Difficulty: Intermediate | Prerequisites: EM wave basics (field directions, intensity), trigonometry (Snell's law uses sine).
Tags: polarisation, polarization, Malus's law, polarising filter, polariser, unpolarised light, Snell's law, refraction, reflection, index of refraction, total internal reflection, critical angle, PHYS 212, University Physics, optics
Polarisation and refraction describe what happens when electromagnetic waves interact with materials. Polarisation filters select one component of the electric field oscillation, and the resulting intensity follows precise cosine-squared rules. Reflection and refraction at an interface follow from the boundary conditions on EM fields, condensed into Snell's law. These topics bridge the "waves" part of the course to geometrical optics. If you are comfortable with Malus's law and Snell's law, most exam problems in this section reduce to careful angle bookkeeping.
Unpolarised light through a polariser loses half its intensity. Each subsequent polariser reduces intensity by cos²θ, where θ is the angle between successive transmission axes. For refraction, n₁ sin θ₁ = n₂ sin θ₂. The geometry of reflected and refracted rays at an interface is fully determined by the two indices and the incident angle.
Unpolarised light
Light whose electric field oscillates in all directions perpendicular to the direction of travel, with no preferred orientation. Sunlight and most artificial light sources produce unpolarised light. In simple terms, the electric field vibrates randomly in every transverse direction.
Polarised light
Light whose electric field oscillates in a single plane. Produced by passing unpolarised light through a polarising filter, or by reflection at certain angles. Think of it as shaking a rope up and down only, rather than in random directions.
Malus's law
When polarised light of intensity I₀ passes through a polariser whose transmission axis makes an angle θ with the light's polarisation direction, the transmitted intensity is I = I₀ cos²θ. In simple terms, the closer the filter's axis is to the light's polarisation, the more light gets through.
Unpolarised light through a polariser
The transmitted intensity is exactly half: I = I₀/2. The output is now polarised along the filter's transmission axis. This is a special case, not Malus's law directly. Malus's law applies to already-polarised light.
Index of refraction (n)
A dimensionless number indicating how much slower light travels in a medium compared to vacuum. n = c/v. For vacuum, n = 1. For air, n ≈ 1. For glass, typically n = 1.3 to 1.9. Think of it as the "slowdown factor" for light entering a material.
Snell's law
n₁ sin θ₁ = n₂ sin θ₂, where θ₁ is the angle of incidence (measured from the normal) and θ₂ is the angle of refraction. In simple terms, light bends towards the normal when entering a denser medium (higher n) and away from the normal when entering a less dense medium.
Law of reflection
The angle of incidence equals the angle of reflection: θ_i = θ_r. Both are measured from the normal to the surface.
Total internal reflection
When light travels from a denser medium to a less dense one (n₁ > n₂), there is a critical angle θ_c = arcsin(n₂/n₁) beyond which all light is reflected and none is transmitted. Think of it as the angle where Snell's law would require sin θ₂ > 1, which is impossible, so the light has nowhere to go but back.
The exam worksheet describes three filters with specific transmission axis angles. Here is the general method:
Step 1: First filter (unpolarised light in)
Intensity out: I_A = I₀ / 2.
Polarisation direction: aligned with this filter's transmission axis.
Step 2: Second filter
Find θ, the angle between the first filter's axis and the second filter's axis.
Intensity out: I_B = I_A cos²θ = (I₀/2) cos²θ.
Polarisation direction: now aligned with the second filter's axis.
Step 3: Third filter
Find φ, the angle between the second filter's axis and the third filter's axis.
Intensity out: I_C = I_B cos²φ.
For the specific case in the worksheet (angles 0°, 75°, 90° relative to vertical, reading from the figure):
First filter at some reference angle. Light after first filter: I₀/2, polarised along that axis.
Second filter at 75° from the first: I_B = (I₀/2) cos²75°.
Third filter at 90° from the first, so 15° from the second: I_C = I_B cos²15°.
Without the middle filter, the first and third filters are 90° apart.
I_C = (I₀/2) cos²90° = 0.
This is the classic result: crossed polarisers transmit nothing. The middle filter at an intermediate angle actually lets some light through by "rotating" the polarisation partway.
If the first filter rotates by 15° counter-clockwise, the angle between the first and second filters changes, and the angle between the second and third also changes. Recompute each step with the updated angles.
The intensity at C depends on the product cos²(new angle 1→2) × cos²(new angle 2→3).
For three filters with the first and third fixed at 90° apart, intensity at C is maximised when the middle filter is at 45° to each. This gives I_C = (I₀/2) cos²45° × cos²45° = (I₀/2)(1/2)(1/2) = I₀/8.
More generally, if the outer filters are at angle Δ apart, the optimum middle-filter angle is Δ/2 from each.
To maximise output by rotating the first filter (with the others fixed), you need to find the angle that maximises the product of the cosine-squared terms through the chain. This is a calculus or symmetry argument.
