PHY132H1/Unit 9

Unit 9 · Introduction to Physics II

Wave Optics

Where rays fail. When light passes an obstacle comparable to its wavelength it spreads, and when two beams overlap they can cancel. Neither has any explanation in the ray picture, and both are what settled the argument about what light is.

By the end you can

• Relate path difference to constructive and destructive interference.
• Analyse the double slit and locate its fringes.
• Explain single-slit diffraction and the resolution limit.
• Account for thin-film colours, including the phase flip on reflection.

Why it matters & where it comes from

Why it exists

Two beams of light producing darkness is impossible for particles and natural for waves. Interference is the decisive evidence, and it also sets a hard limit on what any optical instrument can resolve.

In the real world

Anti-reflective coatings on lenses and glasses, the colours of soap bubbles and oil films, CD and DVD readout, and the diffraction gratings in every spectrometer.

Story & open questions

Young's double-slit experiment of 1801 was ridiculed for contradicting Newton's particle view. A century later the same experiment done one photon at a time still produced fringes, which is the central puzzle of quantum mechanics and remains one.

Interference

Rule Path difference decides everything

Two coherent sources, and a point where their waves arrive having travelled different distances $\Delta$:

Constructive: $\Delta = m\lambda$ for integer $m$ — whole wavelengths, so the waves arrive in step.

Destructive: $\Delta = \left(m + \tfrac12\right)\lambda$ — half wavelengths, so crest meets trough.

Coherent means the sources keep a fixed phase relationship. Ordinary lamps do not, which is why interference needs a laser or a single source split in two.

Example Why two lamps show no fringes

Independent sources shift phase randomly and constantly, so the pattern moves faster than any eye or detector can follow.

Splitting one source into two paths keeps them locked together, which is precisely what a double slit does. The fringes then stand still and can be seen.

The double slit

Rule Fringe positions

Two slits separated by $d$, with a screen at distance $L$:

Bright fringes at $d\sin\theta = m\lambda$.

Dark fringes at $d\sin\theta = \left(m + \tfrac12\right)\lambda$.

For small angles, the spacing on the screen is $\Delta y = \dfrac{\lambda L}{d}$.

Example Measuring a wavelength

Slits $0.1$ mm apart, screen $2$ m away, fringes $1.26$ cm apart.

$\lambda = \dfrac{\Delta y\, d}{L} = \dfrac{(0.0126)(10^{-4})}{2} = 6.3 \times 10^{-7}$ m.

That is $630$ nm — red light. Measuring a distance of less than a millionth of a metre using only a ruler is the point.

Note the inverse dependence: closer slits give wider fringes.

Single-slit diffraction

Rule One slit also produces a pattern

A single slit of width $a$ gives dark fringes at

$a\sin\theta = m\lambda$ for $m = 1, 2, 3, \dots$ — note this is the condition for darkness, the reverse of the double slit.

The central bright band is twice as wide as the others and much brighter.

Narrower slit, wider pattern

Squeezing the slit spreads the light more. This is counterintuitive and is the whole content of diffraction: the more you confine a wave, the more it spreads. It is also, in a different guise, the uncertainty principle.

Example The resolution limit

Every aperture diffracts, so a point source images as a small disc rather than a point. Two sources closer than about

$\theta \approx 1.22\dfrac{\lambda}{D}$

cannot be told apart, where $D$ is the aperture diameter.

This is why large telescopes resolve better — not only more light, but a smaller diffraction limit.

It is also why electron microscopes exist: electrons have a far shorter wavelength than light, so the limit is far finer.

No amount of engineering beats this. It is set by the wave nature of light itself.

Thin films

Rule Two reflections interfere

Light hitting a thin film partly reflects off the top surface and partly off the bottom. The two reflections travel different distances and interfere.

Extra path in the film: $2t$ for near-normal incidence.

The wavelength inside the film is $\lambda/n$, so the path counts in those shortened wavelengths.

Reflection off a denser medium flips the phase

Light reflecting from a medium of higher refractive index gains half a wavelength of phase. Reflecting from a lower index does not.

A soap film in air has a flip at the top surface only, so the two reflections start half a wavelength out of step.

That is why a very thin soap film looks black just before it bursts: with $t$ nearly zero the only difference is the flip, and the reflections cancel.

Forgetting the flip inverts every answer in this section, and it is the most common error in the unit.

Example Anti-reflective coating

A lens coated with a film of index between air and glass.

