Homework #14 (chapter 44): Polarization

Question 7:

Since Polaroid sunglasses preferentially absorb horizontally polarized light, glare – which is mostly horizontally polarized – is removed more efficiently than by ordinary sunglasses. Correspondingly, if unwanted light is somehow vertically polarized, Polaroid sunglasses may be undesirable.

Question 8:

Yes. Before, the angle θ for the second polarizer was 90°, such that no light will be transmitted. Introducing the second sheet turns the direction of the E vector for the polarized light incident on the third polarizer, such that θ differs from 90° and light will be transmitted. (see also ConcepTest C33)

Exercise 4:

Sheet 1 halves the intensity to I₀/2. Thus I₀/2 is incident on sheet 2.

For sheet 2 with I = 1/3 I₀: I = I₀/2 cos²θ = I₀/3 è cos²θ = 2/3 è θ = 35.3°

Exercise 7:

Before sheet 1: I₀

After sheet 1: I₀/2

After sheet 2: I₀/2 cos²30°

After sheet 3: I₀/2 cos²30° cos²30° = I₀/2 cos⁴30°

After sheet 4: I₀/2 cos⁴30° cos²30° = I₀/2 cos⁶30° = 0.21 I₀

Exercise 11:

a) θ_P = tan^-1 n₂/n₁ = tan^-1 (1.33/1.00) = 53.1°

b) Yes, since n_water = n₂ is wavelength dependent.

Problem 4:

a) Put sheet 1 at an angle 0° < θ₁ < 90°, and the second sheet at an angle of θ₂ = 90°, both with respect to the polarization direction of the light incident on sheet 1.

b) For N sheets with evenly spaced angles between them,

I = I₀(cos²θ)^N = I₀cos^2Nθ è I/I₀ = cos^2Nθ > 0.95

è for N = 10: I/I₀ = 0.7805

è for N = 20: I/I₀ = 0.8838

è for N = 48: I/I₀ = 0.9499

è for N = 49: I/I₀ = 0.9509

Therfore, 49 sheets are needed.