Assignment 6, solutions:

Problem 1: (C)

We have constructive interference when the difference in optical path length is D = ml/2.
Here D = nd - d = 40*l/2.  n - 1 = 40*500*10^-9/(2*0.05) = 0.0002

Problem 2: (E)

If you took a picture of the interference pattern with an exposure time t << (1/500)s, the picture would show an interference pattern.  But when concentrating on any spot on the screen it goes through the cycle bright (constructive interference) -> dark (destructive interference) -> bright 500 times per second.

Problem 3: (B)

We have constructive interference when the difference in optical path length is D = ml/2.
Here D = d₁ - d₂ = 100000*l_green/2 = 85865*l_red/2.
l_green = 0.85865*l_red = 0.85865*632.82 nm = 543.37 nm.

Problem 4:

The function f(x) is a periodic function with period L = l.  We can express it in terms of its Fourier series,

Here k_n = n2p/L, Dk = k_n+1 - k_n = 2p/L.  The coefficients C_m are given by

.

Using e^ikx = cos(kx) + i sin(kx) we can also write

,

,  ,
.  

We have A_n = (C_n + C_-n), B_n = i(C_n - C_-n), A₀ = 2C₀,  n > 0.

The function f(x) is an even function, so all the coefficients B_n are zero.
We have C_n = C_-n, A_n = 2C_n.

A₀ = (2/l)f₀(2l/a) = 4f₀/a.  For n > 0 we have

.

For example, let us pick f₀ = 1 and a = 4.  
Then A₀ = 1, A_n = [2/(np)]sin(np/2), A_n = 0 if n is even.
The figure below plots A_n for n from 0 to 10, |A_n|², and the square of the absolute value of the first 10 coefficients obtained using Excel's FFT function, normalized so that A₀ = 1.

Problem 5:

Let us follow the example in the notes for calculating the diffraction pattern for a single slit, but let us assume that we have a finite number N of point sources instead of an infinite number of point sources.  The perpendicular distance of point s from the optical axis is sd, -N/2 < s < N/2.

According to the Huygen-Fresnel principle, the total field at a point y on the screen is the superposition of wave fields from a finite number of point sources.  Each point is the source of a spherical wave.  A distance r from a point the electric field is due to a point sources is 

dE = (A₀/r)exp(i(kr-wt)).

If r₀ is the distance from the point s = 0 on the optical axis to a point y on the screen, then the contribution dE to the total amplitude on the screen from the point at s = 0 is

dE(y) = (A₀/r₀)exp(i(kr₀-wt)).

For off-axis points for which s ¹ 0, the distance is longer or shorter than r₀ by an amount D.  The contribution dE(y) to the total amplitude on the screen from an off-axis point (s ¹ 0) is

dE(y) = (A₀/(r₀+D(s)))exp(i(k(r₀+D(s))-wt)).

To find the total amplitude E(y) we have to add up the contributions from all points. 

E(y) = S[(A₀/(r₀+D(s)))exp(i(k(r₀+D(s))-wt))].

The sum is over all s.  We define sinq = D/sd.  Since r₀>>D, we approximate 1/(r₀+D) with 1/r₀.  However we cannot drop the D inside the cosine function, since kD(s) is not necessarily much smaller than 2p.

We then have

E(y) = S[(A₀/r₀)exp(i(kr₀-wt)exp(i s ksinq d)] 
= (A₀/r₀)exp(i(kr₀-wt)Sexp(i s ksinq d) = E₀S[exp(isd)].
Here E₀ = (A₀/r₀)exp(i(kr₀-wt) and d = ksinq d.  E₀ is the electric field a distance r₀ from a single source.

To evaluate the sum E₀S[exp(isd)] we can use the figure below as a guide.

Using trigonometry we write:

E₀/2R = sin(d/2).
E² = R² + R² -2R²cosNd = 2R²(1-cosNd) = 4R²sin²(Nd/2).
But 4R² = E₀²/sin²(d/2), so
E² = E₀²sin²(Nd/2)/sin²(d/2) = E₀²sin²(Nksinq d/2)/sin²(ksinq d/2).
The intensity is proportional to the square of the field amplitude, so 

I = I₀sin²(Nksinq d/2)/sin²(ksinq d/2)

gives the intensity pattern as a function of q for N equally-spaced sources.

Problem 6:

A function f(x) and its Fourier transform f(k) are related by

.

Here f(x) = 0 for  x < -L and x > L,  f(x) = 1 - |x|/L otherwise.  We therefore have

If you use the Fourier series of a periodic function with period l >> L to approximate the Fourier transform f(k) then the coefficients A_n (n > 0) are given below.

For n = 0 we have A₀/2 = L/l.

Excels Fourier transform function yields the same normalized coefficients |A_n|.

The figure below shows the (normalized) squares of the first 30 coefficients, |A_n|² and also shows the (normalized) first 30 coefficients obtained using Excels FFT function.

Problem 7:

I' is the convolution of the functions I_g and T.

Let the maximum value of I_g and T be 1 and let T(a) = T(x'-x'') be the rectangular function.
T(a) = 1 0 £ a £ d, T(a) = 0 otherwise.
For  0 £ x'-x'' £ d  or  x'' £ x' £ d+x''  T(x'-x'') is not zero.  For x' outside this range T(x'-x'') is zero and therefore I_g(x')T(x'-x'') is zero.

I_g(x'') = 1-2x''/d  for 0 £ x'' £ d/2.  The limits of the integral over x'' may be taken to be 0 and d/2.  For I'(x') to not be zero we therefore need 0 £ x' £ 3d/2.

Let x' lie in the region 0 £ x' £ d/2:

Let x' lie in the region d/2 £ x' £ d:

Let x' lie in the region d £ x' £ 3d/2:

Below is a plot of I_g(x') versus x' for d = 100.