Interference - Young's Double Slit & Thin Films

Master Young's double slit experiment, interference conditions, and thin film interference for JEE

10 min read

Interference - Creating Light and Dark Patterns

The Movie Hook 🎬

Ever seen oil spills creating rainbow colors on water? Or soap bubbles showing swirling colors? Or the holographic security strips on currency notes? These are all interference patterns - nature’s way of creating colors without pigments! Even the beautiful patterns in Doctor Strange’s mirror dimension follow interference principles!

The Big Picture

Interference: Redistribution of light intensity due to superposition of two or more coherent light waves.

Key applications:

  • Thin film colors (soap bubbles, oil films)
  • Anti-reflection coatings (camera lenses, eyeglasses)
  • Holography (3D images)
  • Interferometers (precision measurements)

Connection Alert: Builds on Wave Optics, relates to Diffraction.

Core Concepts

1. Conditions for Sustained Interference

  1. Coherent sources: Same frequency, constant phase difference
  2. Same amplitude or comparable amplitudes (for good contrast)
  3. Waves should be in same state of polarization
  4. Monochromatic light (single wavelength)

Why monochromatic?

  • Different wavelengths β†’ different fringe patterns
  • Overlapping patterns β†’ loss of clarity

2. Young’s Double Slit Experiment (YDSE)

The most important experiment in wave optics!

Setup:

  • Single source S (for coherence)
  • Two slits S₁ and Sβ‚‚ separated by distance d
  • Screen at distance D from slits
  • Point P on screen at distance y from center O

Path difference:

$$\boxed{\Delta x = S_2P - S_1P \approx \frac{yd}{D}}$$

(For D » d and D » y)

3. Position of Bright Fringes (Maxima)

Condition: Path difference = nΞ»

$$\boxed{y_n = \frac{n\lambda D}{d}}$$

Where:

  • n = 0, 1, 2, 3, … (order of fringe)
  • n = 0: Central bright fringe
  • n = 1, 2, 3, …: 1st, 2nd, 3rd order bright fringes

Central maximum is brightest (n = 0)

4. Position of Dark Fringes (Minima)

Condition: Path difference = (n + 1/2)Ξ» = (2n+1)Ξ»/2

$$\boxed{y_n = \frac{(2n+1)\lambda D}{2d}}$$

Where n = 0, 1, 2, 3, …

First dark fringe: n = 0 β†’ y = Ξ»D/(2d)

5. Fringe Width (Ξ²)

Fringe width: Distance between consecutive bright (or dark) fringes

$$\boxed{\beta = \frac{\lambda D}{d}}$$

Key properties:

  • All fringes are equally spaced
  • Ξ² ∝ Ξ» (red light β†’ wider fringes than violet)
  • Ξ² ∝ D (farther screen β†’ wider fringes)
  • Ξ² ∝ 1/d (closer slits β†’ wider fringes)

Memory Trick: “Bigger Wavelength = Bigger Width” (Ξ² ∝ Ξ»)

6. Angular Width

Angular fringe width:

$$\boxed{\theta = \frac{\beta}{D} = \frac{\lambda}{d}}$$

Independent of D!

Angular position of nth bright fringe:

$$\boxed{\theta_n = \frac{n\lambda}{d}}$$

7. Intensity Distribution

For two coherent sources of equal intensity Iβ‚€:

$$\boxed{I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)}$$

Where phase difference:

$$\phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi yd}{\lambda D}$$

Or:

$$\boxed{I = 4I_0 \cos^2\left(\frac{\pi yd}{\lambda D}\right)}$$

At maxima (Ξ”x = nΞ»):

$$I_{max} = 4I_0$$

At minima (Ξ”x = (2n+1)Ξ»/2):

$$I_{min} = 0$$

Average intensity over screen:

$$I_{avg} = 2I_0$$

Variations of YDSE

1. One Slit Wider

If one slit provides intensity I₁ and other Iβ‚‚:

$$\boxed{I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2}$$ $$\boxed{I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2}$$

Visibility (contrast):

$$\boxed{V = \frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}}$$

Maximum visibility = 1 (when I₁ = Iβ‚‚)

2. Thin Film in Front of One Slit

If film of thickness t and refractive index n is placed in front of one slit:

Additional path difference introduced:

$$\boxed{\Delta x_{extra} = (n-1)t}$$

Shift in fringe pattern:

$$\boxed{y_{shift} = \frac{(n-1)tD}{d}}$$

Pattern shifts towards the slit with film.

