PHYS 101 Introductory Physics, Mechanics • McGill University, Montreal

Wave optics and interference: what costs marks in PHYS 101

Q: How do I know whether a formula gives the bright or the dark fringes?

By the apparatus, never by the algebra, because the two equations look identical. If the letter multiplying the sine is the distance between two slits or between two grating lines, the formula gives the bright directions. If it is the width of a single opening, the same formula gives the dark ones, and the centre of the pattern is then the brightest point. Write the letter and its meaning on the paper before you substitute any number.

Q: Why does a narrower slit make a wider diffraction pattern?

Because the pattern is governed by the ratio of the wavelength to the width of the opening. The first dark direction has a sine equal to that ratio, so shrinking the opening raises the ratio and pushes the first zero further out. The central bright band is twice the wavelength times the screen distance divided by the slit width, so halving the slit doubles the band. It is the same effect that makes a pinhole camera blurry past a certain smallness.

Q: What changes when the whole double slit experiment is put under water?

The frequency of the light does not change, because it is fixed by the source. The speed drops to the speed in vacuum divided by the index, so the wavelength drops in the same ratio. Every fringe spacing therefore shrinks by the index: with water at one point three three, a spacing of five point zero six millimetres becomes three point eight one millimetres. Answering that nothing changes because no distance moved is the classic loss of a full mark.

Q: Why does a soap film look black just before it bursts?

Because of the half wave flip at the first reflection. As the film drains, its top gets far thinner than a wavelength, so the extra distance travelled inside it goes to zero. What remains is the flip at the air to soap face, worth half a wavelength, and it is not cancelled by anything at the soap to air face. The two reflected waves are then exactly out of step for every colour at once, so nothing comes back and the top of the film looks black.

Q: Why can a light microscope not show a virus even at high magnification?

Because the limit is diffraction, not the lenses. Light entering a round opening spreads into a small disc, and two details closer than about half a wavelength produce discs that overlap and cannot be separated. With an oil immersion objective that limit is around two hundred and forty nanometres, and a virus is about one hundred. Magnification does not appear anywhere in that limit, so turning it up only gives a bigger blur, which is why the term empty magnification exists.

Q: Does a double slit under water show more fringes in its central band?

No, the same number. Under water the wavelength is divided by the index of the water, so the fringes close up, but the central diffraction band is made by the same light and closes up by exactly the same factor. The count depends only on the ratio of the slit separation to the slit width, which contains no wavelength. Only a change of the slits themselves can change it, never a change of medium or of colour.

This sheet answers one question: what loses marks on wave optics, and what exactly do you write instead. The lecture notes are already in your hands, so nothing here repeats them.

The chapter reduces to a path difference counted in wavelengths. A whole number is a bright fringe, a half integer a dark one, and every formula below is that sentence written for one particular apparatus. Read the traps first, then the grid of reflection flips: together they cover most of what is actually lost on the final.

The PHYS 101 formula sheet gives you the two slit maxima, the single slit minima with the letter w, the Rayleigh criterion, the speed of light in a medium and refraction written with the wave speeds. It does not tell you which letter is a separation and which a width, nor when a reflection flips the wave: that is what this sheet adds.

Updated on October 25, 2026

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The thread of the chapter

The whole chapter is one question: two paths reach the same point, by how many wavelengths do they differ? A whole number gives a bright fringe, a half integer gives a dark one. The costly part is that the same line $D\sin\theta = m\lambda$ means MAXIMA when $D$ separates two slits and MINIMA when $D$ is the width of one.

This chapter is part of PHYS 101, Introductory Physics - Mechanics (McGill)

Before this chapter, you need

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Huygens principle
Path difference
Constructive and destructive interference
Young double slit experiment
Fringe spacing
Small angle approximation
Single slit diffraction
Diffraction grating
Thin film interference
Phase change on reflection
Malus law
Rayleigh criterion

The essentials

On your formula sheet: this chapter's lines, and when each one holds

• $v=c/n$ : $n\ge 1$ , so light is never faster in a material than in vacuum, and its wavelength there is $\frac{\lambda}{n}$ .
• $d\sin\theta_{m}=m\lambda$ : Two slits: BRIGHT fringes, $m=0,\pm1,\pm2,\dots$ , with $d$ the distance between the centres of the slits.
• $w\sin\theta_{m}=m\lambda$ : One slit: DARK fringes, $m=\pm1,\pm2,\dots$ ( $m=0$ is the bright centre), with $w$ the width of the slit. Same form as the two slits, opposite meaning.
• $\theta=1.22\lambda/D$ : A circular opening of diameter $D$ ; $\theta$ , in radians, is the smallest angle between two points that can still be told apart.

