Waves and Modern Physics 203-SN3-RE • CEGEP, Montreal

Revision sheet: wave optics (203-SN3-RE)

Wave optics is the proof that light is a wave: it bends around obstacles, cancels itself and can be polarized. The formulas of the chapter are short, but they look so alike that they get swapped without anyone noticing.

This sheet gathers the mistakes that cost the most on tests, in a chapter of Waves and Modern Physics, 203-SN3-RE (formerly 203-NYC): confusing the conditions for maxima and minima, the inverted intuition about the width of the spots, the reflection phase shift of thin films, and the two polarization traps, unpolarized light and the three polarizers. It is written for CEGEP students in the Natural Science program (Sciences de la nature).

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The thread of the chapter

Two phenomena look alike and do not have the same formula: interference counts MAXIMA, diffraction counts MINIMA. And in a thin film, an extra half wavelength comes in that appears nowhere in the data of the problem.

This chapter is part of 203-SN3-RE, Waves and Modern Physics

The essentials

Interference: maxima. Diffraction: minima

• Constructive interference if the path difference is $\delta=m\lambda$ ; destructive if $\delta=\left(m+\dfrac{1}{2}\right)\lambda$ .
• Young's slits, separation $d$ , screen at distance $L$ : bright fringes at $y_{m}=\dfrac{m\lambda L}{d}$ , and fringe spacing $i=\dfrac{\lambda L}{d}$ .
• Grating of $N$ lines per millimetre, spacing $d=\dfrac{1}{N}$ : maxima at $d\sin\theta_{m}=m\lambda$ , and the number of orders is limited by $\sin\theta\leq 1$ , so $m_{max}=\left\lfloor\dfrac{d}{\lambda}\right\rfloor$ .
• Diffraction by a slit of width $a$ : $a\sin\theta=m\lambda$ gives the MINIMA, never the maxima. The central maximum extends between the first two minima, over a width $\dfrac{2\lambda L}{a}$ .
• The NARROWER the slit, the WIDER the central maximum: $a$ is in the denominator.

The central maximum extends between the two FIRST MINIMA and is twice the spacing between secondary fringes. Its width is inversely proportional to that of the slit.

The same equation, $\delta=m\lambda$ or $a\sin\theta=m\lambda$ , designates a maximum in one case and a minimum in the other. Writing the word “maxima” or “minima” next to the formula at the moment you set it down avoids the costliest mistake of the chapter.

Thin films: counting half wavelengths

• A reflection from a medium of HIGHER index adds a phase shift of half a wavelength. From a medium of LOWER index, it adds nothing.
• In a film of index $n$ and thickness $t$ , at near-normal incidence, the optical path difference is $2nt$ , to be combined with the phase shifts of the TWO reflections.
• If the two reflections behave alike, the half wavelengths cancel: destructive for $2nt=\left(m+\dfrac{1}{2}\right)\lambda$ . This is the case of an antireflection coating.
• If only one reflection shifts the phase, as for a soap bubble in air: destructive for $2nt=m\lambda$ , which makes the film BLACK at zero thickness.
• Air wedge and Newton's rings: one fringe corresponds to a thickness change of $\dfrac{\lambda}{2}$ , and the radii are $r_{m}=\sqrt{m\lambda R}$ .

The index RISES at both interfaces of the coating: two jumps of

\frac{\lambda}{2}

that cancel. In the bubble, it rises and then falls: a single jump, and a very thin film turns black.

The black contact of a bubble that thins before bursting is the visible proof of the phase shift: without it, zero thickness would give a maximum of light.

Polarization

• A polarizer transmits only the component of the field parallel to its axis.
• UNPOLARIZED light: half passes, $I=\dfrac{I_{0}}{2}$ , with no cosine at all, whatever the orientation of the polarizer.
• Light ALREADY polarized, at an angle $\theta$ to the axis: Malus's law, $I=I_{0}\cos^{2}\theta$ .
• Brewster's angle: $\tan\theta_{B}=\dfrac{n_{2}}{n_{1}}$ ; the reflected ray is then totally polarized perpendicular to the plane of incidence, so horizontally from a horizontal surface.
• This is why polarized glasses have a VERTICAL axis: they block the horizontal glare off roads and water.

Light already polarized:

I=I_{0}\cos^{2}\theta

, zero at

90^{\circ}

. Unpolarized light (dotted): half passes, whatever the angle of the polarizer.