The worksheet gives: light goes from air (n₁ = 1) into glass (n₂ = 1.4). The angle between the reflected ray and the refracted ray is 45°.
Setting up the geometry:
The angle of incidence is x₁ (measured from the normal).
The angle of reflection equals the angle of incidence: θ_r = x₁.
The angle of refraction is x₂ (measured from the normal on the other side).
The reflected ray and refracted ray are on opposite sides of the normal. The angle between them is (180° − x₁ − x₂) if they are on the same side of the surface, but from the figure, the reflected ray goes back into the air and the refracted ray enters the glass. The angle between them (measuring through the interface) is:
x₁ + x₂ = 180° − 45° = 135°? Let's be more careful.
The reflected ray makes angle x₁ with the normal on the air side. The refracted ray makes angle x₂ with the normal on the glass side. The angle between the reflected ray and the refracted ray, measured at the point of incidence, is (x₁ + x₂) if they are on opposite sides of the normal, but that gives the angle "opening up" from the normal. Actually, the angle between them as drawn is: the reflected ray is at angle x₁ from the normal (air side), and the refracted ray is at angle x₂ from the normal (glass side). The normal is a straight line, so the reflected and refracted rays are on opposite sides of the surface. The angle between them is 180° − x₁ − x₂.
Given: 180° − x₁ − x₂ = 45°, so x₁ + x₂ = 135°.
Solving for x₁ and x₂:
Snell's law: n₁ sin x₁ = n₂ sin x₂ → sin x₁ = 1.4 sin x₂.
Constraint: x₁ + x₂ = 135° → x₂ = 135° − x₁.
Substituting: sin x₁ = 1.4 sin(135° − x₁).
Expand: sin x₁ = 1.4 [sin 135° cos x₁ − cos 135° sin x₁] = 1.4 [(√2/2) cos x₁ + (√2/2) sin x₁].
sin x₁ = 1.4(√2/2)(cos x₁ + sin x₁) = 0.9899 (cos x₁ + sin x₁).
sin x₁ = 0.9899 cos x₁ + 0.9899 sin x₁.
sin x₁ (1 − 0.9899) = 0.9899 cos x₁.
0.0101 sin x₁ = 0.9899 cos x₁.
tan x₁ = 0.9899 / 0.0101 ≈ 98.0.
x₁ ≈ arctan(98) ≈ 89.4°.
That result seems extreme, which means the 45° angle is likely measured differently in the figure. Looking at the diagram again: the 45° is between the refracted ray and the normal inside the glass (that is, x₂ = 45°, not the angle between reflected and refracted rays).
Reinterpreting with x₂ = 45°:
Snell's law: sin x₁ = 1.4 × sin 45° = 1.4 × 0.7071 = 0.9899.
x₁ = arcsin(0.9899) ≈ 81.9° ≈ 82°.
Alternatively, if the 45° is the angle between the reflected ray and the refracted ray measured at the surface (not from the normal):
The refracted ray in the glass makes angle x₂ with the normal. The reflected ray makes angle x₁ with the normal on the other side. At the surface, these two rays open an angle of (180° − x₁ − x₂). If that equals 45°, then x₁ + x₂ = 135° (as computed above, giving extreme x₁).
From the figure, the 45° appears to be the angle the refracted ray makes with the interface inside the glass. That would make the angle from the normal = 90° − 45° = 45°, giving x₂ = 45°. This is the most likely reading.
With x₂ = 45°:
x₁ = arcsin(1.4 × sin 45°) = arcsin(0.9899) ≈ 81.9°.
If n₂ decreases (but stays above 1):
From Snell's law: sin x₁ = n₂ sin x₂. If we hold the geometry fixed... but actually, re-read the question. It says x₁ is the angle of incidence. The question asks how x₁ and x₂ change. But x₁ is the angle at which the light arrives, which is set by the light source, not by n₂. Unless the problem is defined so that the 45° angle constraint is maintained while n₂ changes.
If the 45° constraint between the reflected and refracted rays is maintained:
x₁ + x₂ = 135°, and sin x₁ = n₂ sin(135° − x₁).
As n₂ decreases towards 1, x₁ must adjust. For n₂ = 1 (no interface), x₁ = x₂ and x₁ + x₂ = 135° gives x₁ = 67.5°. As n₂ increases, x₁ increases and x₂ decreases. So if n₂ decreases, x₁ decreases and x₂ increases.
Answer: (a) Decrease x₁, increase x₂.
If instead x₁ is fixed and only x₂ changes:
sin x₂ = (n₁/n₂) sin x₁. As n₂ decreases, sin x₂ increases, so x₂ increases. And x₁ is unchanged.
But the answer choices all have x₁ changing, so the 45° constraint interpretation is likely correct.
