# Interference and Diffraction

By Levi Clancy for Student Reader on *updated *

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Interference | Interference occurs when three criteria are met: monochromatic waves (all the same wavelength); coherent waves; and two rays. | |
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Diffraction | ||

Order (m or n) | For interference: m_{int} → maxima/peaks | |

Slit Distance (d or a) | meters | Distance between slits. |

Slit Count (N) | This refers to the number of slits in the grate. | |

Grate Width (w) | meters | This refers to the width of the grate. |

Screen Distance (L) | meters | Distance between slits and viewing screen. |

Path Difference (δ) | The path length difference is δ = L_{2} - L_{1} and is relevant when two waves are travelling in the same medium, like in the double slit experiment. | |

Phase Angle (ΔΦ) | The phase difference (phase angle) is relevant when waves travel in two different mediums like thin film. ΔΦ = Φ | |

Location (x or y) | meters (m) | This refers to the location on the screen. If you know θ and L then: Distance from central axis to location: L · tan θ Regarding the spread, multiplication by 2 is due to symmetry around the central axis. Thus, the total spread for that angle will be 2 · ( distance from central axis ). |

θ | Angle formed between the central axis and the line connecting the central point of the slits and a given fringe. | |

Interference Intensity | Sometimes a scenario can be in-between constructive and destructive. In these cases the intensity approach is used. I = δ / λ = ΔΦ / 2 π Interference: I |

###### Constructive Interference

δ = m λ ( m = 0, ±1, ±2 )

ΔΦ = m 2 π

I_{θ} = I_{0}

###### Destructive Interference

(Off by ½ λ)

δ = ( m + ½ ) ½ λ

Δ Φ = ( 2 m + 1 ) π

I_{θ} = 0

Double-Slit | When a double-slit experiment is presented and no detail is provided on whether to study interference or diffraction, then assume that only interference effects are involved. λ = ( y d ) / ( m L ) Position of constructive (bright) interference: m λ = ( d sin θ ) | |
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Film Blocking Wavelength | With n 2 · t = [ λ (m + ½ ) ] / n t = Thickness of film. We usually choose m = 0, the lowest magnitude, since that has the widest angle for destructive interference. Thus the equation simplifies to: thickness = λ / ( 4 n | |

Film Over Slit | With a film covering one or both slits, there is a change in the ΔΦ. Thus, the individual Φ _{1} and Φ_{2} must be calculated.Φ = 2 · π · n t / λ t = Thickness of what is covering the slit ΔΦ = The difference between the two values of Φ that are obtained. If one slit has nothing covering it, then imagine that it is covered with a film of air with the same thickness as the film. | |

Angular Width (Δθ) | ( 2 λ ) / ( w cos θ )This provides the angular width at a particular angle; to find the angular width at a given order, then first use d sin θ = m λ to find the θ value. |

## Diffraction

m_{diff} → minima

m = 0 is central point, which is bright

m = 1 is first dark point

m = 2 is second dark point

a sin θ_{dark} = m λ

½ Δ Φ = ( π a sin θ_{dark} ) / λ = m π

½ Δ Φ / π = ( Path Difference ) / λ

I_{θ} = I_{0} ( sin^{2} ½ Δ Φ ) / ( ½ Δ Φ )^{2}I_{θ} = I_{0} ( sin^{2} ( [ π d sin θ_{dark} ] / λ ) / ( π d sin θ_{dark} ] / λ )^{2}

Diffraction: I_{θ} = I_{0} ( sin^{2} ½ Δ Φ ) / ( ½ Δ Φ )^{2}

Multiply: I_{θ} = I_{0}' ( cos^{2} ( ½ Δ Φ_{int} ) sin^{2} ( ½ Δ Φ_{diff} ) ) / ( ½ Δ Φ_{diff} )^{2}

I_{0}' = I_{0}^{2}

When you have both diffraction and interference, then you'll have a big curve with the minima, then within the curves you'll have tiny ones filling it out as well corresponding to the interference.

