Wave model of light: Diffraction, interference and polarisation, HSC Physics Module 7

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).

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.

Constructive interference (bright fringe) occurs where the path difference from the two slits is a whole number of wavelengths: $\Delta x = m\lambda$, so the waves arrive in phase.

Destructive interference (dark fringe) occurs where the path difference is a half-odd number of wavelengths: $\Delta x = (m + \tfrac{1}{2})\lambda$, so the waves arrive in antiphase and cancel.

Marks: one for the constructive condition $\Delta x = m\lambda$ (in phase), one for the destructive condition $\Delta x = (m + \tfrac{1}{2})\lambda$ (antiphase).

Use the double-slit maximum condition $d \sin\theta = m\lambda$ with $m = 2$.

$\sin\theta = \dfrac{m\lambda}{d} = \dfrac{2 \times 5.50 \times 10^{-7}}{1.0 \times 10^{-4}} = 1.1 \times 10^{-2}$

$\theta = \sin^{-1}(1.1 \times 10^{-2}) = 0.63^{\circ}$.

Marks: one for correctly substituting into $d\sin\theta = m\lambda$ with $m = 2$, one for $\theta = 0.63^{\circ}$ to two significant figures.

(a) $I = I_0 \cos^2\theta$, where $\theta$ is the angle between the incoming polarisation direction and the analyser's transmission axis.

(b) Fraction transmitted $= \cos^2 60^{\circ} = (0.5)^2 = 0.25$, i.e. $25\%$.

(c) Only a transverse wave has a field oscillating perpendicular to the direction of travel, so a filter can selectively transmit one perpendicular orientation and block others (giving the $\cos^2\theta$ dependence). A longitudinal wave (such as sound) oscillates along its direction of travel and has no perpendicular orientation to select, so it cannot be polarised. That polarisers work on light is therefore direct evidence light is transverse.

Marks: one for stating $I = I_0\cos^2\theta$, one for the correct fraction $0.25$, one for the transverse-wave reasoning (perpendicular oscillation direction can be selected; a longitudinal wave has none).

(a) Rearranging the small-angle fringe-spacing formula $\Delta y = \dfrac{\lambda L}{d}$ for $\lambda$:

$\lambda = \dfrac{\Delta y \, d}{L} = \dfrac{(5.6 \times 10^{-3})(2.0 \times 10^{-4})}{2.0} = 5.6 \times 10^{-7}\ \text{m} = 560\ \text{nm}.$

(b) $560\ \text{nm}$ lies in the visible spectrum between green ($\sim 530\ \text{nm}$) and yellow ($\sim 580\ \text{nm}$), so the light is most likely yellow-green.

Marks: one for correctly rearranging to $\lambda = \Delta y\, d / L$, one for substituting all three values with correct units, one for $\lambda = 5.6 \times 10^{-7}\ \text{m}$ (560 nm) to two significant figures, one for correctly locating this wavelength as yellow-green in the visible spectrum.

(a) Reading the curve at $\theta = 60^{\circ}$ gives $I \approx 2.0\ \text{W m}^{-2}$.

(b) By Malus's law, $I = I_0 \cos^2\theta = 8.0 \times \cos^2 60^{\circ} = 8.0 \times (0.5)^2 = 8.0 \times 0.25 = 2.0\ \text{W m}^{-2}$, which matches the graph reading.

(c) The transmitted intensity is zero at $\theta = 90^{\circ}$ and $\theta = 270^{\circ}$ (only $90^{\circ}$ lies in the stated range $0^{\circ}$-$180^{\circ}$, so within range: $\theta = 90^{\circ}$; by periodicity the curve also touches zero again as $\theta \to 180^{\circ}$ is approached from the pattern of $\cos^2\theta$, but the single zero inside $0^{\circ} < \theta < 180^{\circ}$ is at $90^{\circ}$). At $90^{\circ}$ the analyser's transmission axis is perpendicular ("crossed") to the incoming polarisation direction, so the field component projected onto the axis is $E_0\cos 90^{\circ} = 0$ and no light is transmitted.

Marks: one for reading $I \approx 2.0\ \text{W m}^{-2}$ from the graph at $\theta = 60^{\circ}$, one for the matching Malus's law calculation, one for identifying $\theta = 90^{\circ}$ as the zero within the stated range, one for the crossed-polarisers physical explanation.

Band-6 plan. (1) State what the particle (corpuscular) model predicted and could not explain. (2) Explain how double-slit interference (evenly spaced fringes, $d\sin\theta = m\lambda$) demonstrates constructive/destructive superposition, which only waves do. (3) Explain how single-slit diffraction (spreading into a geometric shadow, broad central maximum) is further wave evidence distinct from two-source interference. (4) Explain why interference/diffraction alone only proves a wave, not specifically a transverse one, and that polarisation is the extra fact that narrows it to a transverse wave. Finish with a synthesising judgement.

