Physics · Ch 6 — Optics

Malus' Law

6.12.3.3

Malus' Law

Malus' law: when a beam of plane polarised light of intensity $I_0$ is incident on an analyser, the transmitted intensity $I$ varies directly as the square of the cosine of the angle $\theta$ between the transmission axes of the polariser and the analyser, discovered by the French physicist E.N. Malus in 1809. $\boxed{I=I_0\cos^2\theta}$ . Proof: if $a$ is the amplitude of the electric vector transmitted by the polariser, it resolves into two components relative to the analyser's own axis, $a\cos\theta$ (parallel, transmitted) and $a\sin\theta$ (perpendicular, blocked); only $a\cos\theta$ survives, and since intensity is proportional to the square of amplitude, $I\propto(a\cos\theta)^2=ka^2\cos^2\theta=I_0\cos^2\theta$ , where $I_0=ka^2$ is the maximum transmitted intensity (at $\theta=0$ ). Special cases: at $\theta=0°$ , $\cos^2\theta=1$ and $ …

Figure 6.76Malus's law

What this figure shows. Unpolarised light of intensity I passes through a polariser, emerging as plane polarised light of intensity I0 = I/2 (half is absorbed by the polariser itself); this beam then strikes an analyser whose transmission axis is tilted at angle theta to the polariser's own axis, and the light finally transmitted through the analyser has intensity I = I0 cos^2(theta) -- the diagram traces this two-stage intensity reduction from the origina …

Figure 6.77Malus' law (amplitude resolution)

What this figure shows. The electric-field amplitude a of the plane polarised light emerging from the polariser is resolved into two perpendicular components relative to the analyser's own transmission axis: a component a cos(theta), lying along the analyser's axis, which passes straight through, and a perpendicular component a sin(theta), lying across the analyser's axis, which is completely blocked. Since transmitted intensity is proportional to the square of the transmitted amplitude, only the a cos(theta) component survives to give I = k(a cos theta)^2 = I0 cos^2(theta), …

Misc Example 6.37Intensity through two polaroids at 30 degrees

Worked out. Two polaroids have their transmission axes inclined at 30 degrees, and unpolarised light of intensity I falls on the first. After the first polaroid, the intensity drops to I0 = I/2. Applying Malus' law with theta = 30 degrees, the intensity emerging from the second polaroid is I' = I0 cos^2(30 degrees) = (I/2) times (root3/2)^2 = (I/2) times 3/4 = 3I/8 -- so only three-eighths of the original unpolarised intensity survives passage through both polaroids together, the loss coming partly from the first polaroid's unavoidable 50 percent absorption and partly from t …

Misc Example 6.38Adding a third polaroid at 45 degrees between two crossed polaroids

Worked out. Two polaroids are crossed (transmission axes at 90 degrees), so ordinarily no light at all would emerge from the second one when unpolarised light of intensity I is incident on the first: after the first, I0 = I/2, and Malus' law at theta=90 degrees gives I' = I0 cos^2(90 degrees) = 0, exactly as expected for crossed polaroids. But inserting a third polaroid P3 between them, tilted at 45 degrees to both, changes everything: applying Malus' law once from P1 to P3 (45 degrees) gives an intermediate intensity of I0 cos^2(45 degrees) = (I/2)(1/2) = I/4, and applying it a second time from P3 to P2 (another 45 degrees) gives a final intensity of (I/4) cos^2(45 degrees) = (I/4)(1/2) = I/8 -- so light that was completely blocked by two crossed polaroids alone can be made to partly reappear simply by inserting a third, intermediate polaroid between them, a genuinely counter-intu …

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