Interference and Diffraction

Young's double slit, single slit diffraction, diffraction grating, thin film interference

🌈 Interference and Diffraction

Wave Nature of Light

Light exhibits wave properties:

Interference (constructive and destructive)
Diffraction (bending around obstacles)
Polarization

These prove light is a wave!

Young's Double Slit

Two narrow slits illuminated by coherent light:

Path difference determines interference:

Bright fringes (constructive): $d \sin\theta = m\lambda$

Dark fringes (destructive): $d \sin\theta = \left(m + \frac{1}{2}\right)\lambda$

where:

d = slit separation
θ = angle from center
m = order (0, ±1, ±2, ...)
λ = wavelength

Small angle approximation (θ << 1 rad): $\sin\theta \approx \tan\theta \approx \frac{y}{L}$

Fringe spacing on screen: $\Delta y = \frac{\lambda L}{d}$

where L = distance to screen

💡 Key: Smaller slit spacing d → larger fringe spacing Δy!

Conditions for Interference

Coherent sources: Constant phase relationship
Monochromatic: Single wavelength (or narrow range)
Similar amplitudes: For clear pattern

Laser light is ideal!

Single Slit Diffraction

Single wide slit creates diffraction pattern:

Dark fringes: $a \sin\theta = m\lambda$

where:

a = slit width
m = ±1, ±2, ... (NOT zero!)

Central maximum: Brightest, twice as wide as other maxima

Width of central max: $w = \frac{2\lambda L}{a}$

💡 Key: Narrower slit a → wider central maximum!

Difference from double slit:

Double slit: Narrow bright fringes (interference)
Single slit: Wide central max, dimmer side maxima (diffraction)

Diffraction Grating

Many equally-spaced slits (100s to 1000s per mm):

Bright fringes (very sharp!): $d \sin\theta = m\lambda$

where d = spacing between slits

Advantages:

Very sharp, bright lines
Separates wavelengths well (spectroscopy!)
Higher orders (larger m) more spread out

Dispersion: Different λ at different angles

Red: longer λ, larger θ
Violet: shorter λ, smaller θ

Resolving power: $R = mN$ (N = number of slits)

Thin Film Interference

Light reflects from top and bottom surfaces of thin film:

Phase change on reflection:

Fast to slow (n_low to n_high): 180° = λ/2 shift
Slow to fast: No phase change

Constructive interference:

If ONE reflection has phase change: $2t = \left(m + \frac{1}{2}\right)\lambda_n$
If ZERO or TWO reflections have phase change: $2t = m\lambda_n$

where:

t = film thickness
$\lambda_n = \lambda/n$ = wavelength in film

Destructive is opposite!

Applications:

Soap bubbles (swirling colors)
Oil slicks on water
Anti-reflection coatings (destructive for reflection)

Polarization

Light wave with E field oscillating in one plane only.

Unpolarized → Polarized: Use polarizer (absorbs one component)

Malus's Law: Intensity through second polarizer $I = I_0 \cos^2\theta$

where θ = angle between polarizers

Crossed polarizers (θ = 90°): No light passes!

Ways to polarize:

Polarizing filter
Reflection (Brewster's angle)
Scattering (why sky is blue and polarized!)

Rayleigh Criterion (Resolution)

Two point sources barely resolved when:

$\theta_{min} = 1.22\frac{\lambda}{D}$

where D = aperture diameter

Smaller θ_min → better resolution

Larger aperture (telescope!) → better
Shorter wavelength (blue light) → better

Problem-Solving Strategy

Double Slit:

Bright: $d \sin\theta = m\lambda$
Use small angle: $\sin\theta \approx y/L$
Fringe spacing: $\Delta y = \lambda L/d$

Single Slit:

Dark: $a \sin\theta = m\lambda$ (m ≠ 0)
Central max width: $w = 2\lambda L/a$

Thin Film:

Count phase changes (reflections)
Use $\lambda_n = \lambda/n$ in film
Path difference = 2t
Add/subtract λ/2 for phase changes

Common Mistakes

❌ Confusing d (slit separation) with a (slit width) ❌ Using m = 0 for single slit dark fringe (m starts at ±1!) ❌ Forgetting λ_n = λ/n in thin films ❌ Not accounting for phase change on reflection ❌ Wrong units (nm vs m, degrees vs radians) ❌ Thinking larger slit spacing → larger fringe spacing (opposite!)

