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Module III: Interference in Thin Films

1. Introduction to Interference

Interference
The phenomenon in which two or more coherent light waves superpose to produce regions of maximum intensity (constructive interference) and minimum intensity (destructive interference).

When light waves from coherent sources meet, they combine according to the principle of superposition. The resultant intensity depends on the phase difference between the waves, which in turn depends on the path difference.

Conditions for Observable Interference
  • Light sources must be coherent (constant phase relationship)
  • Light sources should be monochromatic (single wavelength)
  • The amplitudes of interfering waves should be nearly equal
  • Sources should be narrow and close together
Constructive Interference

Waves meet in phase (crests align with crests)

Path difference = nλ (where n = 0, 1, 2, ...)

Phase difference = 2nπ

Result: Maximum intensity (Bright fringe)

Destructive Interference

Waves meet out of phase (crests align with troughs)

Path difference = (2n+1)λ/2

Phase difference = (2n+1)π

Result: Minimum intensity (Dark fringe)

Path Difference = (Phase Difference / 2π) × λ
Relation between path and phase difference

2. Interference in Thin Film of Uniform Thickness

When light falls on a thin transparent film (like soap bubble, oil film on water, or anti-reflective coating), it gets partially reflected from the top surface and partially from the bottom surface. These two reflected rays, being derived from the same incident ray, are coherent and produce interference.

Reflection from a Thin Film of Uniform Thickness
Air (n₀ = 1) Thin Film (refractive index μ) Air or Substrate Incident Ray i Ray 1 r Ray 2 t Ray 1: Reflected at top surface (undergoes π phase change if μ > n₀) Ray 2: Travels extra path 2μt cos r in the film, then reflects at bottom These two rays interfere when they meet
Phase Change on Reflection (Stokes' Relations)
  • When light reflects from a denser medium (going from lower to higher refractive index), it undergoes a phase change of π (equivalent to path difference of λ/2)
  • When light reflects from a rarer medium (higher to lower refractive index), there is no phase change
  • This is crucial for determining interference conditions in thin films

Optical Path Difference Calculation

Path Difference for Reflected Rays

Consider a thin film of refractive index μ and thickness t, with light incident at angle i:

Geometrical path difference: Ray 2 travels an extra distance inside the film

Geometrical path = 2t / cos r

where r is the angle of refraction

Optical path difference: Multiply by refractive index μ

Optical path in film = 2μt / cos r

But Ray 1 travels extra in air while Ray 2 is in the film. This horizontal component must be subtracted:

Horizontal component = 2t tan r × sin i = 2μt sin²r / cos r

Net optical path difference (geometric):

Δ = 2μt/cos r - 2μt sin²r/cos r = 2μt cos r

Additional phase change: Ray 1 reflects from denser medium → extra λ/2 path difference

Effective path difference = 2μt cos r ± λ/2
Effective Path Difference = 2μt cos r ± λ/2
For reflected light from thin film (± depends on relative refractive indices)

3. Conditions for Maxima and Minima (Reflected System)

For a thin film surrounded by air (or when μ_film > μ_surrounding), Ray 1 experiences a phase change of π on reflection from the top surface, while Ray 2 does not experience phase change when reflecting from the bottom surface (rarer medium below) or experiences another π change (denser below). The most common case is film in air:

For Film in Air (μ > 1) - Reflected Light

Effective path difference = 2μt cos r - λ/2

(The λ/2 is subtracted because only Ray 1 has phase change)

Condition for Constructive Interference (Bright Fringes/Maxima)

Derivation

For constructive interference, path difference = nλ (n = 0, 1, 2, ...)

2μt cos r - λ/2 = nλ
2μt cos r = nλ + λ/2 = (2n + 1)λ/2
2μt cos r = (2n + 1)λ/2
Condition for Maxima (Bright) in Reflected Light (n = 0, 1, 2, ...)