The sheet prints each equation, never the condition under which it holds: copy the condition next to its line on your own sheet before the exam.

The only question of the chapter

• Two paths reach the same point. Compute the extra distance $\delta$ travelled by one of them, then divide by the wavelength: the whole chapter is the decimal part of $\delta / \lambda$ .
• $\delta / \lambda$ a whole number: the crests arrive together, BRIGHT.
• $\delta / \lambda$ a half integer: a crest meets a trough, DARK.
• Anything else: in between, and the closer to a half integer the dimmer. A point with $\delta = 1.38\lambda$ is neither a maximum nor a zero, and saying so is a complete answer.

Left, the two arrivals differ by one whole wavelength and the sum is twice as tall: bright. Right, they differ by half a wavelength and the sum is the flat line: dark. Nothing else in the chapter is different from this.

Converting $\delta$ into wavelengths before doing anything else is the single habit that makes this chapter easy. It also survives into the film problems, where $\delta = 2nt$ .

The four conditions, and what the letter in front of the sine means

• Two slits a distance $d$ apart, or a grating of spacing $d$ : $d\sin\theta = m\lambda$ gives the MAXIMA, $m = 0, \pm 1, \pm 2, \ldots$
• One slit of width $a$ : $a\sin\theta = m\lambda$ gives the MINIMA, with $m = 0$ EXCLUDED, since $\theta = 0$ is the brightest point of the pattern. The PHYS 101 formula sheet writes the same line $w\sin\theta_m = m\lambda$ , $w$ being the width.
• Fringe spacing on a screen at distance $L$ : $\Delta y = \lambda L / d$ , the fringes being evenly spaced.
• Width of the central diffraction band: $2\lambda L / a$ , twice the spacing of the side bands.
• Grating with $N$ lines lit: resolving power $R = \lambda / \Delta\lambda = mN$ , which is why a grating separates what two slits cannot.

Same laser, same screen at

2.00

m. Top, two narrow slits

0.250

mm apart: the equation with

d

lands on every BRIGHT fringe,

5.06

mm apart. Bottom, one slit

0.100

mm wide: the equation with

a

lands on the DARK points, never on the centre, and the central band,

25.3

mm, is twice a side band.

Write the letter with its MEANING on the paper before substituting anything: $d$ between two slits, $a$ the width of one. The algebra is identical, so only that sentence decides.

Light in a medium, and when the small angle form is allowed

• Entering a medium of index $n$ , the frequency does NOT change, the speed becomes $c / n$ , so $\lambda_n = \lambda / n$ . Every fringe spacing shrinks by the factor $n$ .
• Inside a film, the path difference in wavelengths is therefore $2nt / \lambda$ with $\lambda$ measured in air.
• Below about $8^{\circ}$ , $\sin\theta$ , $\tan\theta$ and $\theta$ in radians agree to better than one percent (at $10^{\circ}$ , $\tan\theta$ already exceeds $\sin\theta$ by $1.5$ percent), which is what turns $d\sin\theta = m\lambda$ into $y_m = m\lambda L / d$ .
• At $20^{\circ}$ , a grating angle, the agreement is gone: $\sin\theta = 0.3534$ against $\theta = 0.3612$ rad, two percent apart.
• $\sin\theta$ never exceeds $1$ . An order with $\sin\theta > 1$ does not exist, and that is the answer to write.

\sin\theta

\theta

in radians and

\tan\theta

against the angle in degrees: a single curve up to about

10^{\circ}

, three separate ones by

20^{\circ}

, where a grating works:

0.3534

against

0.3612

20.7^{\circ}

Visible light runs from about $400$ nm to $700$ nm, so $\lambda$ is a few tenths of a micrometre. Any wavelength coming out of a calculation outside that range is an arithmetic error, not a discovery.

Refraction written with the speeds, the form of the formula sheet

• The formula sheet gives $v = c/n$ and $v_1\sin\theta_2 = v_2\sin\theta_1$ . Replace each $v$ by $c/n$ and it becomes $n_1\sin\theta_1 = n_2\sin\theta_2$ : one law, two spellings.
• Huygens proves it: while the far end of a front travels $v_1 t$ to the boundary, the wavelet from the near end grows to $v_2 t$ , and both are sides of right triangles on the same stretch $\ell$ of boundary, $v_1 t = \ell\sin\theta_1$ and $v_2 t = \ell\sin\theta_2$ .
• Direction check: the slower medium has the smaller angle. Light entering water at $30.0^{\circ}$ goes on at $22.1^{\circ}$ , closer to the normal, with $v = 2.26\times 10^{8}$ m/s.
• The fronts are one period apart on both sides, so their spacing, the wavelength, drops in the ratio $v_2 / v_1 = 1/n$ : that is where $\lambda_n = \lambda / n$ comes from.

