JEE Mains Physics · Wave Optics

Polarisation by Polaroids and by Reflection

A polaroid halves unpolarised light and then passes cos²θ of polarised light, θ being the angle from the previous axis; light reflected at Brewster's angle, tan i_B = μ, is completely polarised.

All Wave Optics notes JEE Mains Physics strategy

Why this matters

Nineteen PYQs, fourteen of them multiple choice, and five from 2026. Five send light through two polaroids, one with an optically active solution between them and one with polaroids over the slits of a double slit. Seven chain three or more sheets, most of them a third sheet slipped between two crossed ones. Seven are about Brewster's angle: the angle itself, the angle of refraction, and which way the reflected light vibrates.

Concept 1 of 3: One polaroid after another: Malus' law

Unpolarised light vibrates in every direction across the beam. A polaroid lets through only the part along its axis, which on average is half. The light that comes out is polarised along that axis. A second polaroid at angle θ passes only the component E cos θ of it, and intensity goes as the square: cos²θ.

Definition

Unpolarised light through one polaroid: $I = \dfrac{I_{0}}{2}$ , whatever the axis.
Polarised light of intensity I′ through a polaroid at angle θ to its plane: $I = I'\cos^{2}\theta$ (Malus' law).
Parallel axes pass everything that reached the second sheet; crossed axes pass nothing.
An optically active solution between polariser and analyser turns the plane by an angle α. With the analyser parallel to the polariser, the output is $\dfrac{I_{0}}{2}\cos^{2}\alpha$ .
Polaroids over the two slits of a double slit: find each slit's intensity with Malus first, then combine the beams with the interference formula.

Malus' law

I_{1} = \frac{I_{0}}{2}, \qquad I_{2} = I_{1}\cos^{2}\theta

Worked example

Unpolarised light of intensity 40 W/m² passes through two polaroids whose axes are at

60^{\circ}

. Find the intensity that comes out. Then the second polaroid is turned parallel to the first and a solution that rotates the plane by

60^{\circ}

is put between them. What comes out now?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 July 2022 · Q99Moderate

Example 1 · Wave Optics · Polarisation by Polaroids and by Reflection

An unpolarised light beam of intensity

2I_{0}

is passed through a polaroid

P

and then through another polaroid

Q

which is oriented in such a way that its passing axis makes an angle of

30^{\circ}

relative to that of

P

. The intensity of the emergent light is

The first sheet halves, it does not use cos²

Unpolarised light has no single plane, so the first polaroid always passes half. Malus' law starts only from the second sheet.

Polarised light is not halved

If the light is already polarised, go straight to I cos²θ. Halving it first gives an answer half the right size.

Cos squared, not cos

Intensity goes as the square of the amplitude. At 60° the output is a quarter, not a half.

Drill 4 more on one polaroid after another: malus' law

Concept 2 of 3: Chains of polaroids

Two crossed polaroids pass nothing. Slip a third one between them at an angle and light gets through, because each step now turns the plane only part of the way. Each sheet compares only with the sheet just before it: the light has forgotten every earlier axis.

Definition

Each sheet multiplies by $\cos^{2}$ of the angle between its axis and the previous sheet's axis.
A sheet at θ between crossed polaroids: $I = \dfrac{I_{0}}{2}\cos^{2}\theta\,\sin^{2}\theta = \dfrac{I_{0}}{8}\sin^{2}2\theta$ . It is largest, $I_{0}/8$ , at $\theta = 45^{\circ}$ .
n sheets, each at $45^{\circ}$ to the one before, unpolarised input: $I = \dfrac{I_{0}}{2}\left(\dfrac{1}{2}\right)^{n - 1} = \dfrac{I_{0}}{2^{n}}$ .
Order matters: a sheet placed after a crossed pair receives nothing.
Some angles give the same output in pairs (θ and 90° − θ between crossed sheets); a question may need the smaller.

A sheet between crossed polaroids

I = \frac{I_{0}}{2}\cos^{2}\theta\,\sin^{2}\theta = \frac{I_{0}}{8}\sin^{2}2\theta, \qquad I_{n} = \frac{I_{0}}{2^{n}}\ (45^{\circ}\ \text{steps})

Worked example

Unpolarised light of intensity

I_{0}

meets two crossed polaroids with a third between them, its axis at

37^{\circ}

to the first (

\sin 37^{\circ} = 0.6

\cos 37^{\circ} = 0.8

). Find the output.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q10Moderate

Example 2 · Wave Optics · Polarisation by Polaroids and by Reflection

n

polarizing sheets are arranged such that each makes an angle

45^{\circ}

with the preceding sheet. An unpolarized light of intensity I is incident into this arrangement. The output intensity is found to be I/64. The value of

n

will be:

Angle to the previous sheet

Each cos² uses the angle from the sheet just before, not from the first sheet. A sheet at 37° to the first is 53° from the last of a crossed pair.

