Question

1. Unpolarized light of intensity I₀ is incident on three polarizing filters. The axis of the first is vertical, that of the second is 31° from vertical, and that of the third is horizontal. What light intensity emerges from the third filter?
✕ I₀

2. A 215 mW horizontally polarized laser beam passes through a polarizing filter whose axis is 28° from vertical. What is the power of the laser beam as it emerges from the filter?
mW

3. Two narrow slits 44 µm apart are illuminated with light of wavelength 490 nm. What is the angle of the m = 2 bright fringe in radians?
rad

What is the angle in degrees?
°

4. A double-slit interference pattern is created by two narrow slits spaced 0.15 mm apart. The distance between the first and the fifth minimum on a screen 60 cm behind the slits is 6 mm. What is the wavelength of the light used in this experiment?
nm

Answer 1

1) After the initially unpolarized light passes through the first sheet, the intensity of the light when it emerges from that sheet is Io/2.

Then after it passes through the second, sheet the intensity becomes (Io/2)*(cos(31))^2.

For the last sheet, the angle between the polarized light and the direction of polarization is (90 - 31)degrees. So, the intensity when it emerges from the third sheet is (Io/2)*(cos(31))^2*(cos(59))^2 = 0.0975*Io

2) Start with the relationship between the initial and final intensity of light. I=Io cos(θ)^2
The equation for wave intensity can apply to both the initial and transmitted intensity: I = P Io = Po AA

Assuming the area of the beam does not change, the first equation becomes: P=Po cos(θ)^2
Solve for the power of the beam as it emerges from the filter. Since the filter's axis is 28 degrees from
the horizontal, the angle between the electric field and the filter axis = 90 – 28 = 62 degrees

P=(215×10^−3 W)*(cos62)^2) = 47.4 mW

3) d = 44 um, wavelength = L = 490 nm

For bright fringe, d*sin(theta) = m*L

For m = 2, sin(theta) = 2L/d = (2*490*10^-9)/44*10^-6 = 0.0223

Theta = 1.28 degree = 0.02234 radian

4) d = 0.15 mm
∆y = 6 mm
L = 60 cm = 600 mm
--> d * ∆y/L = 4 λ ( The distance between the first and the fifth minimum = 4λ )

-->0.15 * 6/600 = 4 λ
λ = 3.75 x 10^-4 mm = 375 nm