Polarisation — practice questions

Identify knowns: Initial intensity, $I_{0} = 200 \text{ W m}^{-2}$ Angle, $\theta = 30°$

Recall Malus' Law: $I = I_{0}\cos^{2}\theta$

Substitute the values: $I = 200 \times \cos^{2}(30°)$

Calculate $\cos(30°)$: $\cos(30°) \approx 0.866$

Calculate $\cos^{2}(30°)$: $(0.866)^{2} \approx 0.75$

Calculate the final intensity: $I = 200 \times 0.75 = 150 \text{ W m}^{-2}$

Therefore, the intensity of the light after passing through the filter is $150 \text{ W m}^{-2}$.

Intensity after Polariser A: When unpolarised light passes through the first polariser, its intensity is halved. Initial unpolarised intensity, $I_{un} = 50 \text{ W m}^{-2}$ Intensity after A, $I_A = I_{un} / 2 = 50 / 2 = 25 \text{ W m}^{-2}$ This light is now plane-polarised parallel to the axis of A.

Apply Malus' Law for Analyser B: The light incident on B has intensity $I_A = 25 \text{ W m}^{-2}$. This will be our $I_0$ for Malus' Law. The angle $\theta$ between the axis of A and B is $60°$. The final intensity, $I_B$, is given by: $I_B = I_A \cos^{2}\theta$

Substitute and calculate: $I_B = 25 \times \cos^{2}(60°)$ We know that $\cos(60°) = 0.5$. $I_B = 25 \times (0.5)^{2}$ $I_B = 25 \times 0.25$ $I_B = 6.25 \text{ W m}^{-2}$

Therefore, the final intensity of the light emerging from Analyser B is $6.25 \text{ W m}^{-2}$.