Waves: Question 10

Worked solution

Step 1: Intensity after the first filter

Unpolarised light of intensity $I_0$ passing through a single polarising filter emerges as polarised light of intensity: $I_1 = \frac{1}{2}I_0$

(This is given directly in the question, and is a standard result for unpolarised light.)

Step 2: Apply Malus’s law at the second filter

The light reaching the second filter is now polarised, with intensity $I_1 = \tfrac{1}{2}I_0$. Malus’s law gives the intensity transmitted through a second filter at angle $\theta$ to the first: $I_2 = I_1\cos^2\theta$

With $\theta = 60°$, $\cos 60° = 0.5$, so $\cos^2 60° = 0.25$: $I_2 = \frac{1}{2}I_0 \times 0.25 = \frac{1}{8}I_0$

Step 3: Check by combining the steps directly

Check (single combined expression): $I_2 = I_0 \times \tfrac12 \times \cos^2 60° = I_0 \times 0.5 \times 0.25 = 0.125\,I_0 = \tfrac{1}{8}I_0$, the same result, confirming $I_2 = \boxed{\tfrac{1}{8}I_0}$.

Why the other options are wrong

• A ($\tfrac{1}{16}I_0$): this would result from squaring the $\tfrac12$ factor as well as applying $\cos^2 60°$ (i.e. $\tfrac12 \times \tfrac12 \times 0.25$), incorrectly treating the first filter’s transmission with an extra factor of $\cos^2$.
• C ($\tfrac{1}{4}I_0$): this comes from using $\cos 60° = 0.5$ without squaring it, i.e. $\tfrac12 \times 0.5 = \tfrac14 I_0$.
• D ($\tfrac{1}{2}I_0$): this ignores the second filter entirely, stopping after the first filter’s halving of the intensity.

Final answer

• Fraction of $I_0$ emerging from the second filter $= \boxed{\dfrac{1}{8}I_0}$, option B.