While

B

travels

v_1 t

in air, the wavelet from

A

grows to

v_2 t

in water; the new front

B'T

is its tangent, so the angle in the slower water is smaller and the fronts are closer together.

Swapping the speeds, $v_1\sin\theta_1 = v_2\sin\theta_2$ , sends the light AWAY from the normal into the slower medium: $41.7^{\circ}$ instead of $22.1^{\circ}$ . A thin wall of glass on the way changes nothing, since $\sin\theta / v$ is the same in every layer.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Thin films: count the flips before choosing a formula

Read this grid whenever a question says film, coating, bubble, oil or wedge. A reflection off a HIGHER index flips the wave and is worth an extra $\lambda / 2$ ; a reflection off a lower index does not. The last column is what BRIGHT costs once the flips are counted, and the red line is the rule that does not exist.

Film, outside in	Flips at the two faces	BRIGHT reflection needs
$1.00 \to 1.33 \to 1.00$	one, then none	$2nt = (m + \tfrac{1}{2})\lambda$
Example: Soap in air, green at $550$ nm: $t = \lambda / (4n) = 550 / (4\times 1.33) = 103$ nm.
$1.00 \to 1.38 \to 1.52$	one, then one	$2nt = m\lambda$
Example: Coating on glass: bright at $t = 550 / (2\times 1.38) = 199$ nm, and the anti-reflection thickness is the other one, $99.6$ nm.
$1.00 \to 1.20 \to 1.33$	one, then one	$2nt = m\lambda$
Example: Oil on water: the thinnest non zero film is $t = 550 / (2\times 1.20) = 229$ nm, since $m = 0$ would mean no film at all.
$1.52 \to 1.00 \to 1.52$	none, then one	$2t = (m + \tfrac{1}{2})\lambda$
Example: Air wedge at $589$ nm: DARK at the contact edge where $t = 0$ , first bright line at $t = 589 / 4 = 147$ nm.
any film	not counted	$2nt = m\lambda$ no such rule
Example: A soap film of $t = 207$ nm satisfies $2nt = \lambda$ exactly, since $2\times 1.33\times 207 = 550$ nm, and yet it is DARK. What to do: Count the flips at the two faces first, THEN choose between $m\lambda$ and $(m + \tfrac{1}{2})\lambda$ . The formula alone decides nothing.

The check that costs nothing: a film thinner than a few nanometres has $2nt \to 0$ , so with one flip it must be BLACK in reflection. That black band at the top of a draining soap film is the experimental proof that flips are real.

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. Reading the same equation as the bright directions whatever the apparatus

the whole question: every answer lands exactly where the light is not

What not to write

“The slit is $0.100$ mm wide, so the first bright fringe is at $\sin\theta = \lambda / a$ .”

What to write

“One slit of width $a$ : $a\sin\theta = m\lambda$ gives the DARK directions, $m \neq 0$ . The centre $\theta = 0$ is the brightest point of the pattern.”

Same equation, opposite meaning. On the left

d

is the distance BETWEEN two slits and the equation locates the bright fringes; on the right

a

is the WIDTH of one slit and the same equation locates the dark ones.

Why: The algebra of the two situations is identical, so it cannot decide anything. Only the sentence does: two openings means separation and maxima, one opening means width and minima.

2. Numbering the first dark fringe of Young's experiment as one

2 points, and every dark fringe after it is shifted by one slot

What not to write

“First dark fringe: $d\sin\theta = (1 + \tfrac{1}{2})\lambda$ , so $\theta = 0.218^{\circ}$ .”

What to write

“First dark fringe: $m = 0$ , so $d\sin\theta = \tfrac{\lambda}{2}$ and $\theta = 0.0725^{\circ}$ , exactly half the angle of the first bright fringe.”

Dark fringes sit at half integers of path difference, and the first is

\delta = \lambda / 2

: the counter starts at

m = 0

. The value

(1 + \tfrac{1}{2})\lambda

is the SECOND dark fringe, at

0.218^{\circ}

, three times the angle of the first.

Why: The half integer starts at $\tfrac{1}{2}$ , so the counter starts at zero. The free check is that the first dark fringe always sits at half the angle of the first bright one.