Central maximum is NO longer at center!

3. YDSE in Medium

When entire setup is in medium of refractive index n:

Wavelength in medium:

$$\lambda_m = \frac{\lambda}{n}$$

New fringe width:

$$\boxed{\beta_m = \frac{\lambda_m D}{d} = \frac{\lambda D}{nd} = \frac{\beta}{n}}$$

Fringes become narrower!

4. White Light YDSE

Central fringe: White (all wavelengths constructively interfere)

Other fringes: Colored

  • Inner edge: Violet (Ξ» smallest)
  • Outer edge: Red (Ξ» largest)

After a few orders: Overlapping colors β†’ uniform illumination

Position of nth bright fringe for color Ξ»:

$$y_n = \frac{n\lambda D}{d}$$

Since Ξ»_red > Ξ»_violet:

  • Red fringes are farther from center
  • Violet fringes are closer

The Formula Cheat Sheet

YDSE ESSENTIALS:
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
1. Path difference:      Ξ”x = yd/D
2. Bright fringes:       y_n = nΞ»D/d
3. Dark fringes:         y_n = (2n+1)Ξ»D/(2d)
4. Fringe width:         Ξ² = Ξ»D/d
5. Angular width:        ΞΈ = Ξ»/d
6. Intensity:            I = 4Iβ‚€cosΒ²(Ο†/2)
7. With film:            shift = (n-1)tD/d
8. In medium:            Ξ²_m = Ξ²/n
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Thin Film Interference

1. Interference in Thin Films

Examples: Soap bubbles, oil films, anti-reflection coatings

Two reflections:

  1. From upper surface (air-film)
  2. From lower surface (film-air)

Path difference:

$$\boxed{\Delta x = 2nt \cos r}$$

Where:

  • n = refractive index of film
  • t = thickness of film
  • r = angle of refraction in film

For normal incidence (i = 0, r = 0):

$$\boxed{\Delta x = 2nt}$$

2. Phase Change on Reflection

Critical rule: When light reflects from denser medium, phase change of Ο€ occurs.

For film in air:

  • Reflection from upper surface: phase change Ο€ (rarer to denser)
  • Reflection from lower surface: no phase change (denser to rarer)

Effective path difference:

$$\boxed{\Delta x_{eff} = 2nt \pm \frac{\lambda}{2}}$$

Use +Ξ»/2 if phase change occurs at one surface only.

3. Conditions for Constructive/Destructive

For normal incidence on film in air:

Constructive interference (bright):

$$\boxed{2nt = (m + \frac{1}{2})\lambda}$$

(m = 0, 1, 2, …)

Destructive interference (dark):

$$\boxed{2nt = m\lambda}$$

(m = 0, 1, 2, …)

Note: Opposite to standard YDSE due to phase change!

4. Anti-Reflection Coating

Goal: Minimize reflection, maximize transmission

Condition for minimum reflection:

$$\boxed{t = \frac{\lambda}{4n}}$$

For Ξ» = 550 nm (middle of visible spectrum), n β‰ˆ 1.38 (MgFβ‚‚):

$$t = \frac{550}{4 \times 1.38} \approx 100 \text{ nm}$$

Refractive index of coating:

$$\boxed{n_{coating} = \sqrt{n_{glass}}}$$

For glass (n = 1.5): n_coating β‰ˆ 1.22

5. Newton’s Rings

Setup: Plano-convex lens on glass plate

Air film of variable thickness between lens and plate

Thickness at distance r from center:

$$\boxed{t = \frac{r^2}{2R}}$$

Where R = radius of curvature of lens

Radius of nth dark ring:

$$\boxed{r_n = \sqrt{n\lambda R}}$$

Radius of nth bright ring:

$$\boxed{r_n = \sqrt{(n - \frac{1}{2})\lambda R}}$$

Central spot is dark (due to phase change at glass surface)

Common Traps & Mistakes

Trap #1: Path Difference Formula

❌ Wrong: Using exact formula when approximation is valid βœ… Right: For D » d and D » y, use Ξ”x = yd/d