📚 Practice Problems

1Problem 1hard

❓ Question:

In a double-slit experiment, light with wavelength 600 nm passes through slits separated by 0.10 mm. A screen is 2.0 m away. (a) What is the distance between adjacent bright fringes? (b) What is the angle to the third-order bright fringe?

💡 Show Solution

Solution:

Given: λ = 600 nm = 6.0 × 10⁻⁷ m, d = 0.10 mm = 1.0 × 10⁻⁴ m, L = 2.0 m

(a) Distance between bright fringes: For small angles: Δy = λL/d Δy = (6.0 × 10⁻⁷)(2.0)/(1.0 × 10⁻⁴) Δy = (12 × 10⁻⁷)/(1.0 × 10⁻⁴) Δy = 1.2 × 10⁻² m or 1.2 cm

(b) Angle to third bright fringe (m = 3): d sin θ = mλ sin θ = mλ/d = 3(6.0 × 10⁻⁷)/(1.0 × 10⁻⁴) sin θ = (18 × 10⁻⁷)/(1.0 × 10⁻⁴) = 0.018 θ = 1.03° or 1.0°

Very small angle (small angle approximation valid).

2Problem 2medium

❓ Question:

Light of wavelength 500 nm passes through a single slit of width 0.050 mm. (a) What is the angular width of the central bright fringe? (b) If the screen is 1.5 m away, what is the linear width of the central maximum?

💡 Show Solution

Solution:

Given: λ = 500 nm = 5.0 × 10⁻⁷ m, a = 0.050 mm = 5.0 × 10⁻⁵ m, L = 1.5 m

(a) Angular width of central maximum: First minimum occurs at: a sin θ = λ sin θ = λ/a = (5.0 × 10⁻⁷)/(5.0 × 10⁻⁵) = 0.010 θ = 0.573° (to first minimum on one side)

Angular width = 2θ = 1.15° or 1.1° (center to both first minima)

(b) Linear width on screen: y = L tan θ ≈ L sin θ (small angle) y = (1.5)(0.010) = 0.015 m (to first minimum)

Total width = 2y = 3.0 cm (or 0.030 m)

Central maximum is twice as wide as other maxima in single-slit diffraction.

3Problem 3easy

❓ Question:

In a double slit experiment with 550 nm light, slits are 0.20 mm apart, and screen is 2.0 m away. What is the spacing between bright fringes?

💡 Show Solution

Given:

Wavelength: $\lambda = 550$ nm $= 5.5 \times 10^{-7}$ m
Slit separation: $d = 0.20$ mm $= 2.0 \times 10^{-4}$ m
Screen distance: $L = 2.0$ m

Solution:

Fringe spacing: $\Delta y = \frac{\lambda L}{d} = \frac{(5.5 \times 10^{-7})(2.0)}{2.0 \times 10^{-4}}$ $\Delta y = \frac{1.1 \times 10^{-6}}{2.0 \times 10^{-4}} = 5.5 \times 10^{-3} \text{ m}$ $\Delta y = 5.5 \text{ mm}$

Answer: Fringe spacing = 5.5 mm

Bright fringes are 5.5 mm apart on the screen.

4Problem 4easy

❓ Question:

In a double slit experiment with 550 nm light, slits are 0.20 mm apart, and screen is 2.0 m away. What is the spacing between bright fringes?

💡 Show Solution

Given:

Wavelength: $\lambda = 550$ nm $= 5.5 \times 10^{-7}$ m
Slit separation: $d = 0.20$ mm $= 2.0 \times 10^{-4}$ m
Screen distance: $L = 2.0$ m

Solution:

Answer: Fringe spacing = 5.5 mm

Bright fringes are 5.5 mm apart on the screen.

5Problem 5medium

❓ Question:

A single slit of width 0.050 mm produces a diffraction pattern with 600 nm light on a screen 1.5 m away. What is the width of the central bright maximum?

