Waves: Question 5

Worked solution

Step 1: Recall Malus’s law

For plane-polarised light of intensity $I_0$ incident on a polarising filter (analyser), the transmitted intensity is: $I = I_0\cos^2\theta$

where $\theta$ is the angle between the plane of polarisation of the incident light and the transmission axis of the analyser.

Step 2: Set up the equation for the given fraction

Here $I = 0.36I_0$, so: $0.36I_0 = I_0\cos^2\theta \implies \cos^2\theta = 0.36$

Step 3: Solve for $\theta$

Taking the square root (the angle between two lines is taken between $0°$ and $90°$, so the positive root is used): $\cos\theta = \sqrt{0.36} = 0.6$ $\theta = \cos^{-1}(0.6) \approx 53.13°$

Check (independent method. Pythagorean identity): if $\cos\theta = 0.6$, then $\sin\theta = \sqrt{1-0.6^2} = \sqrt{1-0.36} = \sqrt{0.64} = 0.8$. This is the well-known $3$-$4$-$5$ triangle relationship ($0.6 = 3/5$, $0.8 = 4/5$), which independently confirms $\theta = \cos^{-1}(0.6) \approx 53.1°$.

Why the other options are wrong

• A ($36.9°$): mixes up sine and cosine. This is $\sin^{-1}(0.6)$, the complementary angle, not $\cos^{-1}(0.6)$.
• B ($45°$): the “familiar” angle for which $\cos^2(45°) = 0.5$, not $0.36$.
• D ($60°$): the “familiar” angle for which $\cos^2(60°) = 0.25$, not $0.36$.

Final answer

• Angle between the transmission axis and the plane of polarisation $= \boxed{53.1°}$, option C.