Trap #2: Film Interference Phase Change

❌ Wrong: Forgetting Ο€ phase change in thin films βœ… Right: Add Ξ»/2 to path difference for single phase change

Trap #3: Fringe Width Direction

❌ Wrong: Thinking Ξ² changes with position on screen βœ… Right: All fringes equally spaced (Ξ² constant)

Trap #4: White Light Central Fringe

❌ Wrong: Central fringe is colored βœ… Right: Central fringe is WHITE, others are colored

Trap #5: Shift with Film

❌ Wrong: Pattern shifts away from slit with film βœ… Right: Pattern shifts TOWARDS slit with film

Practice Problems

Level 1: JEE Main Warmup

Problem 1.1: In YDSE, slits are 0.5 mm apart and screen is 1 m away. Light of wavelength 600 nm is used. Find fringe width.

Solution

Given: d = 0.5 mm = 0.5 Γ— 10⁻³ m, D = 1 m, Ξ» = 600 nm = 600 Γ— 10⁻⁹ m

$$\beta = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1}{0.5 \times 10^{-3}}$$ $$\beta = \frac{600 \times 10^{-9}}{0.5 \times 10^{-3}} = 1.2 \times 10^{-3} \text{ m} = 1.2 \text{ mm}$$

Answer: Ξ² = 1.2 mm

Problem 1.2: In the above setup, find the position of 3rd bright fringe.

Solution

For 3rd bright fringe: n = 3

$$y_3 = \frac{n\lambda D}{d} = \frac{3 \times 600 \times 10^{-9} \times 1}{0.5 \times 10^{-3}}$$ $$y_3 = 3.6 \times 10^{-3} \text{ m} = 3.6 \text{ mm}$$

Alternative: y_n = nΞ² = 3 Γ— 1.2 = 3.6 mm

Answer: 3.6 mm from center

Problem 1.3: What should be the thickness of anti-reflection coating (n = 1.4) for Ξ» = 560 nm?

Solution

For minimum reflection:

$$t = \frac{\lambda}{4n} = \frac{560 \times 10^{-9}}{4 \times 1.4} = \frac{560 \times 10^{-9}}{5.6} = 100 \times 10^{-9} \text{ m}$$

Answer: t = 100 nm

Level 2: JEE Main/Advanced

Problem 2.1: In YDSE with d = 1 mm, D = 1 m, Ξ» = 500 nm, a thin glass plate of thickness 0.01 mm and n = 1.5 is placed in front of one slit. Find: (a) Shift in fringe pattern (b) Position of central maximum

Solution

Given: d = 1 mm = 10⁻³ m, D = 1 m, Ξ» = 500 nm = 5 Γ— 10⁻⁷ m t = 0.01 mm = 10⁻⁡ m, n = 1.5

(a) Shift:

$$y_{shift} = \frac{(n-1)tD}{d} = \frac{(1.5-1) \times 10^{-5} \times 1}{10^{-3}}$$ $$y_{shift} = \frac{0.5 \times 10^{-5}}{10^{-3}} = 5 \times 10^{-3} \text{ m} = 5 \text{ mm}$$

(b) Position of central maximum:

Central maximum shifts by 5 mm towards the slit with glass plate.

If glass plate is in front of upper slit, central max is at y = +5 mm If in front of lower slit, central max is at y = -5 mm

Answers:

  • (a) 5 mm shift
  • (b) 5 mm from geometrical center (towards slit with film)

Problem 2.2: In YDSE, 5th bright fringe of one wavelength coincides with 6th bright fringe of another wavelength. If first wavelength is 600 nm, find second wavelength.

Solution

For same position:

$$y_5(\lambda_1) = y_6(\lambda_2)$$ $$\frac{5\lambda_1 D}{d} = \frac{6\lambda_2 D}{d}$$ $$5\lambda_1 = 6\lambda_2$$ $$\lambda_2 = \frac{5\lambda_1}{6} = \frac{5 \times 600}{6} = 500 \text{ nm}$$

Answer: Ξ»β‚‚ = 500 nm

Problem 2.3: In YDSE, intensity at point P is Iβ‚€/4 where Iβ‚€ is maximum intensity. Find path difference at P.

