NSWPhysicsSyllabus dot point

Inquiry Question 1: What is light?

Analyse the wave model of light using Young's double-slit experiment, single-slit diffraction and polarisation, and apply Malus's law I = I_0 cos^2 theta to polarised light

Q: 2021 HSC HSC: In a Young's double-slit experiment, two slits 0.25 mm apart are illuminated with monochromatic light of wavelength 590 nm. The screen sits 1.8 m from the slits. Calculate the fringe spacing on the screen, and explain why a single-slit pattern would not show the same equally spaced bright fringes.

Fringe spacing for small angles: $\Delta y = \lambda L / d = (5.90 \times 10^{-7})(1.8) / (2.5 \times 10^{-4}) = 4.25 \times 10^{-3}$ m $= 4.25$ mm. Why a single slit looks different: a single slit produces a diffraction pattern from interference of light from the continuous range of points across the slit width. The central maximum is twice as wide as the side maxima, and side maxima fall off rapidly in intensity (envelope $\propto (\sin \alpha / \alpha)^2$). A double slit gives equally spaced narrow fringes from path-difference interference between the two slits, modulated by the single-slit envelope of each individual slit. So the equally spaced bright fringes come only when there are at least two coherent sources separated by a fixed distance $d$. Markers reward correct fringe-spacing formula and value, plus a clear contrast between two-source interference (equal spacing) and single-slit diffraction (broad central peak with rapidly weakening side peaks).

Q: 2018 HSC HSC: Unpolarised light of intensity 80 W m^-2 passes through two polarising filters whose transmission axes are at 30 degrees to each other. Calculate the intensity transmitted by the second filter, and state what would happen if a third filter were inserted between them at 60 degrees to the first.

After the first filter, unpolarised light is reduced to half its intensity (only one polarisation component passes): $I_1 = I_0 / 2 = 80 / 2 = 40$ W m$^{-2}$. Through the second filter (Malus's law with $\theta = 30^\circ$): $I_2 = I_1 \cos^2 30^\circ = 40 \times (0.866)^2 = 40 \times 0.75 = 30$ W m$^{-2}$. With a third filter inserted at $60^\circ$ to the first (so $30^\circ$ to the second on the entry side, and $30^\circ$ to the original second filter on the exit side): After the third filter: $I_1 \cos^2 60^\circ = 40 \times 0.25 = 10$ W m$^{-2}$. After the original second filter (now $30^\circ$ from the third): $10 \times \cos^2 30^\circ = 10 \times 0.75 = 7.5$ W m$^{-2}$. Surprisingly, inserting an extra filter increases the final intensity from $0$ when the original two are crossed at $90^\circ$, and changes the answer here too. Markers reward the half-intensity step, correct application of Malus's law, and recognition that polarisation order matters.

A focused answer to the HSC Physics Module 7 dot point on the wave model of light. Young's double-slit interference with d sin theta = m lambda, single-slit diffraction, polarisation as evidence light is transverse, and quantitative use of Malus's law.

Generated by Claude OpusReviewed by Better Tuition Academy10 min answerUpdated 2026-05-18

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What this dot point is asking

NESA wants you to use the wave model of light to explain interference, diffraction and polarisation. You should be able to derive and apply the double-slit fringe condition, describe what a single-slit diffraction pattern looks like, explain polarisation as evidence that light is transverse, and use Malus's law quantitatively.

The answer

Why the wave model

Newton's particle (corpuscular) picture of light explained reflection and refraction but failed to predict diffraction and interference. By 1801 Thomas Young's double-slit experiment demonstrated that light produces interference fringes, which only waves can do. The wave model dominated nineteenth-century optics and motivated Maxwell's identification of light as an EM wave.

Young's double-slit experiment

Monochromatic, coherent light passing through two narrow slits separated by $d$ produces alternating bright and dark fringes on a screen at distance $L$ . Bright fringes occur where the path difference equals a whole number of wavelengths:

d \sin \theta = m \lambda, \quad m = 0, \pm 1, \pm 2, \dots

Dark fringes (destructive interference) occur where path difference is a half-odd integer:

d \sin \theta = (m + \tfrac{1}{2}) \lambda

For small angles, $\sin \theta \approx \tan \theta = y / L$ , so the bright-fringe positions on the screen are:

y_m = \frac{m \lambda L}{d}

and the fringe spacing is:

\Delta y = \frac{\lambda L}{d}

Three predictions of the wave model the experiment confirms:

Increasing $\lambda$ widens the fringes (red fringes wider than blue).
Increasing $d$ narrows the fringes.
Increasing $L$ widens the fringes.

Coherence (a fixed phase relationship between the two slits) is necessary, which is why a single source illuminates both slits.

Single-slit diffraction

A single slit of width $a$ produces a broader pattern with a wide central maximum and narrow, rapidly weakening side maxima. The dark fringes occur where:

a \sin \theta = m \lambda, \quad m = \pm 1, \pm 2, \dots

The central maximum spans the angular range $|\sin \theta| < \lambda / a$ , twice the width of each side maximum.

In practice, the double-slit pattern is the product of two factors:

A double-slit interference pattern (equally spaced fringes from the two-slit geometry).
A single-slit diffraction envelope (each slit individually diffracts, modulating intensity).

