
A parallel beam falls at angle *i* on a film of thickness *t* and refractive index μ; the angle of refraction is *r*. Ray 1 is reflected at A on the top surface. Ray 2 refracts at A, reflects at B on the bottom surface, and emerges at C parallel to ray 1.

1. Optical path of ray 2 inside the film: μ(AB + BC)
2. Draw CD perpendicular to the reflected ray 1. The path of ray 1 in air up to D is AD = AC sin i; also μ = sin i / sin r
3. Geometry: AB = BC = t / cos r, and AC = 2t tan r
4. Optical path difference: μ(2t/cos r) − 2t tan r sin i = 2μt/cos r − 2μt sin²r/cos r = **2μt cos r**
5. Reflection at the top surface (rarer to denser) gives an extra phase change of π, equivalent to a path of λ/2. Effective path difference: **Δ = 2μt cos r − λ/2** (or +λ/2).

#### Conditions
- **Bright (constructive)**: 2μt cos r − λ/2 = nλ → **2μt cos r = (2n+1) λ/2**, n = 0, 1, 2 ...
- **Dark (destructive)**: 2μt cos r − λ/2 = (2n+1)λ/2 → **2μt cos r = nλ**

For normal incidence (r = 0): 2μt = (2n+1)λ/2 for bright and 2μt = nλ for dark. A very thin film (t << λ) looks dark in reflected light.

## Check yourself

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

1. Which is essential for two sources to produce sustained interference?
   A. They are independent bulbs of equal brightness
   B. They are coherent: same frequency and constant phase difference
   C. They have different wavelengths
   D. They have random phase changes
   Answer: B. Coherent sources keep a constant phase difference and the same frequency, so the pattern is stable. (Key point: Coherence condition)

2. For a thin film (thickness t, index μ, refraction angle r), what is the optical path difference between the two reflected rays before the phase change correction?
   A. 2μt sin r
   B. μt cos r
   C. 2μt cos r
   D. 2t cos r / μ
   Answer: C. Path difference = μ(AB+BC) − AC sin i = 2μt cos r. (Key point: Optical path difference)

3. Why is λ/2 added to (or subtracted from) the path difference in reflected light?
   A. The film absorbs half a wavelength
   B. Reflection at the top surface (rarer to denser) gives a phase change of π
   C. The ray inside the film slows to half speed
   D. Refraction at the bottom surface adds π
   Answer: B. Reflection from a denser medium at the top surface causes a π phase change, equal to λ/2. (Key point: Phase change on reflection)

4. In reflected light, the film appears bright when:
   A. 2μt cos r = nλ
   B. 2μt cos r = (2n+1)λ/2
   C. 2μt cos r = λ/2 only
   D. 2μt cos r = 2nλ
   Answer: B. Setting 2μt cos r − λ/2 = nλ gives 2μt cos r = (2n+1)λ/2 for brightness. (Key point: Bright condition)

## Related in Engineering Physics

- [Reflected and transmitted interference complementary](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Athin-film-reflected-transmitted-complementary) — Engineering Physics, chapter 4, asked 2×
- [Wedge film fringes and fringe width](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Awedge-film-fringes-fringe-width) — Engineering Physics, chapter 4
- [Refractive index of liquid film by interference](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acoherent-sources-film-refractive-index-method) — Engineering Physics, chapter 4
- [Wedge film fringes, thickness difference](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Awedge-film-fringes-thickness-difference) — Engineering Physics, chapter 4
- [Conditions for constructive and destructive interference](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aconstructive-destructive-path-difference) — Engineering Physics, chapter 4
- [Soap film thickness from dark bands](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asoap-film-thickness-dark-bands) — Engineering Physics, chapter 4

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