3. Putting the order zero into the single slit minimum condition

1 point, plus the shape of the sketch, which is often another point

What not to write

“ $a\sin\theta = 0$ has the solution $\theta = 0$ , so the middle of the pattern is dark.”

What to write

“ $m = 0$ is excluded: at $\theta = 0$ every strip of the slit arrives in step, so the centre is the CENTRAL MAXIMUM, the brightest point of the whole pattern.”

Why: The pairing argument cancels two strips whose paths differ by half a wavelength. With $m = 0$ they differ by nothing, which is the opposite of cancellation.

4. Believing that a narrower slit gives a narrower pattern

2 points, and the wrong conclusion in every design question

What not to write

“I close the slit from $0.100$ mm down to $0.0500$ mm, so the bright spot gets smaller.”

What to write

“The central band is $2\lambda L / a$ wide, so halving $a$ DOUBLES it: from $25.3$ mm to $50.6$ mm on a screen at $2.00$ m.”

Why: The pattern is governed by $\lambda / a$ , so the smaller the opening the bigger the angle. It is the same inversion that makes a pinhole camera blurry past a certain smallness.

5. Leaving the wavelength alone when the light changes medium

2 points, and the same error poisons every thin film answer

What not to write

“The apparatus is lowered into water but nothing has moved, so the spacing is still $5.06$ mm.”

What to write

“In water $\lambda_n = \lambda / n = 633 / 1.33 = 476$ nm, so $\Delta y = 5.06 / 1.33 = 3.81$ mm. What does not change is the FREQUENCY.”

Same slits, same screen: in water the laser is still red, but its wavelength inside drops to

476

nm and the first bright fringe (orange point) moves in from

5.06

3.81

mm.

Why: The source fixes the frequency once and for all; the medium changes the speed to $c / n$ , and $\lambda = v / f$ does the rest. Inside a film that factor $n$ is exactly why the path difference is $2nt$ and not $2t$ .

6. Counting more fringes in the central band once the slits are under water

the whole multiple choice question, and the same slip on any change of colour

What not to write

“Under water the fringes close up by the factor $n$ , so more of them fit inside the central band: seven instead of five.”

What to write

“Under water BOTH lengths shrink by $n$ : the spacing to $\lambda L / (nd) = 3.81$ mm and the central band to $19.0$ mm. Their ratio $2d / a = 5.0$ has no wavelength in it, so the count stays at five.”

Air, then water: everything shrinks by

1.33

and five fringes stay in the band. In red, the false picture, water fringes under the air envelope: seven.

Why: The envelope is the diffraction of the same light, so it obeys the same $\lambda / n$ as the fringes. Only a change of $d$ or of $a$ changes the count; a change of medium or of colour never does.

7. Skipping the half wave flip at a reflection

the whole question: the answer comes out exactly a factor of two too thick

What not to write

“The film is bright when $2nt = m\lambda$ , so the thinnest green soap film is $t = \lambda / (2n) = 207$ nm.”

What to write

“Air to soap the index goes up, one flip; soap to air it goes down, none. Net one flip, so bright needs $2nt = (m + \tfrac{1}{2})\lambda$ and $t = \lambda / (4n) = 103$ nm.”

Count the arrows, not the indices. One flip alone, on the left, forces the half integer condition; two flips, in the middle and on the right, cancel each other and restore the whole number one.

Why: The flip is worth half a wavelength and it does not cancel unless there are two of them. Counting the flips before writing the condition takes five seconds and decides bright against dark.

8. Applying Malus's law at the first polarizer

1 point, and the answer depends on an angle that has no meaning

What not to write

“The light meets the filter at $30^{\circ}$ , so $I = I_0\cos^{2}(30^{\circ}) = 0.75\,I_0$ .”

What to write

“The incoming light is unpolarized, so the first filter transmits exactly $I_0 / 2$ whatever its orientation. Malus applies only from the SECOND filter on: $I = (I_0 / 2)\cos^{2}(30^{\circ}) = 0.375\,I_0$ .”

Fraction transmitted by one filter against its angle. Unpolarized light gives one half at every angle (flat line): there is no angle to put into Malus's law. Only light that is already polarized follows

\cos^{2}\theta

, and its

0.75

30^{\circ}

is exactly the false answer of the trap.

Why: Unpolarized light has no direction to measure an angle from; it is an average over all of them, and that average is one half. The angle only exists once the light has been polarized.

9. Using the small angle form at a grating

1 point per angle, and three or four angles per grating question

What not to write

“ $\theta = m\lambda / d = 0.3534$ rad, that is $20.25^{\circ}$ .”

