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Module III: Interference in Thin Films - Question Solutions

Question 1: Thin Film Appears Black in Reflected Light

Question
Explain why an excessively thin film appears black in reflected light.
Explanation
Reflection from Very Thin Film
Air (n=1) Thin Film (n>1) t ≈ 0 Substrate R₁ (π phase shift) R₂ (no π shift) Phase difference ≈ π (180°) → Destructive interference

When light is incident on a very thin film (thickness t → 0), two reflected rays are produced:

Reflected Rays Analysis:
  • Ray R₁: Reflected from the top surface (air-film interface)
  • Ray R₂: Reflected from the bottom surface (film-substrate interface)
Phase Changes on Reflection
Important Rule:
When light reflects from a medium of higher refractive index, it undergoes a phase change of π (180°). When light reflects from a medium of lower refractive index, there is no phase change.

For a thin film (n > 1) on a substrate:

  • R₁: Reflects at air-film interface (going from lower n to higher n) → Phase change of π
  • R₂: Reflects at film-substrate interface (depends on substrate, typically no additional phase change relative to R₁)
Path Difference Analysis

For a film of thickness t, the path difference between R₁ and R₂ is:

Path difference = 2μt cos r
where μ = refractive index of film, r = angle of refraction

When t → 0 (excessively thin film):

Path difference ≈ 0
Effective Path Difference

Including the phase change of λ/2 (equivalent to π phase shift):

Effective path difference = 2μt cos r + λ/2 ≈ λ/2 (when t → 0)
Condition for Destructive Interference

A path difference of λ/2 corresponds to destructive interference.

When the two reflected rays are exactly out of phase (180° or λ/2 path difference), they cancel each other completely.

Conclusion
An excessively thin film appears black in reflected light because:
  1. The geometric path difference (2μt cos r) is nearly zero
  2. The only path difference comes from the π phase change at the air-film interface
  3. This λ/2 effective path difference causes complete destructive interference
  4. No light is reflected, making the film appear black

Practical Example: This is why soap bubbles appear black just before they burst - at that point, the film becomes extremely thin.

Question 2: Conditions for Maxima and Minima in Thin Films

Question
Obtain the condition for maxima and minima of the light reflected from a thin transparent film of uniform thickness.
Interference in Thin Film of Uniform Thickness
Air (n₀=1) Film (μ) Substrate Incident Ray i R₁ r R₂ t
Derivation of Path Difference

Consider a thin film of uniform thickness t and refractive index μ. A ray of light incident at angle i is partially reflected (R₁) and partially refracted at angle r. The refracted ray reflects from the bottom surface, giving ray R₂.

Calculating Optical Path Difference

Step 1: The path traveled by ray R₂ inside the film is BC + CD.

From geometry: BC = CD = t/cos r

Step 2: Optical path in film = μ(BC + CD) = μ × 2t/cos r = 2μt/cos r

Step 3: Path traveled by R₁ in air = BN (where N is on the extended R₁ path)

BN = BD × sin i, and BD = 2t × tan r

Therefore, BN = 2t × tan r × sin i

Step 4: Using Snell's law: sin i = μ sin r

BN = 2t × tan r × μ sin r = 2μt × sin²r/cos r

Step 5: Path difference:

Δ = 2μt/cos r - 2μt sin²r/cos r = 2μt(1 - sin²r)/cos r = 2μt cos r
Including Phase Change

Ray R₁ undergoes a phase change of π (equivalent to λ/2 path difference) upon reflection from denser medium.

Effective Path Difference = 2μt cos r ± λ/2
Conditions for Maxima (Bright Fringes)
Constructive Interference (Maxima)

For bright fringes, effective path difference = nλ (where n = 0, 1, 2, ...)

2μt cos r + λ/2 = nλ
2μt cos r = (2n - 1)λ/2
Condition for maxima (bright fringes) in reflected light

Or: 2μt cos r = (n - ½)λ, where n = 1, 2, 3, ...

Conditions for Minima (Dark Fringes)
Destructive Interference (Minima)

For dark fringes, effective path difference = (2n + 1)λ/2

2μt cos r + λ/2 = (2n + 1)λ/2
2μt cos r = nλ
Condition for minima (dark fringes) in reflected light

where n = 0, 1, 2, 3, ...

Summary Table
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

Question 3: Newton's Rings - Refractive Index Determination

Question
With the experimental set up of Newton's ring, explain the determination of refractive index of a liquid.
Newton's Rings Setup
Newton's Rings Experimental Setup
Flat Glass Plate Plano-convex Lens (Radius R) Air Film / Liquid Incident Light Concentric Circular Fringes (Newton's Rings)
Principle

Newton's rings are formed due to interference between light reflected from the lower surface of the plano-convex lens and the upper surface of the flat glass plate. The air film between them has varying thickness, creating concentric circular interference fringes.