Count the halving once

Only the first sheet halves unpolarised light. With n sheets at 45°, the factor is ½ for the first and ½ for each of the other n − 1: I₀/2ⁿ in all.

Two angles can give the same output

Between crossed sheets, θ and 90° − θ pass the same intensity. If the question wants one angle, check which one it means.

Drill 6 more on chains of polaroids

Concept 3 of 3: Polarisation by reflection: Brewster's law

At one angle of incidence the reflected and refracted rays leave at right angles. The vibrations that would send light along the reflected ray are then pointing along that ray, and light cannot vibrate along its own direction. So the reflected light keeps only vibrations perpendicular to the plane of incidence: it is completely polarised.

Definition

Brewster's angle: $\tan i_{B} = \dfrac{\mu_{2}}{\mu_{1}}$ (from medium 1 into medium 2). For dielectrics with $\mu_{r} = 1$ , $\mu = \sqrt{\varepsilon_{r}}$ , so $\tan i_{B} = \sqrt{\varepsilon_{2}/\varepsilon_{1}}$ .
At $i_{B}$ the reflected and refracted rays are perpendicular: $r = 90^{\circ} - i_{B}$ .
The reflected light is completely polarised, vibrating perpendicular to the plane of incidence. The refracted light is only partly polarised.
Going the other way (glass to air) the Brewster angle is $\tan^{-1}\dfrac{\mu_{1}}{\mu_{2}} = 90^{\circ} - i_{B}$ .
Light already polarised in the plane of incidence is not reflected at all at Brewster's angle.

Brewster's law

\tan i_{B} = \frac{\mu_{2}}{\mu_{1}}, \qquad i_{B} + r = 90^{\circ}

Worked example

Unpolarised light falls from air on water of refractive index 4/3. Find Brewster's angle, the angle of refraction at that incidence, and the Brewster angle for light going from water into air.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q19Moderate

Example 3 · Wave Optics · Polarisation by Polaroids and by Reflection

When an unpolarized light falls at a particular angle on a glass plate (placed in air), it is observed that the reflected beam is linearly polarized. The angle of refracted beam with respect to the normal is

\_\_\_\_

\left( \tan^{- 1}(1.52) =57.7^{\circ} \right.

, refractive indices of air and glass are 1.00 and 1.52 , respectively)

Tangent, not sine

Brewster's law is tan i_B = μ. Snell's sine law gives the angle of refraction, but the Brewster angle itself comes from the tangent.

Refraction is the complement

At Brewster's angle r = 90° − i_B. There is no need to use Snell's law; the two angles add to a right angle.

Reverse direction, reverse ratio

From glass to air the ratio is 1/μ, giving 90° − i_B. Using tan⁻¹ μ again for the glass-to-air side is the usual slip.

Drill 6 more on polarisation by reflection: brewster's law

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

One polaroid after another: Malus' law
Malus' law
$I_{1} = \frac{I_{0}}{2}, \qquad I_{2} = I_{1}\cos^{2}\theta$
Chains of polaroids
A sheet between crossed polaroids
$I = \frac{I_{0}}{2}\cos^{2}\theta\,\sin^{2}\theta = \frac{I_{0}}{8}\sin^{2}2\theta, \qquad I_{n} = \frac{I_{0}}{2^{n}}\ (45^{\circ}\ \text{steps})$
Polarisation by reflection: Brewster's law
Brewster's law
$\tan i_{B} = \frac{\mu_{2}}{\mu_{1}}, \qquad i_{B} + r = 90^{\circ}$

Watch out for (9)

The first sheet halves, it does not use cos²
→ One polaroid after another: Malus' law
Polarised light is not halved
→ One polaroid after another: Malus' law
Cos squared, not cos
→ One polaroid after another: Malus' law
Angle to the previous sheet
→ Chains of polaroids
Count the halving once
→ Chains of polaroids
Two angles can give the same output
→ Chains of polaroids
Tangent, not sine
→ Polarisation by reflection: Brewster's law
Refraction is the complement
→ Polarisation by reflection: Brewster's law
Reverse direction, reverse ratio
→ Polarisation by reflection: Brewster's law

Test yourself on Wave Optics

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Wave Optics chapter test Drill all 19 questions