Two crossed polarizers let nothing through, but a third slipped BETWEEN them at $45^{\circ}$ restores $\dfrac{I_{0}}{8}$ . It is not a paradox: each polarizer reprojects the light and gives it a new direction.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Taking the diffraction condition for a condition for maxima

the whole question, 6 marks

What not to write

“The slit gives maxima at $a\sin\theta=m\lambda$ , like a grating.”

What to write

“For a single slit, $a\sin\theta=m\lambda$ gives the MINIMA, with $m=1,\ 2,\ 3\ldots$ and never $m=0$ . The central maximum lies between the first two, at $\theta=0$ .”

Why: The two formulas look alike because they count the same thing, a path difference, but one compares two slits and the other compares the two halves of a single slit, which cancel in pairs.

2. Expecting a narrow spot from a narrow slit

3 marks, and an inverted conclusion in the discussion

What not to write

“We narrow the slit, so the spot of light on the screen narrows too.”

What to write

“The width of the central maximum is $\dfrac{2\lambda L}{a}$ : $a$ is in the DENOMINATOR. Halving the slit DOUBLES the width of the spot.”

Blue:

a=0.10

mm, spot of

22

mm. Purple:

a=0.05

mm, spot of

44

mm. A slit half as wide, a spot twice as WIDE (profiles scaled to the same maximum).

Why: This is the most counter-intuitive result of the chapter, and the most characteristic of a wave: the more you try to constrain it, the more it spreads. The same principle sets the resolution limit of instruments.

3. Forgetting the reflection phase shift in a thin film

the whole question, 8 marks

What not to write

“To extinguish the reflection, we need $2nt=\left(m+\dfrac{1}{2}\right)\lambda$ , always.”

What to write

“It all depends on the TWO reflections. On an antireflection coating, both shift the phase by half a wavelength and cancel: the destructive condition is indeed $2nt=\left(m+\dfrac{1}{2}\right)\lambda$ . On a soap bubble in air, only one shifts: the destructive condition becomes $2nt=m\lambda$ .”

Why: The phase shift appears in no numerical datum of the problem: it is deduced solely from the order of the indices. That is why it gets forgotten, and why it must be written down first.

4. Confusing the position of a fringe with the fringe spacing

3 marks

What not to write

“The third bright fringe is at $i=\dfrac{\lambda L}{d}$ from the centre.”

What to write

“ $i$ is the SPACING between two neighbouring fringes; the third bright fringe is at $y_{3}=\dfrac{3\lambda L}{d}=3i$ . With $d=0.25$ mm, $L=2.0$ m and $\lambda=550$ nm, $i=4.4$ mm and $y_{3}=13.2$ mm.”

i=4.4

mm separates two NEIGHBOURING fringes; the third bright fringe is three spacings farther out, at

y_{3}=3i=13.2

mm.

Why: The fringe spacing is constant, which makes the position of any fringe immediate once $i$ is known. Confusing the two amounts to dividing every position by its order.

5. Announcing more orders than a grating can give

3 marks, often the last question of the problem

What not to write

“The $600$ lines/mm grating lit at $550$ nm gives orders $1$ to $5$ .”

What to write

“ $\sin\theta_{m}=\dfrac{m\lambda}{d}$ cannot exceed $1$ : with $d=1.67$ µm, $m_{max}=\left\lfloor\dfrac{1.67}{0.550}\right\rfloor=3$ . Only orders $1$ , $2$ and $3$ exist.”

We plot

\sin\theta_{m}=\frac{m\lambda}{d}

on the circle of radius

1

0.33

0.66

and

0.99

find a point there,

1.32

passes above it. Three orders, not four.

Why: A sine greater than $1$ is not a calculation error but a piece of information: that order is diffracted beyond $90^{\circ}$ , so it does not exist. The same reasoning is used for total internal reflection.

6. Applying Malus's law to unpolarized light

3 marks, and a factor that drags through the whole question

What not to write

“Sunlight arrives at $30^{\circ}$ to the axis of the polarizer, so $I=I_{0}\cos^{2}30^{\circ}=0.75\,I_{0}$ .”

What to write

“Unpolarized light has no preferred direction: an angle has no meaning for it. The first polarizer always transmits HALF of it, $I=\dfrac{I_{0}}{2}$ , and only after that does Malus apply.”

