--- title: "Wedge film fringes, thickness difference" subject: "Engineering Physics" chapter: 4 key: "semester_1/engineering_physics:wedge-film-fringes-thickness-difference" source: https://www.bctengineer.com/q/semester_1/engineering_physics/wedge-film-fringes-thickness-difference site: BCT Engineering --- # Wedge film fringes, thickness difference Engineering Physics · Chapter 4: Physical Optics - Asked in: 2067 Asar, 5 marks - Video: [Wedge shaped film (Part 1)](https://www.youtube.com/watch?v=COupe538OYI) — Physics Jessy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/wedge-film-fringes-thickness-difference - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4 ## Question Explain how interference fringes are formed by a thin wedge shaped film, when examined by normally reflected light. How will you estimate the difference of film thickness between two points? ## Short answer Interference fringes in a thin wedge film arise from path difference 2μt plus a λ/2 phase change at the lower surface, giving straight, equally spaced fringes with a dark edge. The thickness difference between two points is Δt = Nλ/(2μ), where N is the number of fringes between them. ## Answer #### Formation of fringes A thin wedge is an air (or other transparent) film between two plane glass plates inclined at a very small angle θ and touching at the edge. When it is lit by monochromatic light and viewed by normally reflected light, the ray reflected from the top surface of the film and the ray reflected from the bottom surface are coherent (they come from the same incident ray) and interfere. ```text incident / reflected (normal) | | _____|_|______________ upper plate ~~~~~~~~~~~~~~~~~~~~~~~ film, thickness t = xθ edge -----> x ------> lower plate 0 t ``` 1. At a distance *x* from the edge the thickness is t = xθ. 2. Path difference = 2μt cos r. For normal incidence, cos r = 1, so 2μt. 3. Reflection at the lower surface (rarer to denser) gives an extra phase change of π, i.e. an extra λ/2. 4. **Bright fringe:** 2μt = (2n + 1)λ/2, n = 0, 1, 2, ... **Dark fringe:** 2μt = nλ, n = 0, 1, 2, ... 5. At the edge t = 0, so the edge is **dark**. 6. Fringes are straight, parallel to the edge, and equally spaced. 7. Fringe width: for two successive dark fringes, 2μ(x₂ − x₁)θ = λ, so **β = λ / (2μθ)**. #### Estimating the thickness difference between two points Moving from one dark (or bright) fringe to the next changes the thickness by λ/2μ. If *N* fringes lie between the two points, **Δt = N λ / (2μ)** If the distance *L* between the points is known and the fringe width β is measured, N = L / β, so Δt = Lλ / (2μβ) = Lθ. For air, μ = 1 and Δt = Nλ/2. **Example:** with λ = 589 nm and 10 fringes between two points in an air wedge, Δt = 10 × 589 nm / 2 ≈ 2.95 µm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For normal incidence on a wedge film of index μ and thickness t, what is the path difference between the two reflected rays? A. μt B. 2μt C. 2μt cos 2r D. μt/2 Answer: B. The ray in the film travels down and back, so path = 2μt cos r = 2μt at normal incidence. (Key point: Path difference) 2. Why is the edge of the wedge (t = 0) dark in reflected light? A. No light reaches the edge B. Reflection at the lower surface adds an extra phase change of π (λ/2), giving destructive interference C. The two rays are not coherent at the edge D. Both reflections add λ/2 so they cancel exactly Answer: B. Only one reflection (rarer to denser) gives a λ/2 shift, so at t = 0 the rays are out of phase. (Key point: Dark edge) 3. What do the fringes of a wedge look like? A. Concentric circles, unequally spaced B. Straight, parallel to the edge, equally spaced C. Straight, perpendicular to the edge, unequally spaced D. Hyperbolas near the edge Answer: B. t = xθ grows linearly with x, so fringes are straight, parallel to the edge and equally spaced. (Key point: Fringe pattern) 4. If N fringes lie between two points of an air wedge, what is the thickness difference? A. Nλ B. Nλ/2 C. 2Nλ D. Nλ/4 Answer: B. Each fringe shift changes thickness by λ/2μ, so Δt = Nλ/2 for air. (Key point: Thickness difference) ## Related in Engineering Physics - [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 - [Thin film interference by reflected light](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Athin-film-reflected-interference-theory) — Engineering Physics, chapter 4, asked 2× - [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 angle from fringe spacing](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Awedge-angle-fringe-spacing) — 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 - [Fringe width of air wedge numerical](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Awedge-fringe-width-wire-numerical) — Engineering Physics, chapter 4