What to write

“ $\sin\theta = m\lambda / d = 0.3534$ , so $\theta = \arcsin(0.3534) = 20.695^{\circ}$ : past $10^{\circ}$ the approximation is worth nothing.”

Why: Gratings work at large angles by design, since that is what separates the colours. The small angle form belongs to the double slit experiment, where the fringes sit within a fraction of a degree of the axis.

10. Answering for an order that the apparatus cannot produce

1 point, and it is the free point of the question

What not to write

“Third order of sodium on a $600$ lines per millimetre grating: $\sin\theta = 1.06$ , so $\theta = 1.06$ rad $= 60.7^{\circ}$ .”

What to write

“ $\sin\theta = 3\lambda / d = 1.06 > 1$ , which has no solution: this grating does not produce a third order at $589$ nm. The highest order is $m = 2$ .”

Why: A sine above one is not a rounding complaint, it is a physical impossibility. Checking $\sin\theta \le 1$ before taking any arcsine catches the mistake in two seconds.

Which method to choose

Which condition, read from the apparatus and not from the letters

Before any number, find the sentence in the statement that says what the light met. That sentence, and nothing else, chooses the formula.

If two narrow slits a distance $d$ apart, or a grating of $N$ lines per millimetre → $d\sin\theta = m\lambda$ gives the MAXIMA, $m = 0, \pm 1, \pm 2, \ldots$

Example: $600$ lines per mm gives $d = 1.67\times 10^{-6}$ m and $\theta_1 = 20.7^{\circ}$ at $589$ nm

for a grating, $d$ is one divided by the number of lines per unit length, in metres
If one opening of width $a$ , or a beam clipped by an aperture → $a\sin\theta = m\lambda$ gives the MINIMA, with $m = 0$ excluded

Example: $a = 0.100$ mm at $633$ nm puts the first zero at $0.363^{\circ}$

the central band is $2\lambda L / a$ wide, twice the side bands
If a screen at a distance $L$ and angles under a degree → positions instead of angles: $\Delta y = \lambda L / d$ between fringes

Example: $d = 0.250$ mm and $L = 2.00$ m give $5.06$ mm per fringe

check the angle afterwards; past $10^{\circ}$ go back to $y = L\tan\theta$
If a film, a coating, a bubble or an air wedge of thickness $t$ → count the flips, then $2nt = m\lambda$ or $2nt = (m + \tfrac{1}{2})\lambda$

Example: soap in air, one flip, green at $t = 103$ nm

the $n$ is already inside $2nt$ , it is not divided out a second time
If two polarizers, or glare off a horizontal surface → one half for unpolarized light, then $I = I_0\cos^{2}\theta$ ; glare is total at $\tan\theta_B = n$

Example: water $n = 1.33$ gives $\theta_B = 53.1^{\circ}$ from the normal

the reflected light is polarized parallel to the surface, so the filter axis is vertical
If two objects, an instrument, and the word resolve or distinguish → Rayleigh: $\theta_{\min} = 1.22\lambda / D$ , resolved while $s / R > \theta_{\min}$

Example: a $3.0$ mm pupil at $550$ nm gives $2.24\times 10^{-4}$ rad

magnification does not appear, which is why it cannot rescue a microscope

If no branch fits, go back to the fil: what are the two paths that reach this point, and what is the extra distance between them? Every formula above is that question already answered.

Counting the flips of a film

Write the three indices in order, from the outside in. Then cross the two faces one at a time.

If the index goes UP across the face → one flip, worth an extra $\lambda / 2$

Example: air to soap, $1.00$ to $1.33$
If the index goes DOWN across the face → no flip at all

Example: soap to air, $1.33$ to $1.00$
If one flip in total over the two faces → bright is $2nt = (m + \tfrac{1}{2})\lambda$ , dark is $2nt = m\lambda$

Example: soap in air, green reflection at $t = 103$ nm

a film far thinner than the wavelength is then BLACK in reflection
If two flips, or none at all → they cancel: bright is $2nt = m\lambda$ , dark is $2nt = (m + \tfrac{1}{2})\lambda$

Example: $1.00$ to $1.38$ to $1.52$ , reflection killed at $t = 99.6$ nm

Nothing here depends on the thickness, only on the order of the three indices. Do the count before touching the calculator, and write flip or no flip under each face on the paper: markers give the point for that line alone.

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.

Answering a double slit question the way a marker expects

When to use it: The statement gives two slits, a wavelength, a screen distance, and asks for an angle or a position.