Why: Malus describes the projection of one direction onto another: it assumes a starting direction. Unpolarized light is a superposition of all directions, whose average of $\cos^{2}$ is exactly $\dfrac{1}{2}$ .

7. Believing that one more polarizer can only darken

the whole question, 5 marks

What not to write

“Two crossed polarizers let nothing through; adding a third in the middle will change nothing, or darken even more.”

What to write

“A third polarizer at $45^{\circ}$ between the two restores some light: $\dfrac{I_{0}}{2}\times\cos^{2}45^{\circ}\times\cos^{2}45^{\circ}=\dfrac{I_{0}}{8}$ . Each polarizer REPROJECTS the light into a new direction.”

Without the middle polarizer, the two outer ones are crossed and nothing passes. With it, the light is reprojected at

45^{\circ}

and one eighth of the unpolarized intensity comes out.

Why: A polarizer is not a filter that removes rays: it projects the field onto its axis. After the second, the light is no longer vertical but at $45^{\circ}$ , and so it has a horizontal component that the third can transmit.

8. Forgetting the factor of 2 of the Michelson interferometer

3 marks, a factor of 2 on the measurement

What not to write

“The mirror moves by $\Delta d$ , so $\Delta d=N\lambda$ .”

What to write

“The light makes the ROUND TRIP along this arm: $2\Delta d=N\lambda$ . For $\Delta d=0.15$ mm and $\lambda=550$ nm, we count $545$ fringes, not $273$ .”

Why: The same factor of 2 appears in every measurement by reflection, from the laser rangefinder to sonar. A gas cell of length $L$ likewise gives $2L(n-1)=N\lambda$ , for the same reason.

9. Forgetting that the medium shortens the wavelength

3 marks

What not to write

“We immerse Young's setup in water: the fringe spacing does not change, since neither $d$ nor $L$ changes.”

What to write

“In a medium of index $n$ , the wavelength becomes $\dfrac{\lambda}{n}$ : the fringe spacing shrinks accordingly, $i'=\dfrac{i}{1.33}$ , that is $3.3$ mm instead of $4.4$ mm.”

Why: It is the wavelength IN the medium that governs interference, and it is divided by the index while the frequency does not change. The same effect explains why a thin film is calculated with $2nt$ and not $2t$ .

Which method to choose

Interference or diffraction, and what is being counted

The number and size of the openings described by the problem

If TWO narrow slits separated by $d$ → interference: maxima at $y_{m}=\dfrac{m\lambda L}{d}$ , fringe spacing $\dfrac{\lambda L}{d}$

Example: $d=0.25$ mm, $L=2.0$ m, $550$ nm: $i=4.4$ mm
If MANY slits, a grating of $N$ lines per mm → maxima at $d\sin\theta=m\lambda$ with $d=\dfrac{1}{N}$ , and $m$ limited by $\sin\theta\leq 1$

Example: $600$ lines/mm at $550$ nm: three orders at most
If A SINGLE slit of width $a$ → diffraction: $a\sin\theta=m\lambda$ gives the MINIMA, central maximum of width $\dfrac{2\lambda L}{a}$

Example: $a=0.10$ mm: spot of $22$ mm
If two slits of NONNEGLIGIBLE width → the interference fringes are modulated by the diffraction envelope, and the multiples of $\dfrac{d}{a}$ are MISSING orders

Example: $\dfrac{d}{a}=4$ : orders $4$ , $8$ , $12$ disappear
If a plate, a film, a bubble, an air wedge → thin films: $2nt$ plus the balance of the reflection phase shifts

Example: quarter-wave antireflection coating, black contact of an air wedge
If a movable mirror, a gas cell → Michelson: $2\Delta d=N\lambda$ or $2L(n-1)=N\lambda$

Example: the factor 2 comes from the round trip

A problem that gives both $d$ and $a$ always announces missing orders: it is the only reason it would supply both. Spotting this extra datum steers the end of the problem before you even begin.

