Homework #11 (chapter 41): Interference

Question 1:

Both – diffraction from each slit and interference of the waves coming from the two slits.

Question 18:

The two beams are incoherent, because each has as its source an incandescent light bulb, so that there is no particular phase relationship between light from each bulb (short coherence length). Thus, no interference pattern is visible, since coherence is a necessary condition for interference patterns.

Exercise 3:

Δy = λ D / d = 5.12*10^-7m 5.4m / 1.2*10^-3m = 2.3 mm

Exercise 7 :

y_m = mλD/d

y_m⁽²⁾ – y_m⁽¹⁾ = (λ₂ – λ₁)mD/d

è y₃⁽²⁾ – y₃⁽¹⁾ = (6.12-4.80)10^-7m * 3 *1.36m / 5.22*10^-3 m = 0.103 mm

Exercise 28 :

2d = (m+1/2)λ_film = (m+1/2)λ_air / n_film for maxima

2d = mλ_film = mλ_air / n_film for minima

λ_max = 600 nm = 2 d n / (m+1/2)

λ_min = 450 nm = 2 d n / m

λ_max / λ_min = m / (m+1/2) = 4/3 è m = -2

Using |m|= 2

d = λ_min |m| / (2n) = ¾ 450 nm = 338 nm

Exercise 40:

Each fringe corresponds to one-half wavelength.

Thus, 5cm*(n_air-1) = 60 (5.00*10^-5cm) / 2 è n_air = 6*5.00*10^-5 + 1 = 1.0003

Problem 2 :

The optical pathlength difference (n-1)t corresponds to a change in λ of 7λ.

è (n-1)t = 7λ è t = 7λ/(n-1) = 7*5.50*10^-7m / 0.58 = 6.64 μm

Problem 4:

Here (n₁-1)t – (n₂-1)t = (5-0) λ

è (n₁-n₂)t = 5 λ

è t = 5λ / (n₁-n₂) = 5*4.80*10^-7m / (1.7-1.4) = 8.00 μm

Problem 5:

Using Φ = 2π d sinθ/ λ so ΔΦ = 2π d Δ(sinθ)/ λ = 2π d cosθ Δθ / λ

Choose first minimum, so that m=1, since the width of the fringes is the same.

Then 2π d cosθ Δθ / λ = π è Δθ = λ / (2 d cosθ)

For small θ: cosθ ≈ 1 è Δθ = λ / (2 d)