Chapter

Interference and Diffraction

Blinn College - Physics 1402 - Terry Honan

. - Simple Interference

As an electromagnetic radiation passes some position the electric field varies sinusoidally with time.

A cos(ω t+ϕ).

where the amplitude A is the peak electric field, A=(Subscript[E, max]). When we combine waves from two sources, we add the electric fields for each wave; this is known as the principle of superposition. Suppose we combine waves from two sources with the same amplitude (Subscript[A, 0]) and frequency, but with a different relative phase angle ϕ.  To do this we just add the two functions of time

(Subscript[A, 0]) cos ω t+(Subscript[A, 0]) cos(ω t+ϕ).

If the two waves are in phase ϕ=0 we get constructive interference and

(Subscript[A, 0]) cos ω t+(Subscript[A, 0]) cos ω t=2 (Subscript[A, 0]) cos ω t

and the resulting amplitude of the combined wave is  A=2 (Subscript[A, 0]). If the two waves are totally out of phase ϕ=π we get destructive interference.  In the destructive case, since cos(ω t+π)=-cos ω t, we get

(Subscript[A, 0]) cos ω t+(Subscript[A, 0]) cos(ω t+π)=0;

the resulting combined amplitude is A=0.

The more general result is a bit more complicated; we will just state the result and not use it beyond here. The essential fact is that constructive and destructive interference are not the only possibilities; they are the limiting cases of something more general.

(Subscript[A, 0]) cos ω t+(Subscript[A, 0]) cos(ω t+ϕ)=A cos(ω t+(ϕ/2))  where A=2(Subscript[A, 0]) cos (ϕ/2)

The intensity is proportional to the square of the peak electric field and this is just the amplitude.

I=(1/2)(Subscript[ε, 0])c (Subscript[E, max]^2)     ⟹    I∝(A^2)

If two flashlights, each with intensity (Subscript[I, 0]) shine at the same point the resulting intensity is 2 (Subscript[I, 0]); this is incoherent mixing, meaning that the phase of light of one source is not related to the phase of the other.  The key to interference is coherence; the phases of the two sources are related.  The following table summarizes the amplitudes and intensities for the special cases of constructive and destructive interference, for the general case and for incoherent mixing.  In all cases the amplitude and intensity of the uncombined waves are (Subscript[A, 0]) and (Subscript[I, 0]).

({{,    Amplitude, Intensity   }, {Constructive Interference
ϕ=0, 2 (Subscript[A, 0]), 4 (Subscript[I, 0])}, {Destructive Interference
ϕ=π, 0, 0}, {General Case
(Any ϕ), 0≤A≤2 (Subscript[A, 0]), 0≤I≤4 (Subscript[I, 0])}, {Incoherent Mixing 
 (two flashlights) , -, 2 (Subscript[I, 0])}})

What does this look like? Constructive interference corresponds to bright spots, high intensity. Destructive interference has zero intensity and it this a dark spot. If the phase difference is ϕ=m 2π, where m is any integer then the sinusoidal functions are in phase and that is constructive interference. If the phase difference is ϕ=m 2π+π=(m+1/2) 2π then they destructively interfere as well.

Interference applies to any type of wave. Suppose you have sound from two speakers that leaves each speaker in phase and with the same amplitude. If you are a distance (Subscript[r, 1]) from one speaker and (Subscript[r, 2]) from the other then even though the sound starts off in phase, they travel different distances and end up out of phase. If (Subscript[r, 1])=(Subscript[r, 2]) then the waves are in phase and that then is a loud spot, where loud is high intensity. If (Subscript[r, 2])-(Subscript[r, 1])=λ/2, where λ is the sound wavelength, then the waves destructively interfere and there is no sound there.

