Polarization of light — Malus law and Brewster angle calculation

Question

(a) Unpolarized light of intensity $I_0$ passes through two polaroids. The axis of the second polaroid makes an angle of 60° with the first. Find the intensity of the transmitted light.

(b) Find the Brewster angle for light going from air to glass ( $n = 1.5$ ).

(JEE Main 2022, similar pattern)

Solution — Step by Step

When unpolarized light passes through a polaroid, the intensity is halved (the polaroid transmits only one component):

I_1 = \frac{I_0}{2}

The transmitted light is now plane polarized along the axis of the first polaroid.

Malus’s law: when polarized light of intensity $I_1$ passes through a polaroid whose axis makes angle $\theta$ with the polarization direction:

I_2 = I_1 \cos^2\theta

With $\theta = 60°$ :

I_2 = \frac{I_0}{2} \cos^2 60° = \frac{I_0}{2} \times \left(\frac{1}{2}\right)^2 = \frac{I_0}{2} \times \frac{1}{4}

\boxed{I_2 = \frac{I_0}{8}}

At Brewster’s angle, reflected light is completely polarized. The condition is:

\tan\theta_B = n = \frac{n_2}{n_1}

\theta_B = \tan^{-1}(1.5)

\boxed{\theta_B \approx 56.3°}

At this angle, the reflected and refracted rays are perpendicular to each other ( $\theta_B + \theta_r = 90°$ ).

Why This Works

Malus’s law follows from the fact that a polaroid only transmits the component of the electric field along its axis. If the field amplitude is $E_0\cos\theta$ , the intensity (proportional to $E^2$ ) is $I_0\cos^2\theta$ .

Brewster’s angle has a beautiful physical explanation: at this angle, the reflected and refracted rays are perpendicular. The oscillating dipoles in the glass surface cannot radiate along their oscillation direction (they radiate perpendicular to it). Since the reflected ray is along the dipole oscillation direction, the component that would be reflected is suppressed — only the perpendicular component reflects, giving fully polarized reflected light.

Alternative Method

For three or more polaroids in series, apply Malus’s law sequentially. If polaroids have axes at angles $\theta_1 = 0°$ , $\theta_2 = 30°$ , $\theta_3 = 60°$ :

$I = \frac{I_0}{2}\cos^2 30° \cdot \cos^2 30° = \frac{I_0}{2} \cdot \frac{3}{4} \cdot \frac{3}{4} = \frac{9I_0}{32}$

Adding an intermediate polaroid actually increases the final intensity — a counterintuitive result that JEE loves to test.

For JEE, the three most tested results: (1) unpolarized through one polaroid gives $I_0/2$ , (2) Malus’s law $I = I_0\cos^2\theta$ , (3) Brewster angle $\tan\theta_B = n$ . These three formulas cover 90% of polarization questions.

Common Mistake

When applying Malus’s law, the angle $\theta$ is between the pass axes of the two polaroids, not between the polaroid axis and the incident beam. Also, Malus’s law applies only to already-polarized light. For unpolarized light hitting the first polaroid, use the $I_0/2$ rule — not $\cos^2\theta$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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