1 Name the letters before substituting anything: $d$ is the distance BETWEEN the slits, $L$ the slits to screen distance, and $a$ would be the width of one slit, which is not what is asked here.
2 Write the condition together with the word it belongs to: bright, $d\sin\theta = m\lambda$ ; dark, $d\sin\theta = (m + \tfrac{1}{2})\lambda$ with the first dark fringe at $m = 0$ .
3 Convert every length into metres on one line, powers of ten included, and keep them in that form to the end.
4 Solve for $\sin\theta$ , check that it does not exceed $1$ , and only then take the arcsine.
5 Turn the angle into a screen position with $y = L\tan\theta$ , and say in one clause whether the small angle form $y_m = m\lambda L / d$ was used and why it is allowed.

Concluding sentence

“The third bright fringe is at $y_3 = 3\lambda L / d = 15.2$ mm from the centre, and the small angle form is justified since $\theta = 0.435^{\circ}$ stays well under $10^{\circ}$ .”

The trap: Skipping the first line, the one that names $d$ . Most of the marks lost on this chapter are lost there and not in the arithmetic.

Marking: 1 point for the named condition, 1 for the substitution in metres, 1 for the angle, 1 for the position, 1 for the justification of the approximation.

Answering a thin film question

When to use it: The statement gives a film, a thickness or a colour, and the words reflected, seen in reflection or coating.

1 Write the three indices in order from the outside in, including the one of the substrate, which the statement often hides in a single word such as glass or water.
2 Cross the two faces one at a time and write flip or no flip under each, with the two indices that justify it.
3 Add them: one flip adds $\lambda / 2$ , two flips or none add nothing.
4 Write the condition actually asked for, bright or dark, then solve for $t$ with the smallest whole $m$ that gives a positive thickness, rejecting $m = 0$ out loud when it means no film at all.

Concluding sentence

“The two faces give one flip in total, so a bright reflection needs $2nt = (m + \tfrac{1}{2})\lambda$ , and the thinnest film is $t = \lambda / (4n) = 103$ nm.”

The trap: Treating the index twice: $n$ is already inside the path difference $2nt$ , so the wavelength in the condition is the one measured in AIR and is never divided by $n$ again.

Marking: 2 points for the flips written face by face, 2 for the condition, 1 for the arithmetic and the rejection of $m = 0$ .

Check before you hand in

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

The sine never exceeds one

Before taking any arcsine, look at the value. If $m\lambda / d$ or $m\lambda / a$ comes out above $1$ , that order simply does not exist, and the sentence to write is that the apparatus does not produce it.

Third order of sodium on a $600$ lines per millimetre grating gives $1.06$ : no third order, and the calculator returning an error is the instrument agreeing with you.
The order of magnitude on a laboratory bench

A fringe spacing is a few millimetres, a film thickness a few hundred nanometres, a visible wavelength between $400$ and $700$ nm. An answer in metres or in micrometres means a power of ten was dropped in a conversion.

A spacing coming out at $5.06$ m says that $d$ was entered in millimetres instead of metres; a spacing of $5.06$ micrometres says the same about $L$ .
Put the answer back and recover the wavelength

Take the position you found and run the formula backwards: $\lambda = \Delta y\, d / L$ . If the wavelength that comes back is not inside the visible range, the answer is wrong and no further checking is needed.

$\Delta y = 5.06$ mm with $d = 0.250$ mm and $L = 2.00$ m gives $633$ nm, a red laser. Getting $63.3$ nm would be ultraviolet, which no helium neon laser emits.
The contact edge of an air wedge is dark

Whatever the algebra says, the line where the two plates touch has zero thickness and is DARK, because the single flip survives when the path difference goes to zero. If your condition makes it bright, you dropped the flip.

$2t = m\lambda$ with $t = 0$ and $m = 0$ : a dark line. The same check works on a soap film, whose draining top goes black before it bursts.
Degrees or radians, decided by the number itself

A double slit angle is a fraction of a degree, a grating angle is tens of degrees, a Rayleigh angle is $10^{-4}$ radians or less. Reading the size of your answer tells you at once whether the calculator was in the wrong mode.

A first order double slit angle of $0.145$ is fine in degrees and absurd in radians, since $0.145$ rad would be $8.3^{\circ}$ , thirty times too far out.
A formula in letters passes three tests

Before using a formula you wrote in letters, check its dimensions, put in a value for which you already know the answer, and check that it varies the right way. A formula that fails one test is wrong, whatever its algebra, and a limit where it gives nonsense marks the edge of its domain.