Example . - A Middle-C Note from Two Speakers

λ=((Subscript[v, sound])/f)=1.31(QuantityUnit)

Solution

(Subscript[r, 2])-(Subscript[r, 1])=m λ=m 1.31(QuantityUnit)  for  m=0,±1,±2,±3,…  
or  (Subscript[r, 2])-(Subscript[r, 1])=0,±1.31(QuantityUnit),±2.61(QuantityUnit),±3.92(QuantityUnit),…

Solution

(Subscript[r, 2])-(Subscript[r, 1])=(m+(1/2)) λ=(m+(1/2)) 1.31(QuantityUnit)  for  (m+(1/2))=±(1/2),±(3/2),±(5/2),…  
or  (Subscript[r, 2])-(Subscript[r, 1])=±1.96(QuantityUnit),±3.27(QuantityUnit),±4.57(QuantityUnit),…

. - Young's Double-slit Experiment

Consider monochromatic light of wavelength λ normally incident on a pair of narrow vertical slits each of width W and separated by d, the distance between the centers of the slits.  We will make the simplifying assumption of narrow slits, or W≪d.  We will take L to be the distance from the slits to some distant screen; by distant we mean that L is much larger than both d and λ.  A point P on the screen can be labeled by θ, the angle of deflection or by y, the distance along the screen, where y=0 and θ=0 describe undeflected rays.  θ and y are related by

tan θ=(y/L).

The distance from one slit to P is (Subscript[r, 1]) and (Subscript[r, 2]) is the distance from the other slit to P.  Each r is much larger than d but their difference δ is on the same order.  Using trig and the approximation that d≪r we get δ=d sin θ

δ=(Subscript[r, 2])-(Subscript[r, 1])=d sin θ.

Constructive and Destructive Interference

Huygen's principle states that every point on a wave front behaves as a separate point source.  When a plane wave hits a pair of narrow slits then each slit represents a different source.  The key point is that each source is in phase with the other; this coherence is the key to interference.  If the two distances (Subscript[r, 1]) and (Subscript[r, 2]) are equal (so δ=0) then the light rays hitting P are in phase and there is constructive interference.  Moreover, if the path difference δ is an integer number of wavelengths there is constructive interference.  If the path difference is half a wavelength then the rays are out of phase and there is destructive interference.

Using that the path difference is d sin θ and taking m to be any integer we can write the conditions for constructive and destructive interference.

d sin θ= m λ                     (constructive interference)

d sin θ= (m+(1/2))  λ           (destructive interference)

where m=0,±1, ±2,...  and  m+(1/2)=±(1/2),±(3/2),±(5/2),… .

Graph of Intensity

Example . - Young’s Double-slit Experiment

Solution

d=0.0165×(10^(-3))(QuantityUnit) ,  L=3.5(QuantityUnit)  and  λ=632.8×(10^(-9))(QuantityUnit)

(1^st) dark fringe:   θ=(sin^(-1))((1/2)(λ/d))=1.10°  ⟹  y=L tan θ=0.067(QuantityUnit)

(2^nd) dark fringe:   θ=(sin^(-1))((3/2)(λ/d))=3.30°  ⟹  y=L tan θ=0.202(QuantityUnit)

(1^st) bright fringe:   θ=(sin^(-1))(1(λ/d))=2.20°  ⟹  y=L tan θ=0.134(QuantityUnit)

(2^nd) bright fringe:   θ=(sin^(-1))(2(λ/d))=4.40°  ⟹  y=L tan θ=0.269(QuantityUnit)

Solution

(Subscript[λ, red])=700×(10^(-9))(QuantityUnit)  and  (Subscript[λ, violet])=400×(10^(-9))(QuantityUnit)

(Subscript[θ, red])=(sin^(-1))(((Subscript[λ, red])/d))=2.43°  and  (Subscript[θ, violet])=(sin^(-1))(((Subscript[λ, violet])/d))=1.39°  ⟹  (Subscript[θ, red])-(Subscript[θ, violet])=1.04°

. - Many Slits and Diffraction Gratings

Now consider the case of N narrow slits.  The following graphs show the intensity graphs.  The graphs on the right is an enlargement of one period of the graph where the intensity is multiplied by 10.  Note that the condition for constructive interference is the same but as the number of slits increases the intensity goes to zero at all positions in between.

A diffraction grating is the limit as N becomes large.  The interference conditions for a diffraction grating is the same as for the double slit for constructive interference but everywhere else there is destructive interference. There is a crisp pattern of lines representing the bright fringes.

d sin θ= m λ  for constructive interference

and destructive interference elsewhere.

