Question 11:
Radio wavelengths are about the same order of magnitude as buildings, whereas wavelengths of visible light are many orders of magnitude shorter. Thus, diffraction effects are noticeable for the former, but negligible for the latter.
Question 25:
a) To collect more light, which gives brighter images.
b) To improve image resolution by reducing diffraction effects.
Exercise 5:
Using a sinθ = m λ for minima:
a) λa = a sinθ, λb = (a/2) sinθ è λa = 2 λb
b) Since
a sinθ is fixed, the mth
minimum for λa will always correspond
to the 2mth minimum for λb.
Exercise 8 :
b) y = mλD/a è Δy = Δm λD/a è a = Δm λD/ Δy =
4*546*10-9m*0.413m/0.35*10-3m=2.58mm
a) θ≈sinθ=mλ/a = 1*546*10-9m/2.58*10-3m
= 2.11*10-4 rad
Exercise 14 :
Rayleigh criterion: s = l θR with l = 163 km and θR = 1.22 λ / d with d being the pupil diameter.
b) s = 1.22 λ l / d = 1.22*540*10-9m*1.63*105m / 0.0049m = 21.9 m
a) θR
= 1.22 λ / d = 1.22*540*10-9m / 0.0049m = 1.34*10-4 rad
Exercise 21:
S = R θR = R 1.22
λ / d è
R = s d / (1.22λ) = 2*10-3m*4.4*10-3m / (1.22*475*10-9m)
= 15.2 m