22: Two-Slit Interference

INTRO:
As Richard Feynman stated in his book on quantum mechanics,

Interference contains the heart and soul of quantum mechanics.

In fact, interference is a phenomenon of classical waves, easily perceived with sound or light waves. (It contains the soul of quantum mechanics only after you swallow the preposterous notion that particles in motion are described by a wave equation rather than the laws of Newtonian mechanics.)

In this problem, you will look at a classic wave interference problem involving electromagnetic waves. Young's double-slit experiment provided an irrefutable demonstration of the wave nature of light and is certainly one of the most elegant experiments in physics (because it demonstrates the important concept of interference so simply). For the purposes of this problem, we assume that two long parallel slits extending along the z axis (out of the plane shown in the figure) are separted by a distance d. They are illuminated coherently, that is, in phase, by light with a wavelength λ, for example by a laser beam polarized in the z direction. (Lacking a laser, Young used an intense source diffracted by a slit to produce coherent illumination of his double slits.)

The key point is that the electric field far downstream from the slit (e.g., at a large positive x value) is the sum of the electric fields emanating from each of the two slits. Hence, the relative phase of these electric fields at some observation point O determines whether they add in phase (constructively) or out or phase (destructively).(Figure 1)

To refresh your memory about traveling waves, the electric field E(x,t) that is incident on the double slits from the left is a function of x and t. Let us assume that it has amplitude Eleft. We will also assume a cosine trigonometric function with the arbitrary phase set equal to zero (i.e., at the point x=0, t=0 you find that E=E_left). Then
E(x,t)=E_leftcos[2π/λ (x−ct)].

The argument of the cosine function is the phase Φ(x,t). It can be written as Φ(x,t)=kx−ωt, where ω is the angular frequency (ω=2πf) and k is the wave number defined by k=2π/λ. The phase increases by 2π each time the distance increases by λ. Moreover, the phase is constant for an observer moving in the positive x direction at the speed of light, i.e., for whom x = ct.

Figure 1

Figure 2

PART A:
(Figure 2) Now consider the electric field observed at a point O that is far from the two slits, say at a distance r from the midpoint of the segment connecting the slits, at an angle θ from the x axis. Here, far means that r≫d, a regime sometimes called Fraunhofer diffraction.

The critical point is that the distances from the slits to point O are not equal; hence the waves will be out of phase due to the longer distance traveled by the wave from one slit relative to the other. Calculate the phase Φ_lower(O,t) of the wave from the lower slit that arrives at point O.

Express your answer in terms of d, θ, λ, c, r, t, and constants like π.

SOLUTION:
Φ_lower(O, t) = 2π/λ (r−ct) + π/λ dsin(θ)

Part B:
Now calculate the phase Φ_upper(O,t) of the wave from the upper slit that arrives at point O.
Express your answer in terms of d, θ, λ, c, r, t, and constants such as π.

SOLUTION:
Φupper(O, t) = 2π/λ (r−ct) - π/λ dsin(θ)

NOTE:
In order to make the math as simple as possible, we will define two phases:
ϕ = 2π/λ (r−ct) and δϕ=π/λ dsin(θ).
Then Φ_lower= ϕ + δϕ and Φ_upper= ϕ − δϕ.

Part C:
Assuming that the maximum amplitude of the field at point O for a wave from midway between the slits is E(r), now find the magnitude of the combined field E at O due to the two slits. You may ignore variations in the maximum amplitude and consider only variations in phase of the waves emerging from the slits.
Express your answer in terms of E(r), ϕ, and δϕ.

SOLUTION:
so, begin by finding both of the magnitudes of the electric field due to the slits
since E(x,t)=E_leftcos[Φ(x,t)] & Φlower(O, t) = ϕ + δϕ
Elower(O, t) = E(r) cos[ϕ + δϕ]
& with Φupper(O, t) = ϕ - δϕ,
Eupper(O, t) = E(r) cos[ϕ - δϕ]
To get the magnitude of the combined field at O, simply add the two separate fields
E = E(r) cos[ϕ + δϕ] + E(r) cos[ϕ - δϕ]
= E(r)[cos(ϕ + δϕ)+cos(ϕ - δϕ)]
If we use the identity cos(A+B)+cos(A-B)=2cos(A)cos(B),
we're able to fully simplify the answer to E = 2E(r)cos(ϕ)cos(δϕ)

Part D:
The key aspect of two-slit interference is the dependence of the total intensity at point O on the angle θ. Find this intensity I(θ).
The formula for intensity is

I=ϵ₀c(amplitude of E)²⋅½.

Express your answer in terms of I_max, θ, d, and λ, where I_max=2ε₀cE(r)². Note: cos²x should be coded as cos(x)^2.

SOLUTION:
Recall that E can be written as the product of the amplitude and the cosine of the phase, where the phase depends on time. In the above expression for E, the cos(ϕ) term is a function of time, whereas the rest of the variables/functions are not. Therefore, the amplitude of E is 2E(r)cos(δϕ). 
To find the dependence of I on θ, recall that δϕ = π/λ dsin(θ) and substitute for that in the equation
this results in 2E(r)cos(π/λ dsin(θ))

we have a formula in the book that states that I_{double slit}=4I_maxcos²[πdy⋅1/(λL)] and also one that states I_{double slit}=4I_maxcos²(θ)

using these we are able to determine that I(θ) = I_maxcos²(π/λ dsinθ)

Part E:
Two-slit interference is usually observed at small angles, and thus sin(θ) can be replaced by just θ. In this limit, the important observable is the spacing between successive minima (or maxima) of the interference pattern. Find the angular spacing Δθ of the interference pattern.
Express your answer in terms of d, λ, and any needed constants.

SOLUTION:
The intensity has the form Imaxcos2(π/λ dsinθ)
Since this depends on a cosine squared, the phase difference between two maxima or minima is π, not 2π, as it would be for a simple cosine function. Thus, to find the angular separation, you must set the difference in phase between two different directions equal to π:
(π/λ dsinθ₂)−(π/λ dsinθ₁)=π.
so
(πdsinθ₂)−(πdsinθ₁)=πλ
→dsinθ₂−dsinθ₁=λ
→sinθ₂−sinθ₁=λ/d

Note that once you make the substitution sin(θ)=θ for both phases, you will have a relatively simple expression involving Δθ, where Δθ=θ₂−θ₁.
sinθ₂−sinθ₁=λ/d 
→θ₂−θ₁=λ/d 
Δθ=λ/d