Both surfaces now reflect off a denser medium, so both reflections flip and the flips cancel out.

Choosing $2t = \lambda/(2n)$, so $t = \lambda/(4n)$, makes the two reflections destructive and kills the reflection.

It works exactly at one wavelength only, usually chosen in the green where the eye is most sensitive. The residual purple tint on coated lenses is the red and blue that were not cancelled.

Common pitfalls

• Swapping the bright and dark conditions between single and double slit.
• Forgetting the half-wavelength phase flip in thin films.
• Using the vacuum wavelength inside a film. Divide by $n$.
• Expecting fringes from two independent lamps. They are not coherent.
• Thinking a narrower slit gives a narrower pattern.

Practice

Work each fully before opening the answer. Original exercises written for this guide in the style of the course's problem sets.

Practice 1 · Path difference

Two waves arrive with a path difference of $1.5\lambda$. Constructive or destructive?

Answer

Destructive — it is a half-integer number of wavelengths.

Practice 2 · Fringe spacing

Slits $0.2$ mm apart, screen $1.5$ m away, $500$ nm light. Find the spacing.

Answer

$\Delta y = \lambda L/d = (5 \times 10^{-7})(1.5)/(2 \times 10^{-4}) = 3.75$ mm.

Practice 3 · Closer slits

What happens to the fringes if the slits are moved closer together?

Answer

They spread further apart. The spacing goes as $1/d$.

Practice 4 · Coherence

Why does a double slit use one source rather than two lamps?

Answer

Independent sources have randomly shifting phase, so the pattern washes out. Splitting one source keeps the two beams locked in phase.

Practice 5 · Single slit

A $0.05$ mm slit with $600$ nm light. Find the angle of the first dark fringe.

Answer

$\sin\theta = \lambda/a = (6 \times 10^{-7})/(5 \times 10^{-5}) = 0.012$, so $\theta = 0.69$ degrees.

Practice 6 · Resolution

Why does a bigger telescope resolve finer detail?

Solution

The diffraction limit $\theta \approx 1.22\lambda/D$ shrinks as the aperture $D$ grows. More light is a separate benefit; the resolution gain is purely a wave effect.

Practice 7 · Black soap film

Why does a soap film look black just before bursting?

Answer

The thickness has gone to nearly zero, so the only difference between the two reflections is the half-wavelength flip at the top surface. They cancel and no light is reflected.

Practice 8 · Coating thickness

Find the thinnest anti-reflective coating for $550$ nm light with $n = 1.38$.

Solution

$t = \lambda/(4n) = (5.5 \times 10^{-7})/(4 \times 1.38) = 99.6$ nm — about a hundred nanometres, which is why such coatings are deposited a few molecular layers at a time.

Expanded topic checklist

Course coverage. This unit is part of the complete PHY132H1 topic map. It should cover the following items before you move on:

• Interference. Path difference conditions and what coherence requires.
• Double slit. Fringe positions and the inverse dependence on separation.
• Diffraction. Single-slit minima and the resolution limit.
• Thin films. The phase flip, the wavelength in the medium, and anti-reflective coatings.

Textbook/source alignment

The calendar's "light" and "optics" topics include wave optics. Check the syllabus on Quercus for assigned problems; the double slit is usually a laboratory experiment as well.

Core idea map

Waves that meet add, and whether they reinforce or cancel depends only on path difference. Slits, films and apertures are all the same calculation with different geometry. Unit 10 turns to what light's constant speed does to space and time.

Study routine. Count wavelengths of path difference, and remember to add half a wavelength for each reflection off a denser medium. Getting that bookkeeping right is the whole unit.

Deep practice extensions

Use these after the main practice set. They push on definitions and edge cases rather than routine calculation.

Extension 1 · Constructive condition

What path difference gives constructive interference?

Guide

A whole number of wavelengths.

Extension 2 · Slit width

What happens to a diffraction pattern as the slit narrows?

Guide

It spreads out. Greater confinement means greater spreading.

Extension 3 · The flip

When does reflection add half a wavelength?

Guide

When reflecting off a medium with a higher refractive index.

Extension 4 · Hard limit

Can better lenses beat the diffraction limit?

Guide

No. It follows from the wave nature of light and the aperture size. Shorter wavelengths are the only way around it, which is what electron microscopy does.

← Unit 08Geometric optics Unit 10 →Special relativity