Condition for Destructive Interference (Dark Fringes/Minima)

Derivation

For destructive interference, path difference = (2n + 1)λ/2

2μt cos r - λ/2 = (2n + 1)λ/2
2μt cos r = (2n + 1)λ/2 + λ/2 = (n + 1)λ = nλ

(where we relabel n+1 as n)

2μt cos r = nλ
Condition for Minima (Dark) in Reflected Light (n = 1, 2, 3, ...)
For Normal Incidence (r = 0°)

When light falls perpendicular to the film surface, cos r = 1:

  • Maxima: 2μt = (2n + 1)λ/2
  • Minima: 2μt = nλ
Example: Thin Film in Reflected Light
A soap film (μ = 1.33) in air appears bright when viewed at normal incidence with light of wavelength 589 nm. Calculate the minimum thickness of the film.
Step 1: Identify the condition

For bright film (maxima) in reflected light at normal incidence:

2μt = (2n + 1)λ/2
Step 2: For minimum thickness, take n = 0
2μt = (1)λ/2 = λ/2
t = λ/(4μ)
Step 3: Substitute values
t = 589 nm / (4 × 1.33)
t = 589 / 5.32 = 110.7 nm
Minimum thickness = 110.7 nm
Condition Reflected Light Transmitted Light
Maxima (Bright) 2μt cos r = (2n+1)λ/2 2μt cos r = nλ
Minima (Dark) 2μt cos r = nλ 2μt cos r = (2n+1)λ/2
Why Colors in Thin Films?

When white light falls on a thin film, different wavelengths satisfy the maxima/minima conditions at different thicknesses. At any point with specific thickness t:

  • Some wavelengths show constructive interference (appear bright)
  • Other wavelengths show destructive interference (get suppressed)
  • The remaining wavelengths mix to produce the observed color

This explains the colorful patterns seen in soap bubbles and oil films on water.

4. Wedge-Shaped Film (Qualitative)

Wedge-Shaped Film
A thin film of non-uniform thickness that increases linearly from one end to another, forming a wedge angle θ between the two surfaces. The thickness varies as t = x tan θ ≈ xθ (for small angles).
Wedge-Shaped Air Film
Glass Plate (Bottom) Glass Plate (Top) Air Film (n=1) θ Contact Edge Monochromatic Light (Normal Incidence) t₁ t₂ t₃ Observed Fringe Pattern (equally spaced)

Formation of Fringes

When monochromatic light falls normally on a wedge-shaped film, interference occurs between rays reflected from the top and bottom surfaces of the air film. Since the thickness varies along the wedge:

Fringe Width (β) = λ / (2θ)
For air wedge at normal incidence (θ in radians)
Important Observations
  • Dark fringe at contact edge: At t = 0, path difference = λ/2 (due to phase change), giving destructive interference
  • Fringe spacing depends on θ: Smaller wedge angle → wider fringes
  • With white light: Colored fringes appear, with edge being dark
Example: Wedge Fringe Width
Two glass plates are in contact at one edge and separated by a thin wire at the other edge 10 cm away. When illuminated with light of wavelength 600 nm, 20 dark fringes are observed. Calculate the diameter of the wire.
Step 1: Calculate fringe width

Total distance = 10 cm, Number of fringes = 20

Fringe width β = 10/20 = 0.5 cm = 5 mm
Step 2: Use β = λ/(2θ) to find θ
θ = λ/(2β) = (600 × 10⁻⁹)/(2 × 5 × 10⁻³)
θ = 6 × 10⁻⁵ radians
Step 3: Calculate wire diameter

Diameter d = L × θ (where L = 10 cm)

d = 0.1 × 6 × 10⁻⁵ = 6 × 10⁻⁶ m = 6 μm
Diameter of wire = 6 μm

5. Newton's Rings

Newton's Rings
A pattern of concentric circular interference fringes formed when a plano-convex lens is placed on a flat glass plate. The air film between them varies in thickness, creating circular fringes.
Newton's Rings Setup
Glass Plate Plano-convex Lens Air Film Contact Point R Center of curvature t r Top View: Concentric Circular Fringes (Dark center spot)

Relation Between Thickness and Radius

Derivation of t in terms of r and R

For the geometry of a spherical surface touching a flat plate:

Using the property of a chord in a circle:

r² = R² - (R - t)² = 2Rt - t²

Since t << R, we can neglect t²:

r² ≈ 2Rt
t = r²/(2R)
t = r²/(2R)
Thickness of air film at radius r from contact point

Conditions for Bright and Dark Rings

For normal incidence on air film, using the thin film interference conditions:

Dark Rings: 2t = nλ → r² = nλR
r_n = √(nλR) for n-th dark ring
Bright Rings: 2t = (2n-1)λ/2
r_n = √[(2n-1)λR/2] for n-th bright ring
Why is the Center Dark?
At the point of contact, t = 0. The path difference is only due to the phase change of π (λ/2) at the lower glass surface. This gives destructive interference, so the center is dark in reflected light.