💡 Show Solution

Given:

Slit width: $a = 0.050$ mm $= 5.0 \times 10^{-5}$ m
Wavelength: $\lambda = 600$ nm $= 6.0 \times 10^{-7}$ m
Screen distance: $L = 1.5$ m

Solution:

Width of central maximum: $w = \frac{2\lambda L}{a} = \frac{2(6.0 \times 10^{-7})(1.5)}{5.0 \times 10^{-5}}$ $w = \frac{1.8 \times 10^{-6}}{5.0 \times 10^{-5}} = 0.036 \text{ m}$ $w = 36 \text{ mm} = 3.6 \text{ cm}$

Answer: Central maximum width = 3.6 cm

This is the distance between the first dark fringes on either side of center.

6Problem 6medium

❓ Question:

A single slit of width 0.050 mm produces a diffraction pattern with 600 nm light on a screen 1.5 m away. What is the width of the central bright maximum?

💡 Show Solution

Given:

Slit width: $a = 0.050$ mm $= 5.0 \times 10^{-5}$ m
Wavelength: $\lambda = 600$ nm $= 6.0 \times 10^{-7}$ m
Screen distance: $L = 1.5$ m

Solution:

Answer: Central maximum width = 3.6 cm

This is the distance between the first dark fringes on either side of center.

7Problem 7hard

❓ Question:

A soap film (n = 1.33) is 450 nm thick. What wavelength of visible light (in air) is most strongly reflected? Assume normal incidence.

💡 Show Solution

Given:

Film index: $n = 1.33$
Thickness: $t = 450$ nm
Air on both sides (n = 1.00)

Solution:

Step 1: Count phase changes.

Top surface: air (1.00) → film (1.33) = phase change (λ/2)
Bottom surface: film (1.33) → air (1.00) = no phase change

One phase change total

Step 2: Constructive interference (for ONE phase change): $2t = \left(m + \frac{1}{2}\right)\lambda_n = \left(m + \frac{1}{2}\right)\frac{\lambda}{n}$ $\lambda = \frac{2tn}{m + 1/2}$

Step 3: Try m = 0, 1, 2... to find visible wavelength (400-700 nm)

$m = 0$ : $\lambda = \frac{2(450)(1.33)}{0.5} = 2394 \text{ nm}$ (infrared, too long)

$m = 1$ : $\lambda = \frac{2(450)(1.33)}{1.5} = 798 \text{ nm}$ (infrared, too long)

$m = 2$ : $\lambda = \frac{2(450)(1.33)}{2.5} = 479 \text{ nm}$ (blue-green, visible!) ✓

Answer: λ = 479 nm (blue-green light)

This is why soap bubbles appear colored!

8Problem 8hard

❓ Question:

A soap film (n = 1.33) is 450 nm thick. What wavelength of visible light (in air) is most strongly reflected? Assume normal incidence.

💡 Show Solution

Given:

Film index: $n = 1.33$
Thickness: $t = 450$ nm
Air on both sides (n = 1.00)

Solution:

Step 1: Count phase changes.

Top surface: air (1.00) → film (1.33) = phase change (λ/2)
Bottom surface: film (1.33) → air (1.00) = no phase change

One phase change total

Step 2: Constructive interference (for ONE phase change): $2t = \left(m + \frac{1}{2}\right)\lambda_n = \left(m + \frac{1}{2}\right)\frac{\lambda}{n}$ $\lambda = \frac{2tn}{m + 1/2}$

Step 3: Try m = 0, 1, 2... to find visible wavelength (400-700 nm)

$m = 0$ : $\lambda = \frac{2(450)(1.33)}{0.5} = 2394 \text{ nm}$ (infrared, too long)

$m = 1$ : $\lambda = \frac{2(450)(1.33)}{1.5} = 798 \text{ nm}$ (infrared, too long)

$m = 2$ : $\lambda = \frac{2(450)(1.33)}{2.5} = 479 \text{ nm}$ (blue-green, visible!) ✓

Answer: λ = 479 nm (blue-green light)

This is why soap bubbles appear colored!

🎴

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