Missing orders appear when a double-slit interference maximum coincides with a single-slit minimum.

Polarisation

Light is a transverse EM wave: $\vec{E}$ and $\vec{B}$ are perpendicular to the direction of propagation. Unpolarised light contains $\vec{E}$ vibrating in all directions perpendicular to the wave; a polarising filter passes only the component along its transmission axis.

Key observations only the transverse-wave model explains:

A polarising filter reduces unpolarised light to half its intensity (each direction averages to one component).
Two filters crossed at $90^\circ$ transmit zero intensity.
Reflection off a non-metallic surface at Brewster's angle gives strongly polarised reflected light.

Longitudinal waves (such as sound) cannot be polarised, so polarisation is direct evidence that light is transverse.

Malus's law

If polarised light of intensity $I_0$ encounters a second polariser whose transmission axis makes angle $\theta$ with the first:

\boxed{I = I_0 \cos^2 \theta}

This follows from the projection $E = E_0 \cos \theta$ of the electric field onto the new axis, then squaring (intensity is proportional to $E^2$ ).

For unpolarised input, the first polariser halves the intensity, then any subsequent polariser follows Malus's law from there.

Worked example: two polarisers

Unpolarised light at $120$ W m $^{-2}$ enters two polarisers whose axes are at $45^\circ$ to each other.

After polariser 1: $I_1 = I_0 / 2 = 60$ W m $^{-2}$ .

After polariser 2: $I_2 = I_1 \cos^2 45^\circ = 60 \times 0.5 = 30$ W m $^{-2}$ .

If the second polariser is rotated to $90^\circ$ , transmitted intensity drops to zero ( $\cos^2 90^\circ = 0$ ).

Common traps

Using $d \sin \theta = m \lambda$ for single-slit minima. The double-slit condition gives maxima; for a single slit of width $a$ the same equation form gives minima.

Forgetting the half-intensity step for unpolarised light. Always halve first, then apply Malus's law on subsequent polarisers.

Saying interference proves light is a wave but not a particle. Interference proves light has wave-like behaviour. The photoelectric effect later proves it also has particle-like behaviour. Both are needed.

Confusing path difference with phase difference. Path difference of $m \lambda$ equals phase difference of $2 \pi m$ .

Treating diffraction and interference as different phenomena. They are the same underlying superposition. Diffraction is interference from a continuous range of source points; "double-slit interference" is interference from two discrete sources.

In one sentence

The wave model accounts for double-slit interference fringes at $d \sin \theta = m \lambda$ with spacing $\Delta y = \lambda L / d$ , single-slit diffraction with a broad central maximum, and polarisation governed by Malus's law $I = I_0 \cos^2 \theta$ which establishes light as a transverse wave.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2021 HSC5 marksIn a Young's double-slit experiment, two slits 0.25 mm apart are illuminated with monochromatic light of wavelength 590 nm. The screen sits 1.8 m from the slits. Calculate the fringe spacing on the screen, and explain why a single-slit pattern would not show the same equally spaced bright fringes.

Show worked answer →

Fringe spacing for small angles:

$\Delta y = \lambda L / d = (5.90 \times 10^{-7})(1.8) / (2.5 \times 10^{-4}) = 4.25 \times 10^{-3}$ m $= 4.25$ mm.

Why a single slit looks different: a single slit produces a diffraction pattern from interference of light from the continuous range of points across the slit width. The central maximum is twice as wide as the side maxima, and side maxima fall off rapidly in intensity (envelope $\propto (\sin \alpha / \alpha)^2$ ). A double slit gives equally spaced narrow fringes from path-difference interference between the two slits, modulated by the single-slit envelope of each individual slit. So the equally spaced bright fringes come only when there are at least two coherent sources separated by a fixed distance $d$ .

Markers reward correct fringe-spacing formula and value, plus a clear contrast between two-source interference (equal spacing) and single-slit diffraction (broad central peak with rapidly weakening side peaks).

2018 HSC3 marksUnpolarised light of intensity 80 W m^-2 passes through two polarising filters whose transmission axes are at 30 degrees to each other. Calculate the intensity transmitted by the second filter, and state what would happen if a third filter were inserted between them at 60 degrees to the first.

Show worked answer →

After the first filter, unpolarised light is reduced to half its intensity (only one polarisation component passes):

$I_1 = I_0 / 2 = 80 / 2 = 40$ W m $^{-2}$ .

Through the second filter (Malus's law with $\theta = 30^\circ$ ):

$I_2 = I_1 \cos^2 30^\circ = 40 \times (0.866)^2 = 40 \times 0.75 = 30$ W m $^{-2}$ .

With a third filter inserted at $60^\circ$ to the first (so $30^\circ$ to the second on the entry side, and $30^\circ$ to the original second filter on the exit side):

After the third filter: $I_1 \cos^2 60^\circ = 40 \times 0.25 = 10$ W m $^{-2}$ .

After the original second filter (now $30^\circ$ from the third): $10 \times \cos^2 30^\circ = 10 \times 0.75 = 7.5$ W m $^{-2}$ .

Surprisingly, inserting an extra filter increases the final intensity from $0$ when the original two are crossed at $90^\circ$ , and changes the answer here too. Markers reward the half-intensity step, correct application of Malus's law, and recognition that polarisation order matters.