Central band under water, $W = 2\lambda L / (na)$ : a length; $n = 1$ gives back $2\lambda L / a$ ; a larger $n$ or $a$ narrows it. As $a$ shrinks to $\lambda / n$ the exact first zero goes to $90^{\circ}$ while the formula still gives $2L$ : the small angle form has left its domain.

The typical problem, taken apart

Counting the fringes inside the diffraction envelope

A helium neon laser, $\lambda = 633$ nm, lights a double slit whose two openings are $a = 0.100$ mm wide and whose centres are $d = 0.250$ mm apart. The screen is $L = 2.00$ m away.

How many bright fringes are visible inside the central band of the diffraction pattern? This is the exercise that crosses the two halves of the chapter, because it needs $d$ and $a$ in the same question and the two letters do opposite things.

Two letters on the same apparatus:

d

is the distance between the two slit centres and sets the fringes,

a

is the width of each slit and sets the broad band they sit inside. Confusing them is the whole difficulty.

Step 1

Fringes, from $d$ alone: $\Delta y = \lambda L / d = (633\times 10^{-9}\times 2.00) / 2.50\times 10^{-4} = 5.06$ mm.

Why

Start with the letter that is given first in the statement and never mix the two. This line uses $d$ and only $d$ , so nothing about the slit width can contaminate it.

Step 2

Envelope, from $a$ alone: $\sin\theta_1 = \lambda / a = 6.33\times 10^{-3}$ , so $y_1 = L\tan\theta_1 = 12.66$ mm.

Why

The same algebra as step 1 with a different letter, and it gives a MINIMUM instead of a maximum. Writing the word minimum next to it on the paper is what stops the two from being swapped three lines later.

Step 3

Width of the central band: $2y_1 = 25.3$ mm, from $-12.66$ mm to $+12.66$ mm.

Why

The band is symmetric about the centre, so the half width found in step 2 doubles. This is the interval the count will be made in, and it must be written down before any counting starts.

Step 4

Fringes inside: $y_m = m\times 5.06$ mm with $|y_m| < 12.66$ mm, so $|m| < 2.5$ and $m = -2, -1, 0, 1, 2$ . That is FIVE bright fringes.

Why

Turning the condition into an inequality on $m$ rather than listing positions is what makes the count safe: the strict inequality is what excludes $m = \pm 3$ at $15.19$ mm, outside the band.

Step 5

Missing orders: a fringe disappears when it falls exactly on a diffraction zero, that is when $d / a$ is a whole number. Here $d / a = 0.250 / 0.100 = 2.5$ , so no fringe is killed.

Why

The examiner is looking for this sentence, because it is the only place where the two letters are compared instead of used separately. A ratio of $2$ or $3$ would have removed a fringe from the count.

Step 6

Check: the band holds $2d / a = 5.0$ fringe spacings, and $2\lfloor d/a \rfloor + 1 = 5$ fringes, which agrees with the list.

Why

A second route to the same number, built from the ratio rather than from the positions. Two independent arrivals at $5$ is what lets you leave the question without rereading it.

The conclusion, written out

“The fringes are $5.06$ mm apart and the central diffraction band runs from $-12.66$ mm to $+12.66$ mm, so the orders $m = -2$ to $m = 2$ are visible inside it, that is FIVE bright fringes. Since $d / a = 2.5$ is not a whole number, no order is suppressed by a diffraction zero.”

The classic mistake on this problem: Computing $\Delta y$ with $a$ instead of $d$ , which gives $12.7$ mm and then a single fringe in the band. The two numbers are close enough to look plausible, which is exactly why the letters have to be named in the first line.

Learn by heart

• Bright when the path difference is a WHOLE number of wavelengths, dark when it is a HALF INTEGER number. Everything else on this sheet is that sentence applied to one apparatus.
• Two slits or a grating: $d\sin\theta = m\lambda$ gives the MAXIMA. One slit of width $a$ : $a\sin\theta = m\lambda$ gives the MINIMA, and $m = 0$ is excluded.
• On a screen: $\Delta y = \lambda L / d$ between fringes, $2\lambda L / a$ for the width of the central band. The narrower the slit, the WIDER the pattern.
• In a medium of index $n$ , the frequency is unchanged and $\lambda_n = \lambda / n$ , so every spacing shrinks by $n$ and a film carries $2nt$ .
• A reflection off a HIGHER index flips the wave, worth $\lambda / 2$ . Two flips or none cancel; one flip alone swaps bright and dark.
• Polarizers: one half for unpolarized light, then $I = I_0\cos^{2}\theta$ . Glare off a surface is fully polarized at $\tan\theta_B = n$ , parallel to the surface.
• Rayleigh: $\theta_{\min} = 1.22\lambda / D$ , and the smallest detail a light microscope shows is about half a wavelength, roughly $0.2$ micrometre, whatever the magnification.
• Formula sheet: $v = c/n$ and $v_1\sin\theta_2 = v_2\sin\theta_1$ , the slower medium having the smaller angle; the sheet writes the single slit width $w$ .

