Student Reader

Interference and Diffraction

InterferenceInterference occurs when three criteria are met: monochromatic waves (all the same wavelength); coherent waves; and two rays.
Diffraction
Order (m or n)For interference: mint → maxima/peaks
Slit Distance (d or a)metersDistance between slits.
Slit Count (N)This refers to the number of slits in the grate.
Grate Width (w)metersThis refers to the width of the grate.
Screen Distance (L)metersDistance between slits and viewing screen.
Path Difference (δ)The path length difference is δ = L2 - L1 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 = Φ2 - Φ1 = (2 π d / λ ) sin θ

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
Spread for given circumstance: 2 · L · tan θ

Tangent is used because tan θ = y / L 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 ).
θAngle formed between the central axis and the line connecting the central point of the slits and a given fringe.
Interference IntensitySometimes a scenario can be in-between constructive and destructive. In these cases the intensity approach is used.

I = δ / λ = ΔΦ / 2 π

Interference: Iθ = I0 ( cos2 ½ Δ Φint )

Constructive Interference

δ = m λ ( m = 0, ±1, ±2 )
ΔΦ = m 2 π
Iθ = I0

Destructive Interference

(Off by ½ λ)
δ = ( m + ½ ) ½ λ
Δ Φ = ( 2 m + 1 ) π
Iθ = 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 )
Position of destructive (dark) interference: ( m + ½ ) λ = ( d sin θ )

Film Blocking Wavelength

With nfilm as the index of refraction of the film. You want to find the thickness of a thin film coating that eliminates a certain wavelength of light from appearing.

2 · t = [ λ (m + ½ ) ] / nfilm

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 nfilm )

Film Over SlitWith a film covering one or both slits, there is a change in the ΔΦ. Thus, the individual Φ1 and Φ2 must be calculated.

Φ = 2 · π · n t / λair

t = Thickness of what is covering the slit
n = Index of refraction for what is covering the slit
λair = Wavelength of the light in air

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.
And of course, the index of refraction of air is n = 1.

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

mdiff → minima
m = 0 is central point, which is bright
m = 1 is first dark point
m = 2 is second dark point

a sin θdark = m λ
½ Δ Φ = ( π a sin θdark ) / λ = m π
½ Δ Φ / π = ( Path Difference ) / λ
Iθ = I0 ( sin2 ½ Δ Φ ) / ( ½ Δ Φ )2
Iθ = I0 ( sin2 ( [ π d sin θdark ] / λ ) / ( π d sin θdark ] / λ )2

Diffraction: Iθ = I0 ( sin2 ½ Δ Φ ) / ( ½ Δ Φ )2

Multiply: Iθ = I0' ( cos2 ( ½ Δ Φint ) sin2 ( ½ Δ Φdiff ) ) / ( ½ Δ Φdiff )2

I0' = I02

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: m0 is a maximum at the middle, but m1, m-1, m2, m-2, etc are all minima. The distance from the central axis (m0) to m1 is y1 and so on. ym = 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
d = spacing between atoms
Φ = angle of ray to surface, not to normal to surface.
If λ / Δ R > R then you are unable to distinguish the 2 wavelengths

Invisible
Very short wavelength: 10-2 to 10 nanometers
No diffraction or interference using ordinary gratings
X-ray diffraction by crystal

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)
Distance away = θ · do

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

How many lines per centimeter does a grating have if the fourth order occurs at a 25.0 angle for 600 nm light?

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.

Light of wavelength 700 nm passes through a slit 2.0 µm wide and a single-slit diffraction pattern is formed vertically on a screen 40 cm away. Determine the light intensity I at 30 cm above the central maximum, expressed as a fraction of the central maximum's intensity.

½ Δ Φ = ( π a sin θ ) / λ
Iθ / I0 = ( sin2 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 = ( .402 + .302 )½ = .50
sin θ = opposite / hypotenuse = .30 / .50 = .60

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

Iθ / I0 = ( sin2 5.38558741 ) / ( 5.38558741 )2 = 0.0210746392 = 2.1%

White light containing wavelengths from 450 nm to 750 nm falls on a grating with 8000 lines per centimeter. How wide is the first-order spectrum on a screen 4 m away?

sin θ = λ m / d

Distance from central axis = L tan θ

λ1 = 450 · 10-9 m
λ2 = 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 λ1 and λ2. 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 θ1 = ( 450 · 10-9 ) / ( 1.25 · 10-6 ) = .360 → θ = 21.1°
sin θ2 = ( 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

A diffraction grating has 4 · 105 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 · 105 ) = 2.5 · 10-6 meters
λ1 = 720 · 10-9 meters
λ2 = 450 · 10-9 meters
magnitude = 2

sin θ1 = 720 · 10-9 · 2 / ( 2.5 · 10-6 ) = .288 → θ = 16.73825676°
sin θ2 = 440 · 10-9 · 2 / ( 2.5 · 10-6 ) = .176 → θ = 10.13685717°

Δθ = 16.73825676 - 10.13685717 = 6.60°

Let 500 nm light be incident normally on a diffraction grating for which d = 1200 nm. How many orders (principal maxima) are present?

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.

Continuing the problem above, if the grating is 2.00 cm wide, what is the full angular width of each principal maximum?
When green light of 510 nm falls on a diffraction grating, its first-order peak on a screen 50.0 cm away falls 4.00 cm from the central peak. How many lines are on the grating? Another source produces a line 5.00 cm from the central peak. What is its wavelength?

d sin θ = m λ

m = 1
λ1 = 510 · 10-9 m
λ2 = ?
θ1 = ?
θ2 = ?
d = ?

First find sin θ with a Pythagorean approach: hypotenuse = ( .042 + .502 )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 θ2 with a Pythagorean approach: hypotenuse = ( .052 + .502 )½ = 0.502493781
sin θ2 = opposite / hypotenuse = .05 / 0.502493781 = 0.099503719

6.39536746 · 10-6 · 0.099503719 = 1 · λ2
λ2 = 6.36362847 · 10-7 meters = 636 nm

How far away can a human eye distinguish two truck headlights 3.0 m apart? Consider only diffraction effects and assume a pupil diameter of 5.0 mm and a wavelength of 500 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

Two stars 20 light-years away are barely resolved by a 50 cm (mirror diameter) telescope. How far apart are the stars? Assume λ = 500 nm and that the resolution is limited by diffraction.

Angular distance = θ = 1.22 λ / Diameter
Diameter = .50 meters
λ = 500 × 10-9
do = 20 light years = 1.89210568 × 1017 meters

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

Distance apart = θ do = 1.22 × 10-6 × 1.89210568 × 1017 = 2.30836893 × 1011 meters

What is the minimum angular separation (in arcminutes) an eye could resolve when viewing two stars? Consider only diffraction effects and assume a pupil diameter of 5.0 mm and a wavelength of 500 nm.
A 5000 line/cm diffraction grating is 4.00 cm wide. If light with wavelengths near 500 nm falls on the grating, how close can two wavelengths be if they are to be resolved in any order? What order gives the best resolution?

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

In a double-slit experiment, if the central diffraction peak contains 10 interference fringes, how many fringes are contained within each secondary diffraction peak (between ±1 and ±). Assume the first diffraction minimum occurs at an interference minimum.
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