• 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

1. Coherent sources: Constant phase relationship
2. Monochromatic: Single wavelength (or narrow range)
3. 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:

1. Polarizing filter
2. Reflection (Brewster's angle)
3. 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:

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

Single Slit:

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

Thin Film:

1. Count phase changes (reflections)
2. Use $\lambda_n = \lambda/n$ in film
3. Path difference = 2t
4. 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!)

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.

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.

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.

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.

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!

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!