Frequently asked questions

How do I know whether a formula gives the bright or the dark fringes?

By the apparatus, never by the algebra, because the two equations look identical. If the letter multiplying the sine is the distance between two slits or between two grating lines, the formula gives the bright directions. If it is the width of a single opening, the same formula gives the dark ones, and the centre of the pattern is then the brightest point. Write the letter and its meaning on the paper before you substitute any number.

Why does a narrower slit make a wider diffraction pattern?

Because the pattern is governed by the ratio of the wavelength to the width of the opening. The first dark direction has a sine equal to that ratio, so shrinking the opening raises the ratio and pushes the first zero further out. The central bright band is twice the wavelength times the screen distance divided by the slit width, so halving the slit doubles the band. It is the same effect that makes a pinhole camera blurry past a certain smallness.

What changes when the whole double slit experiment is put under water?

The frequency of the light does not change, because it is fixed by the source. The speed drops to the speed in vacuum divided by the index, so the wavelength drops in the same ratio. Every fringe spacing therefore shrinks by the index: with water at one point three three, a spacing of five point zero six millimetres becomes three point eight one millimetres. Answering that nothing changes because no distance moved is the classic loss of a full mark.

Why does a soap film look black just before it bursts?

Because of the half wave flip at the first reflection. As the film drains, its top gets far thinner than a wavelength, so the extra distance travelled inside it goes to zero. What remains is the flip at the air to soap face, worth half a wavelength, and it is not cancelled by anything at the soap to air face. The two reflected waves are then exactly out of step for every colour at once, so nothing comes back and the top of the film looks black.

Why can a light microscope not show a virus even at high magnification?

Because the limit is diffraction, not the lenses. Light entering a round opening spreads into a small disc, and two details closer than about half a wavelength produce discs that overlap and cannot be separated. With an oil immersion objective that limit is around two hundred and forty nanometres, and a virus is about one hundred. Magnification does not appear anywhere in that limit, so turning it up only gives a bigger blur, which is why the term empty magnification exists.

Does a double slit under water show more fringes in its central band?

No, the same number. Under water the wavelength is divided by the index of the water, so the fringes close up, but the central diffraction band is made by the same light and closes up by exactly the same factor. The count depends only on the ratio of the slit separation to the slit width, which contains no wavelength. Only a change of the slits themselves can change it, never a change of medium or of colour.

Practise it

Corrected exercises: Wave optics, interference and diffraction, PHYS 101

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

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Refraction and dispersion, with a look ahead to lenses

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/phys101-wave-optics. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Wave optics and interference: what costs marks in PHYS 101

The essentials

On your formula sheet: this chapter's lines, and when each one holds

The only question of the chapter

The four conditions, and what the letter in front of the sine means

Light in a medium, and when the small angle form is allowed

Refraction written with the speeds, the form of the formula sheet

The rules in table form

Thin films: count the flips before choosing a formula

The mistakes that cost marks

1. Reading the same equation as the bright directions whatever the apparatus

2. Numbering the first dark fringe of Young's experiment as one

3. Putting the order zero into the single slit minimum condition

4. Believing that a narrower slit gives a narrower pattern

5. Leaving the wavelength alone when the light changes medium

6. Counting more fringes in the central band once the slits are under water

7. Skipping the half wave flip at a reflection

8. Applying Malus's law at the first polarizer

9. Using the small angle form at a grating

10. Answering for an order that the apparatus cannot produce

Which method to choose

Which condition, read from the apparatus and not from the letters

Counting the flips of a film

How the answer is expected to be written

Answering a double slit question the way a marker expects

Answering a thin film question

Check before you hand in

The typical problem, taken apart

Counting the fringes inside the diffraction envelope

Learn by heart

Frequently asked questions

How do I know whether a formula gives the bright or the dark fringes?

Why does a narrower slit make a wider diffraction pattern?

What changes when the whole double slit experiment is put under water?

Why does a soap film look black just before it bursts?

Why can a light microscope not show a virus even at high magnification?

Does a double slit under water show more fringes in its central band?

Corrected exercises: Wave optics, interference and diffraction, PHYS 101

See also

Stuck on wave optics before the final?