Formula for Ring Diameter

For the nᵗʰ dark ring with air between lens and plate:

Dₙ² = 4nλR
where Dₙ = diameter of nᵗʰ dark ring, λ = wavelength, R = radius of curvature

When a liquid of refractive index μ is placed between the lens and plate:

D'ₙ² = 4nλR/μ
where D'ₙ = diameter with liquid
Method to Determine Refractive Index
Step 1: Measure with Air
Set up Newton's rings apparatus with air between lens and plate. Measure the diameters of various dark rings (say nᵗʰ and mᵗʰ rings).
Dₙ² - Dₘ² = 4(n - m)λR
Step 2: Introduce Liquid
Carefully introduce the liquid whose refractive index is to be determined between the lens and plate. The rings will contract because the optical path increases.
Step 3: Measure with Liquid
Measure the diameters of the same rings (nᵗʰ and mᵗʰ) with liquid:
D'ₙ² - D'ₘ² = 4(n - m)λR/μ
Step 4: Calculate Refractive Index
Divide the two equations:
(Dₙ² - Dₘ²)/(D'ₙ² - D'ₘ²) = μ
Final Formula for Refractive Index
μ = (Dₙ² - Dₘ²)/(D'ₙ² - D'ₘ²)
where D = diameter with air, D' = diameter with liquid
Advantages of this Method
  • Does not require knowledge of wavelength λ or radius R
  • Only ratio of squared diameters is needed
  • Can be used for transparent liquids
  • High accuracy with proper measurements

Question 4: Newton's Rings Numerical - Radius of Curvature

Question
In Newton's rings experiment the diameter of 5ᵗʰ ring was 0.336 cm and the diameter of 15ᵗʰ ring was 0.590 cm. Find the radius of curvature of plano convex lens if the wavelength of light used is 5890 Å.
Given Data:
  • Diameter of 5ᵗʰ dark ring, D₅ = 0.336 cm = 0.336 × 10⁻² m
  • Diameter of 15ᵗʰ dark ring, D₁₅ = 0.590 cm = 0.590 × 10⁻² m
  • Wavelength, λ = 5890 Å = 5890 × 10⁻¹⁰ m = 5.89 × 10⁻⁷ m
  • n = 5, m = 15
Formula Used:
For Newton's rings (dark rings):
Dₙ² = 4nλR

Therefore:

Dₘ² - Dₙ² = 4(m - n)λR
R = (Dₘ² - Dₙ²) / [4(m - n)λ]
Calculation:

Calculate D₁₅² - D₅²:

D₁₅² = (0.590 × 10⁻²)² = 0.3481 × 10⁻⁴ m²
D₅² = (0.336 × 10⁻²)² = 0.1129 × 10⁻⁴ m²
D₁₅² - D₅² = (0.3481 - 0.1129) × 10⁻⁴ = 0.2352 × 10⁻⁴ m²

Calculate 4(m - n)λ:

4(m - n)λ = 4 × (15 - 5) × 5.89 × 10⁻⁷
= 4 × 10 × 5.89 × 10⁻⁷ = 235.6 × 10⁻⁷ m

Calculate R:

R = (0.2352 × 10⁻⁴) / (235.6 × 10⁻⁷)
R = (0.2352 × 10⁻⁴) / (2.356 × 10⁻⁵)
R = 0.998 m
Radius of curvature R ≈ 1.0 m (or 100 cm)

Question 5: Newton's Rings with Liquid - Numerical

Question
In a Newton's ring arrangement, a drop of liquid of refractive index 1.33 is placed in between the lens and plate. The diameter of 9ᵗʰ dark ring is found to be 0.58 cm. Obtain the radius of curvature, if the wavelength of light is 6000 Å.
Given Data:
  • Refractive index of liquid, μ = 1.33
  • Diameter of 9ᵗʰ dark ring, D₉ = 0.58 cm = 0.58 × 10⁻² m
  • Wavelength, λ = 6000 Å = 6000 × 10⁻¹⁰ m = 6 × 10⁻⁷ m
  • n = 9
Formula Used:
For Newton's rings with a liquid of refractive index μ:
Dₙ² = 4nλR/μ

Solving for R:

R = (μ × Dₙ²) / (4nλ)
Calculation:

Calculate Dₙ²:

D₉² = (0.58 × 10⁻²)² = 0.3364 × 10⁻⁴ m²

Calculate 4nλ:

4nλ = 4 × 9 × 6 × 10⁻⁷ = 216 × 10⁻⁷ m

Calculate R:

R = (1.33 × 0.3364 × 10⁻⁴) / (216 × 10⁻⁷)
R = (0.4474 × 10⁻⁴) / (2.16 × 10⁻⁵)
R = (4.474 × 10⁻⁵) / (2.16 × 10⁻⁵)
R = 2.071 m
Radius of curvature R ≈ 2.07 m (or 207 cm)
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