Young's double slit, single slit diffraction, diffraction grating, thin film interference
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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Interference and Diffraction is part of the AP Physics 2 course on Study Mondo, specifically in the Optics section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Interference and Diffraction?
dsinθ=(m+21)λ
where:
d = slit separation
θ = angle from center
m = order (0, ±1, ±2, ...)
λ = wavelength
Small angle approximation (θ << 1 rad):
sinθ≈tanθ≈Ly
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:
asinθ=mλ
where:
a = slit width
m = ±1, ±2, ... (NOT zero!)
Central maximum: Brightest, twice as wide as other maxima
Width of central max:
w=a2λL
💡 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!):
dsinθ=mλ
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=(m+21)λn
If ZERO or TWO reflections have phase change:
2t=mλn
where:
t = film thickness
λn=λ/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=I0cos2θ
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:
θmin=1.22Dλ
where D = aperture diameter
Smaller θ_min → better resolution
Larger aperture (telescope!) → better
Shorter wavelength (blue light) → better
Problem-Solving Strategy
Double Slit:
Bright: dsinθ=mλ
Use small angle: sinθ≈y/L
Fringe spacing: Δy=λL/d
Single Slit:
Dark: asinθ=mλ (m ≠ 0)
Central max width: w=2λL/a
Thin Film:
Count phase changes (reflections)
Use λn=λ/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!)
=5.5×10−7
Slit separation: d=0.20 mm =2.0×10−4 m
Screen distance: L=2.0 m
Solution:
Fringe spacing:
Δy=dλL=2.0×10−4(5.5×10−7)(2.0)Δy=2.0×10−41.1×10Δy=5.5 mm
Answer: Fringe spacing = 5.5 mm
Bright fringes are 5.5 mm apart on the screen.
2Problem 2medium
❓ 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 =5.0×10−5 m
Wavelength: λ=600 nm =6.0×10−7 m
Screen distance: L=1.5 m
Solution:
Width of central maximum:
w=a2λL=
Answer: Central maximum width = 3.6 cm
This is the distance between the first dark fringes on either side of center.
3Problem 3hard
❓ 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=(m+21)λ
Step 3: Try m = 0, 1, 2... to find visible wavelength (400-700 nm)
m=0:
λ=0.52(450)(1.33) (infrared, too long)
m=1:
λ=1.52(450)(1.33) (infrared, too long)
m=2:
λ=2.52(450)(1.33) (blue-green, visible!) ✓
Answer: λ = 479 nm (blue-green light)
This is why soap bubbles appear colored!
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Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.