Solution
$$I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$$

Given I = Iβ‚€/4:

$$\frac{I_0}{4} = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$$ $$\cos^2\left(\frac{\phi}{2}\right) = \frac{1}{16}$$ $$\cos\left(\frac{\phi}{2}\right) = \pm \frac{1}{4}$$

Taking positive:

$$\frac{\phi}{2} = \cos^{-1}(1/4) \approx 75.5Β°$$ $$\phi = 151Β° = 151 \times \frac{\pi}{180} = 2.64 \text{ rad}$$

Path difference:

$$\Delta x = \frac{\lambda}{2\pi} \times \phi = \frac{\lambda}{2\pi} \times 2.64 = 0.42\lambda$$

Answer: Ξ”x β‰ˆ 0.42Ξ» (or using exact calculation)

Alternative approach:

$$\cos^2(\phi/2) = 1/16$$

This gives Ο†/2 = Β±75.5Β° β†’ Ο† β‰ˆ Β±151Β°

For first occurrence: Ξ”x = (151/360)Ξ» β‰ˆ 0.42Ξ»

Level 3: JEE Advanced

Problem 3.1: In Newton’s rings experiment with R = 100 cm and Ξ» = 600 nm, find: (a) Radius of 10th dark ring (b) Radius of 10th bright ring

Solution

Given: R = 100 cm = 1 m, Ξ» = 600 nm = 6 Γ— 10⁻⁷ m

(a) 10th dark ring (n = 10):

$$r_n = \sqrt{n\lambda R} = \sqrt{10 \times 6 \times 10^{-7} \times 1}$$ $$r_{10} = \sqrt{6 \times 10^{-6}} = 2.45 \times 10^{-3} \text{ m} = 2.45 \text{ mm}$$

(b) 10th bright ring:

$$r_n = \sqrt{(n - 1/2)\lambda R} = \sqrt{9.5 \times 6 \times 10^{-7} \times 1}$$ $$r_{10} = \sqrt{5.7 \times 10^{-6}} = 2.39 \times 10^{-3} \text{ m} = 2.39 \text{ mm}$$

Answers:

  • (a) 2.45 mm
  • (b) 2.39 mm

Problem 3.2: In YDSE, one slit provides 4 times intensity of other. Find ratio of maximum to minimum intensity.

Solution

Let Iβ‚‚ = I, then I₁ = 4I

$$I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{4I} + \sqrt{I})^2 = (2\sqrt{I} + \sqrt{I})^2 = 9I$$ $$I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 = (2\sqrt{I} - \sqrt{I})^2 = I$$ $$\frac{I_{max}}{I_{min}} = \frac{9I}{I} = 9$$

Answer: 9:1

Formula:

$$\frac{I_{max}}{I_{min}} = \frac{(\sqrt{r} + 1)^2}{(\sqrt{r} - 1)^2}$$

where r = I₁/Iβ‚‚ = 4

$$= \frac{(2+1)^2}{(2-1)^2} = \frac{9}{1} = 9$$

Quick Revision Cards

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β•‘  INTERFERENCE QUICK FACTS           β•‘
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YDSE:
🎯 β = λD/d (fringe width)
🎯 y_n = nλD/d (bright)
🎯 All fringes equally spaced
🎯 I_max = 4Iβ‚€, I_min = 0
🎯 Film shift = (n-1)tD/d

THIN FILMS:
🎯 Path diff = 2nt (normal incidence)
🎯 Phase change Ο€ (rarerβ†’denser reflection)
🎯 Anti-reflection: t = λ/4n
🎯 Newton's rings: r = √(nλR) (dark)

KEY POINTS:
🎯 Coherence essential
🎯 White light: central white, others colored
🎯 In medium: β becomes β/n

Cross-Topic Connections

Final Tips

  1. Always write d, D, Ξ» clearly - easy to confuse!
  2. Fringe width constant throughout screen
  3. Film shift: Pattern moves TOWARDS film
  4. Thin films: Remember phase change Ο€
  5. White light: Central is white!

Pro Tip: For maximum contrast, I₁ = Iβ‚‚ (equal slit widths)

Next up: Diffraction - bending around obstacles!