The diffraction pattern: m_{0} is a maximum at the middle, but m_{1}, m_{-1}, m_{2}, m_{-2}, etc are all minima. The distance from the central axis (m_{0}) to m_{1} is y_{1} and so on. y_{m} = L sin θ = m λ L / d

Diffraction → only consider path difference = d sin θ = m λ (dark pts)

m = 0 → θ = 0 and is a maxima. So m = ± 1, ± 2 etc etc

X-Ray Diffraction Bragg Diffraction | Path difference: 2 d sin Φ = m λ Invisible | |
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Angular Distance | Sometimes you will be asked about whether something a certain distance away can be resolved. In that event, use these equations. The angular distance is how far apart two objects are, and whether these objects can be distinguished. The distance away is simply how far away these objects can be and still be distinguished at their particular angular distance. Angular distance = θ = 1.22 λ / Diameter ( note that θ will be in radians) | |

Resolving Wavelengths | To determine whether a diffraction grating can distinguish two wavelengths of light, set Δλ as their difference and let m to be the minimum order needed to distinguish the two wavelengths. λ / Δλ ≤ N m To find the best order to resolve the two wavelengths, then simply use d sin θ = m λ and to find the largest order (smallest possible resolvable wavelength) then let sin θ be its maximum, which is sin θ = 1 unless specified otherwise. |

## Practice problems

d sin θ = m λ

d = distance between slits

θ = 25.0°

m → fourth order = 4

λ = 600 · 10^{-9} m

d = 4 · 600 · 10^{-9} / ( sin 25.0 ) = 5.6788838 × 10^{-6} meters

The distance between slits is 5.6788838 × 10^{-6} meters.

There are 1 / ( 5.6788838 × 10^{-6} ) = 176090.942 slits per meter.

There are 1760 slits per centimeter.

½ Δ Φ = ( π a sin θ ) / λ

I_{θ} / I_{0} = ( sin^{2} ½ Δ Φ ) / ( ½ Δ Φ )^{2}λ = 700 · 10^{-9} m

a = 2.0 · 10^{-6} m

L = .40 m

y = .30 m

θ = ?

First find sin θ with a Pythagorean approach: hypotenuse = ( .40^{2} + .30^{2} )^{½} = .50

sin θ = opposite / hypotenuse = .30 / .50 = .60

½ Δ Φ = ( π · 2.0 · 10^{-6} · .60 ) / ( 700 · 10^{-9} ) = 5.38558741

I_{θ} / I_{0} = ( sin^{2} 5.38558741 ) / ( 5.38558741 )^{2} = 0.0210746392 = 2.1%

sin θ = λ m / d

Distance from central axis = L tan θ

λ_{1} = 450 · 10^{-9} m

λ_{2} = 750 · 10^{-9} m

d = .01 meters / 8000 lines = 1.25 · 10^{-6} m

L = 4 m

Generally, whenever you are given two wavelengths then you should find the value in question for each then present the difference as your answer. When asking about the width of the first-order spectra, then you know you must find the locations of the first-order magnitude for λ_{1} and λ_{2}. The answer is the difference between these two locations; this is the width of the overlap of their first-order spectra. The width of the first-order spectrum is equivalent to Δy, the width of the area where there is overlap of first-order spectra for 450 nm and 750 nm.

sin θ_{1} = ( 450 · 10^{-9} ) / ( 1.25 · 10^{-6} ) = .360 → θ = 21.1°

sin θ_{2} = ( 750 · 10^{-9} ) / ( 1.25 · 10^{-6} ) = .540 → θ = 32.7°

4 · tan 21.1° = 1.543471544 meters is distance from central axis at m = 1 for 450 nm

4 · tan 32.7° = 2.567954361 meters is distance from central axis at m = 1 for 750 nm

2.567954361 - 1.543471544 = 1.0 meters is the width of the first-order spectrum

^{5}lines per meter. Find the angular spread in the second-order spectrum between red light of wavelength 720 nm and blue light of wavelength 440 nm.