Quantity | Formula |
|---|---|
Unpolarised light through first polariser | I = I₀ / 2 |
Malus's law | I = I₀ cos²θ |
Snell's law | n₁ sin θ₁ = n₂ sin θ₂ |
Law of reflection | θ_i = θ_r |
Critical angle | θ_c = arcsin(n₂/n₁), valid when n₁ > n₂ |
Speed of light in medium | v = c/n |
Polarising sunglasses work by blocking horizontally polarised glare reflected off roads and water, transmitting only vertically polarised light. LCD screens use two crossed polarisers with a liquid crystal between them that rotates the polarisation when voltage is applied, controlling which pixels appear bright. Snell's law governs how lenses focus light, and total internal reflection is the principle behind fibre-optic cables that carry internet data across oceans.
Students often apply Malus's law to unpolarised light. Malus's law (I = I₀ cos²θ) only works on already-polarised light. The first polariser in a chain always cuts unpolarised light to I₀/2, regardless of its orientation.
When three polarisers are in a chain, students sometimes compute the angle between the first and third filters and skip the middle one. You must step through sequentially: first-to-second, then second-to-third. Each polariser resets the polarisation direction.
Students sometimes think removing a filter always increases the transmitted intensity. With three filters where the outer two are crossed (90° apart), the middle filter actually helps light get through. Removing it drops the output to zero.
In Snell's law problems, students occasionally use the angle from the surface rather than the angle from the normal. Always measure from the normal.
Students confuse the conditions for total internal reflection. It only occurs when light goes from a higher-n medium to a lower-n medium (e.g., glass to air), never the other way round.
⚠️ Polarisation problems always follow the same chain: I₀/2 through the first filter, then Malus's law for each subsequent filter. Practise the chain until it is automatic.
⚠️ Read the figure carefully for filter angles. The angle in Malus's law is always between adjacent transmission axes, not between a filter and some absolute reference.
⚠️ In Snell's law problems, sketch the normal and label all angles from it. Many errors come from misidentifying which angle is which.
⚠️ The "angle between reflected and refracted rays" requires careful geometry. Draw the normal, place both rays, and work out the relationship before substituting into Snell's law.
⚠️ Know the critical angle formula and when it applies. If n₁ < n₂, there is no total internal reflection.
Fill in the blank: Unpolarised light of intensity I₀ passes through an ideal polariser. The transmitted intensity is ___.
True or false: Malus's law gives I = I₀ sin²θ.
True or false: Two crossed polarisers (90° apart) transmit zero light. Adding a third filter between them at 45° allows some light through.
Fill in the blank: Snell's law is n₁ sin θ₁ = n₂ sin ___.
True or false: Total internal reflection can occur when light goes from air into glass.
Answers: 1. I₀/2. 2. False (it is cos²θ). 3. True. 4. θ₂. 5. False (it requires going from a denser medium to a less dense one).
Q: Unpolarised light of intensity I₀ passes through three ideal polarisers. The first is vertical, the second is at 75° from vertical, the third is horizontal (90° from vertical). What is the intensity after each filter?
A: After filter 1: I₀/2 (polarised vertical). After filter 2: (I₀/2) cos²75° ≈ (I₀/2)(0.06699) ≈ 0.0335 I₀. After filter 3: the angle between filter 2 (75°) and filter 3 (90°) is 15°, so I_C = 0.0335 I₀ × cos²15° ≈ 0.0335 I₀ × 0.933 ≈ 0.0313 I₀.
Q: If the middle filter is removed, what is the intensity after the last filter?
A: The first filter transmits I₀/2 (polarised vertical). The remaining filter is horizontal, 90° from vertical. I = (I₀/2) cos²90° = 0. No light gets through.
Q: Light travels from air (n = 1) into glass (n = 1.4). The refracted angle is 45°. What is the angle of incidence?
A: sin x₁ = 1.4 × sin 45° = 1.4 × 0.7071 = 0.9899. x₁ = arcsin(0.9899) ≈ 81.9°.
Q: If n₂ decreases (but stays above 1), and the geometric constraint between reflected and refracted rays is maintained, how do x₁ and x₂ change?
A: x₁ decreases and x₂ increases. The interface bends the light less, so the incidence angle needed to maintain the same angular relationship is smaller, and the refracted ray bends less (closer to the incident direction, meaning a larger angle from the normal).
Q: What is the critical angle for light going from glass (n = 1.4) to air (n = 1)?
A: θ_c = arcsin(1/1.4) = arcsin(0.714) ≈ 45.6°.
Polarisation connects to the vector nature of electromagnetic waves covered in the EM waves section. The E-field direction is the polarisation direction, and all the polariser maths is really about projecting one vector onto another. Snell's law connects to wave behaviour at boundaries, which also appears in impedance matching (partial reflection) for EM waves and transmission lines. Total internal reflection reappears in waveguide and fibre-optic problems if you continue into PHYS 214 or engineering optics.
Related Terms / Search Tags: polarisation, polarization, Malus's law, polariser, polarizer, polarising filter, unpolarised light, unpolarized light, crossed polarisers, intensity, cos squared, Snell's law, law of refraction, law of reflection, index of refraction, refractive index, total internal reflection, critical angle, TIR, optics, PHYS 212, University Physics, Exam 3