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°
Young's double slit, single slit diffraction, diffraction grating, thin film interference
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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!)
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: λ=550 nm =5.5×10−7 m
Slit separation: d=0.20 mm =2.0×10−4 m
Screen distance: L=2.0 m
Solution:
Fringe spacing:
Δy=dλL=
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: λ=550 nm =5.5×10−7 m
Slit separation: d=0.20 mm =2.0×10−4 m
Screen distance: L=2.0 m
Solution:
Fringe spacing:
Δy=dλL=
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 =5.0×10−5 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.
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 =5.0×10−5 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.
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=(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!
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=(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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