Thin films: establishing the condition in three lines

The order of the indices of the three media crossed

If the index RISES at the first interface → first reflection shifted by half a wavelength

Example: air to film, always
If the index also RISES at the second interface → the two phase shifts cancel: destructive for $2nt=\left(m+\dfrac{1}{2}\right)\lambda$

Example: MgF $_{2}$ antireflection coating on glass
If the index FALLS at the second interface → only one phase shift: destructive for $2nt=m\lambda$

Example: soap bubble in air, air wedge
If the MINIMUM thickness is asked for → take the smallest $m$ that gives $t>0$

Example: $m=0$ in the first case gives $t=\dfrac{\lambda}{4n}$
If a colour REINFORCED rather than extinguished is asked for → take the opposite condition, swapping $m\lambda$ and $\left(m+\dfrac{1}{2}\right)\lambda$

Example: the colours of an oil slick
If the film is an air WEDGE or a lens resting on a plate → each fringe corresponds to $\Delta t=\dfrac{\lambda}{2}$ , and $r_{m}=\sqrt{m\lambda R}$ for the rings

Example: counting fringes gives the angle of the wedge

The very first line of a thin-film problem must be the balance of the phase shifts, even before writing $2nt$ . This line is worth two marks and determines all the following ones.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

A thin-film problem, written up for the marker

When to use it: The problem is about an antireflection coating, a soap bubble, an oil film or Newton's rings.

1 List the three indices in the order of traversal, then state the phase shift of EACH reflection: half a wavelength if the index rises, nothing if it falls.
2 Take the balance: if the two reflections behave the same way, the half wavelengths cancel; otherwise, one remains.
3 Write the optical path difference in the film, $2nt$ , and combine it with the previous balance.
4 Set the condition asked for, extinction or reinforcement, and isolate $t$ .
5 Choose the smallest $m$ that gives a strictly positive thickness, and check the order of magnitude, a few hundred nanometres.

Concluding sentence

“The air-fluoride reflection shifts the phase by half a wavelength, and so does the fluoride-glass reflection, since the index rises at both interfaces: the two phase shifts cancel. Extinction therefore requires $2nt=\left(m+\dfrac{1}{2}\right)\lambda$ , so for $m=0$ the minimum thickness is $t=\dfrac{\lambda}{4n}=99.6$ nm.”

The trap: Applying by reflex the condition of a case already seen. An antireflection coating and a soap bubble have OPPOSITE conditions, and only the balance of the phase shifts tells you which to use.

Marking: Generally 2 marks for the balance of the phase shifts, 2 for the condition set down, 2 for the thickness, 1 for the justified choice of $m$ .

Measuring a wavelength with Young's slits

When to use it: The problem gives an interference pattern measured with a ruler and asks for the wavelength.

1 Measure over SEVERAL fringe spacings rather than one: read the distance between fringe $-5$ and fringe $+5$ , then divide by $10$ .
2 Write $i=\dfrac{\lambda L}{d}$ and isolate $\lambda=\dfrac{i\,d}{L}$ , converting all lengths to metres.
3 Check that the result falls in the visible range, between $400$ and $700$ nm, otherwise a conversion is wrong.
4 If the setup is immersed, state that the measured wavelength is the one IN the medium, and get back to the vacuum value by multiplying by $n$ .

Concluding sentence

“Ten fringe spacings span $44.0$ mm, so $i=4.40$ mm. With $d=0.250$ mm and $L=2.00$ m, $\lambda=\dfrac{i\,d}{L}=\dfrac{(4.40\times 10^{-3})(0.250\times 10^{-3})}{2.00}=550$ nm.”

The trap: Measuring a single fringe spacing. The reading uncertainty, of the order of half a millimetre, then represents more than $10\,\%$ of the result, against $1\,\%$ if you measure over ten fringes.

Marking: 1 mark for measuring over several fringes, 2 for the isolated formula, 1 for the conversions, 1 for the check in the visible range.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The word “maxima” or “minima” written next to the formula

Reread each condition set down and check that it is labelled. Two slits give maxima at $m\lambda$ , one slit gives minima at $m\lambda$ .

An $m=0$ used in a diffraction condition: order zero there is the central maximum, it does not exist as a minimum.
The order of magnitude of a film thickness

A thin film has a thickness of the order of a quarter of the wavelength, so $100$ to $300$ nm. A result in micrometres or millimetres signals a lost power of ten.

A thickness of $99.6$ µm instead of $99.6$ nm: the film would be as thick as a hair.
The sine that exceeds 1

On a grating, calculate $\dfrac{m\lambda}{d}$ for the highest order announced: if it exceeds $1$ , that order does not exist and the answer must say so.