Diameter of Newton's Rings

D_n = 2r_n = 2√(nλR) for Dark Rings
Diameter of n-th dark ring

Squaring both sides:

D_n² = 4nλR
Square of diameter of n-th dark ring

Applications of Newton's Rings

Application 1: Determination of Wavelength

If the radius of curvature R of the lens is known, the wavelength of monochromatic light can be determined:

Formula for Wavelength Determination

For the m-th and n-th dark rings (m > n):

D_m² = 4mλR and D_n² = 4nλR
D_m² - D_n² = 4(m-n)λR
λ = (D_m² - D_n²) / [4(m-n)R]
Wavelength of monochromatic light
Example: Wavelength Determination
In a Newton's rings experiment, the diameter of the 5th dark ring is 3.36 mm and that of the 15th dark ring is 5.90 mm. If the radius of curvature of the lens is 100 cm, find the wavelength of light used.
Step 1: Identify given values

D₅ = 3.36 mm = 0.336 cm, D₁₅ = 5.90 mm = 0.590 cm

R = 100 cm, n = 5, m = 15

Step 2: Calculate D² values
D₁₅² = (0.590)² = 0.3481 cm²
D₅² = (0.336)² = 0.1129 cm²
D₁₅² - D₅² = 0.3481 - 0.1129 = 0.2352 cm²
Step 3: Apply formula
λ = (D₁₅² - D₅²) / [4(15-5)R]
λ = 0.2352 / (4 × 10 × 100)
λ = 0.2352 / 4000 = 5.88 × 10⁻⁵ cm
λ = 588 nm
Wavelength λ = 588 nm (Yellow light - sodium D-line region)
Application 2: Determination of Refractive Index of Liquid

When a liquid of refractive index μ is introduced between the lens and plate, the wavelength in the medium becomes λ/μ:

Formula for Refractive Index

For air film: D_air² = 4nλR

For liquid film: D_liquid² = 4nλR/μ

Taking the ratio:

D_air²/D_liquid² = μ
μ = D_air² / D_liquid² = (D_m² - D_n²)_air / (D_m² - D_n²)_liquid
Refractive index of liquid
Example: Refractive Index Determination
In a Newton's rings experiment with air film, the diameter of the 10th dark ring is 5 mm. When a liquid is introduced, the diameter of the 10th dark ring becomes 4.5 mm. Calculate the refractive index of the liquid.
Step 1: Note the given values
D_air = 5 mm, D_liquid = 4.5 mm
Step 2: Apply the formula
μ = D_air²/D_liquid² = (5)²/(4.5)²
μ = 25/20.25 = 1.234
Refractive Index of Liquid μ = 1.234

6. Engineering Applications of Thin Film Interference

Anti-Reflective Coatings
  • Thin film on lens surfaces
  • Thickness = λ/4n for destructive interference
  • Reduces reflection, increases transmission
  • Used in cameras, eyeglasses, solar cells
Optical Filters
  • Multilayer thin film structures
  • Selective transmission of wavelengths
  • Band-pass, high-pass, low-pass filters
  • Used in spectrometers, lasers
Optical Testing
  • Testing flatness of surfaces
  • Measuring small thicknesses
  • Quality control in optics manufacturing
  • Newton's rings for lens testing
Anti-Reflective (AR) Coating Design

For a single-layer AR coating on glass (n_glass = 1.5):

  • Optimal coating refractive index: n_c = √(n_air × n_glass) = √1.5 ≈ 1.22
  • Optical thickness = λ/4 (quarter-wave plate)
  • MgF₂ (n = 1.38) is commonly used
  • Reduces reflection from ~4% to <1%
Why Quarter-Wave Thickness?

When light travels through a coating of thickness t = λ/(4n), the round-trip path difference is 2nt = λ/2. Combined with the π phase change from the coating-glass interface, the total path difference becomes λ. This means the reflected waves from top and bottom of the coating are exactly out of phase, causing destructive interference and minimizing reflection.

Summary: Key Formulas

Path diff = 2μt cos r ± λ/2
Thin Film Path Difference
2μt cos r = (2n+1)λ/2
Maxima (Reflected)
2μt cos r = nλ
Minima (Reflected)
β = λ/(2θ)
Wedge Fringe Width
D_n² = 4nλR
Newton's Rings (Dark)
λ = (D_m² - D_n²)/4(m-n)R
Wavelength from Newton's Rings
μ = D_air²/D_liquid²
Refractive Index