Assignment 5, solutions:

Problem 1:

The intensity of the central maximum increases with the square of the number of sources.  The width decreases, the peak becomes sharper.  As the number of sources increases, the narrower and the more widely spaced are maxima of equal intensity.  The graphs could depict the transition from the double slit to the grating.

Problem 2:

(a)  l = 633 nm, beam diameter d = 2 cm.  k₁ = k₂ = k = 2p/l  = 9.92 10⁶ m^-1.

k_1x = kcosq, k_2x = kcosq, k_1y = ksinq, k_2y = -ksinq.

In the regions where the beams overlap we have E = E₁ + E₂.  Let E = Ek.

E(r,t) = A₁cos(k₁×r-wt) + A₂cos(k₂×r-wt).

E(r,t)² = (A₁cos(k₁×r-wt))² + (A₂cos(k₂×r-wt))² + 2A₁cos(k₁×r-wt)A₂cos(k₂×r-wt).

cosAcosB = (1/2)[cos(A+B) + cos(A-B)]
cos(k₁×r-wt+k₂×r-wt) = cos(2kcosq x - 2wt).
cos(k₁×r-wt-k₂×r+wt) = cos(2ksinq y)

E(r,t)² = (A₁cos(k₁×r-wt))² + (A₂cos(k₂×r-wt))² + A₁A₂[cos(2kcosq x - 2wt)+cos(2k₁sinq y)]

<I> µ <E(r,t)²>
The average values of cos²(k₁×r-wt) and cos²(k₂×r-wt) are 1/2, and cos(2kcosq x - 2wt) is zero, when averaged over a large number of periods.

<E(r,t)²> = A₁²/2 +A₂²/2 + A₁A₂cos(2ksinq y)]

<I> = <I₁> + <I₂> + 2(<I₁><I₂>)^1/2cos(2ksinq y)]

The intensity therefore varies with y as C₁+C₂cos((2p/l')y) = C₁+C₂cos(2ksinq y).

l' = p/ksinq.

When q = 5^o, then l' = 3.63 10^-6 m, we observe (0.02 m)/(3.63 10^-6 m cos5^o) = ~5530 fringes.

(b)  When q = 15^o, then l' = 1.22 10^-6 m,  we observe (0.02 m)/(1.22 10^-6 m cos7.5^o) = ~16535 fringes.  The fringe spacing decreases.

Problem 3:
E
See Thin-Film Interference!

Problem 4:
D
double slit maxima: d sinq = ml,  m = integer
single slit minima: w sinq = nl,  n = integer
d/w = m/n,  nd = mw.
From the picture: d > w.  Only answer D has nd = mw and d > w.

Problem 5:
B
grating maxima: d sinq = ml,  m = integer
m = 1, d = (0.01/2000) m =5*10^-6 m, l =5.2*10^-7 m, sinq ~ 0.1 ~ q.
q ~ 0.1radians ~ 6 degrees.

Problem 6:
A
See Thin-Film Interference!

Problem 7:
A
Resolution limit because of diffraction: ~qD with q ~ l/d.
Resolution limit because of size of pinhole: ~d
We want qD ~ d, lD ~ d², d ~ (lD)^1/2.

Problem 8:
E
2d = (n+1/2)l.
See Thin-Film Interference!