An order $4$ announced with $\sin\theta=1.32$ : you must conclude that it is not observable.
The factor of one half for unpolarized light

Check that a first polarizer receives a factor $\dfrac{1}{2}$ and not a squared cosine, and that the following ones receive Malus with the angle BETWEEN the axes.

A $\cos^{2}$ applied to the first polarizer: the angle used has no physical existence.
The fringe spacing compared with the measurement

Check that the calculated fringe spacing is of the order of a millimetre for a laboratory setup. A result in micrometres or centimetres signals an error on $d$ or on $L$ .

An $i=4.4$ µm: the millimetres of $d$ were not converted to metres.

The typical problem, taken apart

The antireflection coating, from phase shift to thickness

A layer of magnesium fluoride of index $n=1.38$ is deposited on a glass lens of index $n_{g}=1.52$ , in order to extinguish by reflection light of wavelength $\lambda=550$ nm in the middle of the visible range.

Determine the minimum thickness of this layer, then say what the condition would become for a soap bubble of index $1.33$ surrounded by air.

The index rises at BOTH interfaces: the two reflected rays undergo the same half-turn, which cancels in the difference. Only the extra path

2nt

travelled in the film remains.

Step 1

Balance of the phase shifts. Reflection 1, at the air-film interface: the index goes from $1.00$ to $1.38$ , it RISES, so a phase shift of half a wavelength. Reflection 2, at the film-glass interface: from $1.38$ to $1.52$ , it rises again, so a phase shift as well.

Why

This line contains no calculation and yet is worth two marks: it is entirely determined by the order of the indices, a datum the problem supplies without ever commenting on it.

Step 2

The two reflections undergo the SAME phase shift: in the difference between the two rays, the two half wavelengths compensate and disappear.

Why

This is the decisive point. An antireflection coating on glass and a soap bubble in air do not have the same balance, and so not the same condition, even though the geometry is identical.

Step 3

Remaining path difference: ray 2 crosses the thickness of the film twice, in a medium of index $n$ , so $\delta=2nt$ .

Why

It really is $2nt$ and not $2t$ : in the film, the wavelength is $\dfrac{\lambda}{n}$ , so the same thickness contains $n$ times more oscillations.

Step 4

Extinction condition: $2nt=\left(m+\dfrac{1}{2}\right)\lambda$ .

Why

We want the two rays to cancel, so a path difference of a HALF-INTEGER number of wavelengths. The zero balance of the phase shifts lets us write this condition directly.

Step 5

Minimum thickness, for $m=0$ : $t=\dfrac{\lambda}{4n}=\dfrac{550}{4\times 1.38}=99.6$ nm.

Why

We recognize the “quarter-wave” coating: the optical thickness $nt$ is exactly $\dfrac{\lambda}{4}$ , that is $137.5$ nm. That is the name this treatment carries in industry.

Step 6

Check: $2nt=2\times 1.38\times 99.6=275$ nm $=\dfrac{\lambda}{2}$ , which is indeed a half wavelength.

Why

We check the condition on the number obtained, and not on the formula it comes from: it is the only way to catch an error in dividing by $4n$ .

Step 7

The soap bubble case: air, film $n=1.33$ , air. The index rises at the first interface but FALLS at the second: only one phase shift remains. The extinction condition becomes $2nt=m\lambda$ , so a zero thickness extinguishes the light and the bubble turns BLACK just before bursting.

Why

The same setup, with the indices in another order, gives the opposite condition. It is also an observation you can verify by eye, which makes it the best proof of the phase shift.

The conclusion, written out

“The two reflections shift the phase in the same way, so their half wavelengths cancel and extinction requires $2nt=\left(m+\dfrac{1}{2}\right)\lambda$ . The minimum thickness is $t=\dfrac{\lambda}{4n}=99.6$ nm, an optical thickness of a quarter of a wavelength. For a soap bubble in air, only one reflection shifts the phase and the condition becomes $2nt=m\lambda$ , which makes the film black at zero thickness.”

The classic mistake on this problem: Writing $2nt=m\lambda$ directly for the antireflection coating, by analogy with the soap bubble. You then find $t=\dfrac{\lambda}{2n}=199$ nm, a thickness that REINFORCES the reflection instead of extinguishing it: the treatment would turn the lens into a mirror.