sin θ = λ m / d

d = 1 / ( 4 · 10^{5} ) = 2.5 · 10^{-6} meters

λ_{1} = 720 · 10^{-9} meters

λ_{2} = 450 · 10^{-9} meters

magnitude = 2

sin θ_{1} = 720 · 10^{-9} · 2 / ( 2.5 · 10^{-6} ) = .288 → θ = 16.73825676°

sin θ_{2} = 440 · 10^{-9} · 2 / ( 2.5 · 10^{-6} ) = .176 → θ = 10.13685717°

Δθ = 16.73825676 - 10.13685717 = 6.60°

d sin θ = m λ

λ = 500 · 10^{-9} m

d = 1200 · 10^{-9} m

It is a trigonometric property that -1 ≤ sin θ ≤ 1

We let sin θ be its maximum value → sin θ = 1

From here we find the value of m at its maximum

1200 · 10^{-9} · 1 = m · 500 · 10^{-9}m = 2.4

It might seem 2.4 is the greatest order possible, but the order can only be a whole number.

Thus, the greatest order is actually 2, which is the largest whole number not larger than 2.4.

However, the question asks how many orders are present.

Counting them up ( -2, -1, 0, 1, 2 ) the answer is 5.

d sin θ = m λ

m = 1

λ_{1} = 510 · 10^{-9} m

λ_{2} = ?

θ_{1} = ?

θ_{2} = ?

d = ?

First find sin θ with a Pythagorean approach: hypotenuse = ( .04^{2} + .50^{2} )^{½} = 0.501597448

sin θ = opposite / hypotenuse = .04 / 0.501597448 = 0.0797452223

d · 0.0797452223 = 1 · 510 · 10^{-9}d = 6.39536746 · 10^{-6} → 1 / ( 6.39536746 · 10^{-6} ) = 156363.181 lines / meter = 1560 lines / centimeter

Next, find sin θ_{2} with a Pythagorean approach: hypotenuse = ( .05^{2} + .50^{2} )^{½} = 0.502493781

sin θ_{2} = opposite / hypotenuse = .05 / 0.502493781 = 0.099503719

6.39536746 · 10^{-6} · 0.099503719 = 1 · λ_{2}λ_{2} = 6.36362847 · 10^{-7} meters = 636 nm

θ = 1.22 λ / d

d = .006 m

λ = 500 · 10^{-9} m

Z = Max distance of objects from human eye

We are not provided with θ, so in this case we will use the small angle approximation and trigonometry and proceed normally.

θ ≈ sin θ = 3.0 / Z = 1.22 · 500 · 10^{-9} / .005

3.0 / Z = 122 · 10^{-6}Z = 24590.1639 m = 24.6 km

Angular distance = θ = 1.22 λ / Diameter

Diameter = .50 meters

λ = 500 × 10^{-9}d_{o} = 20 light years = 1.89210568 × 10^{17} meters

θ = 1.22 × 500 × 10^{-9} / .50 = 1.22 × 10^{-6} radians

Distance apart = θ d_{o} = 1.22 × 10^{-6} × 1.89210568 × 10^{17} = 2.30836893 × 10^{11} meters

λ / Δλ = N m

d sin θ = m λ

λ = 500 · 10^{-9}N = 5000 * 4.00 = 20,000 lines

d = 1 centimeter / 5000 lines = .01 / 5000 = 2 · 10^{-6} m

Δλ = Distance between wavelengths that can still be resolved.

m = Largest order possible, which is where two wavelengths can be closest and still be resolved.

It is a trigonometric property that -1 ≤ sin θ ≤ 1

We let sin θ be its maximum value → sin θ = 1

From here we find the value of m at its maximum

d · 1 = m λ

2 · 10^{-6} = m · 500 · 10^{-9}m = 4 → this is already an integer, so we do not need to round it down.

Now it is possible to find Δλ which is how close two wavelengths can be and still be distinguished.

500 · 10^{-9} / Δλ = 20,000 · 4

Δλ = .00625 · 10^{-9} = .00625 nm

Finally it is possible to find the optimal order to resolve the wavelengths, by substituting back in Δλ

( 500 · 10^{-9} ) / ( .00625 · 10^{-9} ) = 20,000 · m

80,000 = 20,000 · m

m = 4