Problem 42

  The orientation of the transmission axis for each of the three sheets of polarizing material in the drawing [pg. 757] is labeled relative to the vertical. A beam of light, polarized in the vertical direction, is incident on the first sheet. The intensity of the incident beam is 1550 W/m². Obtain the intensity of the beam transmitted through the three sheets when: (a) $\theta_1 = 0^o$, $\theta_2 = 40^o$, $\theta_3 = 75^o$; (b) $\theta_1 = 30^o$, $\theta_2 = 30^o$, $\theta_3 = 70^o$.

SOLUTION:
(a) Malus's Law sayeth that the intensity transmitted through the first sheet will be

$$S_1 = S_0 cos^2 \theta_1 = 1550 W/m^2$$

and through the second sheet will be

$$S_2 = S_1 cos^2 (\theta_2 - \theta_1) = 9.1 \times 10^2 W/m^2$$

and through the third sheet will be

$$S_3 = S_2 cos^2 (\theta_3 - \theta_2) = 611 W/m^2$$

(b) Repeat (sigh) the above for the new angles, and you'll get

S₁ = 1160 W/m² , S₂ = 1160 W/m²

and

S₃ = 681 W/m²