Learn by heart

• Interference: maxima at $\delta=m\lambda$ . Single-slit diffraction: MINIMA at $a\sin\theta=m\lambda$ .
• Young: $i=\dfrac{\lambda L}{d}$ and $y_{m}=m\,i$ .
• Grating: $d\sin\theta=m\lambda$ with $d=\dfrac{1}{N}$ , and $m$ limited by $\sin\theta\leq 1$ .
• Central diffraction maximum: width $\dfrac{2\lambda L}{a}$ , so the wider the narrower the slit.
• Reflection from a higher-index medium: half a wavelength extra. Otherwise, nothing.
• Thin film: $2nt$ , plus the balance of the two phase shifts. Antireflection: $2nt=\left(m+\dfrac{1}{2}\right)\lambda$ ; bubble: $2nt=m\lambda$ .
• In a medium of index $n$ , $\lambda$ becomes $\dfrac{\lambda}{n}$ and the fringe spacing shrinks accordingly.
• Unpolarized light: half passes. Polarized light: $I=I_{0}\cos^{2}\theta$ .
• Michelson: $2\Delta d=N\lambda$ , because of the round trip.
• Brewster: $\tan\theta_{B}=\dfrac{n_{2}}{n_{1}}$ , glare polarized horizontally.

Frequently asked questions

What is the difference between interference and diffraction?

Interference comes from several openings whose waves superpose, and its condition with a whole number of wavelengths gives maxima. Diffraction comes from a single opening whose halves cancel, and the same writing gives minima there. Confusing the two amounts to putting the bright spots where it is dark.

Why does a narrower slit give a wider spot?

Because the width of the central maximum is proportional to the wavelength and inversely proportional to the width of the slit. The more a wave is confined in space, the more it spreads out on leaving. This is typically wave behaviour, and it also sets the resolution limit of optical instruments.

Why is an antireflection coating a quarter of a wavelength thick?

Because the light crosses the layer twice, which gives half a wavelength of delay, exactly what is needed to cancel the first reflected ray. Since the two reflections are shifted in phase in the same way, their half wavelengths compensate and only this extra path remains.

Why does a soap bubble turn black before it bursts?

Because at almost zero thickness, the two reflected rays no longer have any path difference, but the first has undergone a half-turn that the second has not. They therefore cancel exactly, and the film returns no light at all. It is the visible proof of the phase shift on reflection.

What happens if you slip a third polarizer between two crossed polarizers?

Light reappears, whereas nothing passed before. The middle polarizer reprojects the light into an intermediate direction, which then has a component along the last axis. With an angle of forty-five degrees, one eighth of the original unpolarized intensity comes out.

Why do polarized glasses have a vertical axis?

Because the light reflected by a horizontal surface, a wet road or a body of water, is polarized horizontally, especially near Brewster's angle. A filter with a vertical axis therefore blocks precisely this light, which removes the glare without darkening the rest of the landscape much.

Practise it

Corrected exercises: Wave optics, 203-SN3-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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Revision sheet: wave optics (203-SN3-RE)

The essentials

Interference: maxima. Diffraction: minima

Thin films: counting half wavelengths

Polarization

The mistakes that cost marks

1. Taking the diffraction condition for a condition for maxima

2. Expecting a narrow spot from a narrow slit

3. Forgetting the reflection phase shift in a thin film

4. Confusing the position of a fringe with the fringe spacing

5. Announcing more orders than a grating can give

6. Applying Malus's law to unpolarized light

7. Believing that one more polarizer can only darken

8. Forgetting the factor of 2 of the Michelson interferometer

9. Forgetting that the medium shortens the wavelength

Which method to choose

Interference or diffraction, and what is being counted

Thin films: establishing the condition in three lines

How the answer is expected to be written

A thin-film problem, written up for the marker

Measuring a wavelength with Young's slits

Check before you hand in

The typical problem, taken apart

The antireflection coating, from phase shift to thickness

Learn by heart

Frequently asked questions

What is the difference between interference and diffraction?

Why does a narrower slit give a wider spot?

Why is an antireflection coating a quarter of a wavelength thick?

Why does a soap bubble turn black before it bursts?

What happens if you slip a third polarizer between two crossed polarizers?

Why do polarized glasses have a vertical axis?

Corrected exercises: Wave optics, 203-SN3-RE

See also

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