Model answer. Newton's particle model treated light as a stream of corpuscles travelling in straight lines, which explains reflection and (with an incorrect speed prediction) refraction, but gives no reason for light to bend around obstacles or to reinforce and cancel itself. Young's double-slit experiment (1801) showed that light from two coherent slits produces alternating bright and dark fringes on a screen, with bright fringes where the path difference is a whole number of wavelengths, $d\sin\theta = m\lambda$, and dark fringes where it is a half-odd number of wavelengths. This alternating pattern is exactly the constructive and destructive superposition expected of two overlapping wave sources, and particles travelling in straight lines through two slits could never produce it - only two bright bands directly behind each slit, not a fringe pattern.

Single-slit diffraction adds a second, independent line of wave evidence. Even a single opening of width $a$ produces a spread pattern with a broad central maximum and progressively weaker side maxima, with minima at $a\sin\theta = m\lambda$. A stream of particles passing through one narrow gap should cast a sharp geometric shadow; instead light spreads into the shadow region, which only happens when the width of the opening is comparable to the wavelength and the wavefronts from across the slit interfere with each other.

However, interference and diffraction alone only establish that light behaves as some kind of wave; they do not distinguish a transverse wave (like light) from a longitudinal wave (like sound). Polarisation supplies that missing piece: passing light through a polarising filter, and finding that a second filter can progressively extinguish the light as it is rotated (following $I = I_0\cos^2\theta$, reaching zero when crossed at $90^{\circ}$), shows that light's oscillation has a definite direction perpendicular to its travel that can be selectively blocked. A longitudinal wave has no such perpendicular direction to select and cannot be polarised at all. Together, interference/diffraction establish the wave nature of light, and polarisation establishes that it is specifically a transverse wave - a complete case that the particle model cannot match.

Marker's note: the top band treats double-slit interference AND single-slit diffraction as distinct pieces of evidence (not one description standing in for both), explicitly states why polarisation is necessary in addition to interference (transverse vs longitudinal, not just "more evidence"), and reaches the synthesising judgement that together the three phenomena complete the case for a transverse wave model. A response that describes the experiments without explaining what each rules in or out of the particle model caps in the middle band.

Band-6 plan. (1) State the fringe-spacing relationship $\Delta y = \lambda L / d$ and identify the $d$-dependence. (2) Directly evaluate the claim: increasing $d$ actually narrows fringes, so the claim is factually wrong, but note the confusion (perhaps with $L$). (3) Discuss the large-$d$ limit (fringes crowd together, become unresolvable, pattern approaches two independent single-slit patterns). (4) Discuss the small-$d$ limit (very wide fringes but weak intensity/practical difficulties). (5) Reach a judgement on how to actually make fringes easier to see.

Model answer. The fringe spacing on the screen is $\Delta y = \dfrac{\lambda L}{d}$, so $\Delta y$ is inversely proportional to the slit separation $d$: as $d$ increases, $\Delta y$ decreases. The student's claim is therefore incorrect - increasing $d$ makes the fringes narrower and more closely spaced, not wider. The student may be confusing $d$ with the screen distance $L$ or the wavelength $\lambda$, both of which do widen the fringes when increased, since $\Delta y \propto L$ and $\Delta y \propto \lambda$.

As $d$ becomes very large, $\Delta y \to 0$ and the fringes crowd together faster than the eye or a detector can resolve, so the pattern washes out into what looks like a uniform overlap of the two slits' individual single-slit diffraction envelopes - the two-source interference structure becomes practically invisible even though it is still present. Conversely, as $d$ becomes very small (approaching the wavelength itself), $\Delta y$ becomes very large, spreading only a few, very wide fringes across the screen; but if $d$ becomes too small the two slits stop behaving as well-separated coherent point sources and the simple small-angle formula $\sin\theta \approx \tan\theta = y/L$ also starts to break down, and diffraction spreading from each slit dominates.

In practice, fringes are made easier to see not by increasing $d$ but by decreasing $d$ (within limits), increasing $L$, or using a longer wavelength - all of which increase $\Delta y$ - while keeping $d$ small enough that a useful number of resolvable fringes still fit on the screen.

Marker's note: the top band correctly identifies the inverse relationship (rejecting the claim, not simply agreeing with it), analyses BOTH limits ($d$ very large and very small) rather than only one, and closes with a concrete, physically correct statement of what does widen the fringes. Marks are lost for treating the claim as correct or for discussing only one limiting case.