When a laser, which has monochromatic light, is shot through a diffraction with vertical slits, you observe a series of horizontally-spaced dots representing the constructive interference. The difference from the double slit is that it is dark everywhere else.

Note that the position of the (m^th) maximum varies with wavelength. If white light passes through a diffraction grating then the central fringe (m=0) is the same for all wavelengths and thus is white, but the higher order fringes will break light into its spectrum.  Diffraction gratings are better for spectroscopy (resolving light into its constituent wavelengths) than prisms. With a prism, red is the color that refracts the least. This is reversed with a diffraction grating; since red has the longest wavelength, it diffracts the most.

Example . - The Visible Spectrum and the m=1 Fringe

Solution

d sin θ= m λ   ⟹  sin θ=((m λ)/d)   ⟹   ((m 400(QuantityUnit))/d)≤sin θ≤((m 700(QuantityUnit))/d)

Example . - Diffraction Grating

Solution

d=(1/(230 lines/(QuantityUnit)))=((1(QuantityUnit))/230)=(((10^(-3))(QuantityUnit))/230)=4.348×(10^(-6))(QuantityUnit)

λ=632.8(QuantityUnit)   and   L=5.8(QuantityUnit)

for m=1, θ=(sin^(-1))(1 (λ/d))=8.37°  ⟹ y=L tan θ=0.853(QuantityUnit)

for m=2, θ=(sin^(-1))(2 (λ/d))=16.9°  ⟹ y=L tan θ=1.76(QuantityUnit)

for m=3, θ=(sin^(-1))(3 (λ/d))=25.9°  ⟹ y=L tan θ=2.81(QuantityUnit)

Solution

d sin θ= m λ   ⟹  m (λ/d)=sin θ<1   ⟹   m<(d/λ)=6.87

(Subscript[m, max])=6

total number of fringes=2 (Subscript[m, max])+1=13

. - Single-slit Diffraction

When light passes through a single narrow slit of width W one observes a diffraction pattern.  Here each point in the slit is a separate source.  It is simple to find the condition for destructive interference.

Consider two rays from points separated by W/2 in the slit. The difference of the path lengths δ is given by

δ=(Subscript[r, 2])-(Subscript[r, 1])=(W/2) sin θ

If this distance is half a wavelength.

δ=(W/(2 )) sin θ=(λ/2).

then those two rays destructively interfere. But if this condition is met then the ray leaving any point in the slit will be exactly canceled by the ray from W/2 away and this is the condition for destructive interference for all the rays.

This argument fails if we break the slit into thirds, because there would always be one ray that is not canceled. However if we break the slit into any even number of pieces 2 m, then light from a point (W/(2 m)) away would destructively interfere and all the rays could be canceled

δ=(W/(2 m)) sin θ=(λ/2).

The condition for destructive interference is then

W sin θ= m λ,  where m=±1, ±2, ±3, ... (destructive interference)

Below is the graph of the intensity pattern for a single slit.  Note that the m=0 position is not destructive interference; it is the center of the central bright fringe.

The central bright fringe is from m=-1 to m=1 and has twice the width of the other bright fringes. It is also much brighter.

As an aside, the intensity formula for the above graph is

I=(((Subscript[I, max]) [((sin[(π W/λ) sin θ])/((π W/λ) sin θ))])^2).

Example . - SIngle-slit Diffraction

Solution

W=0.075×(10^(-3))(QuantityUnit) ,  L=3.5(QuantityUnit)  and  λ=632.8(QuantityUnit)

W sin θ=m λ  ⟹  W sin (Subscript[θ, 1])=1 λ  ⟹  (Subscript[θ, 1])=(sin^(-1))((λ/W))=0.483°   ⟹ (Subscript[y, 1])=L tan (Subscript[θ, 1])=0.0295(QuantityUnit)
 ⟹ 2 (Subscript[y, 1])=0.0581(QuantityUnit)

Solution

W sin θ=m λ  ⟹  W sin (Subscript[θ, 5])=5 λ  ⟹  (Subscript[θ, 5])=(sin^(-1))(((5λ)/W))=2.42°   ⟹ (Subscript[y, 5])=L tan (Subscript[θ, 5])=0.148(QuantityUnit)