Corrected exercises: wave optics, interference and diffraction (PHYS 101)

$\lambda = v / f = 340 / 340 = 1.00$ m for the sound. Green light is $\lambda = 550$ nm, and the whole exercise turns on converting that correctly: $550$ nm $= 550\times 10^{-9}$ m $= 5.50\times 10^{-7}$ m. Write the power of ten down once and reuse it; a factor of $10^{9}$ dropped here poisons every later line, and it is the single most common arithmetic loss on this chapter.

For the sound, $\lambda / a = 1.00 / 0.90 = 1.11$. For the light, $\lambda / a = 5.50\times 10^{-7} / 0.90 = 6.11\times 10^{-7}$. The two numbers differ by about six million, and that gap, not any difference in the nature of the two waves, is the whole answer to why one bends and the other does not.

Sound: $\sin\theta = 1.11$ has NO solution, because a sine never exceeds $1$. There is no first missing direction, which means the sound is sent everywhere behind the wall; that is why you hear someone in the corridor without seeing them. Light: $\sin\theta = 6.11\times 10^{-7}$ gives $\theta = 6.11\times 10^{-7}$ rad $= 3.5\times 10^{-5}$ degrees. Over a room, a spread of $3.5\times 10^{-5}$ degrees moves the edge of the beam by a fraction of a micrometre, far below anything an eye can see, so the light draws a sharp shadow and we call it a ray.

We want $\sin\theta = \lambda / a$ with $\theta = 1.0^{\circ}$, so $a = \lambda / \sin 1.0^{\circ} = 5.50\times 10^{-7} / 0.01745 = 3.15\times 10^{-5}$ m, about $32$ micrometres, roughly a third of the thickness of a human hair. This is the practical rule of the chapter: to see light behave as a wave you must build an obstacle whose size is measured in wavelengths, and that is exactly what a slit, a grating or a soap film is.

Huygens: every point reached by a wavefront behaves as a source of a spherical wavelet, and the wavefront an instant later is the surface tangent to all those wavelets. In the middle of a wide opening each wavelet is boxed in by its neighbours and the tangent surface stays flat, so the wave goes straight. At the lips there is no neighbour on the outside, nothing cancels the sideways part of the wavelet, and the tangent surface curls around the edge. The trap here is to answer that the light is blocked at the edge; it is not blocked, it is REDIRECTED, and where it goes is what the rest of this chapter computes.

When the front $AB$ touches the wall at $A$, its far end $B$ still has a distance $v_1 t$ to travel in air before it reaches the wall at $B'$.

In that same time the wavelet born at $A$ grows to a radius $v_2 t$ in the water, and the new front is the tangent $B'T$ drawn from $B'$ to that wavelet.

The two right triangles $ABB'$ and $ATB'$ share the hypotenuse $AB' = \ell$ along the wall, and their angles at $A$ and $B'$ are the angles of the rays with the normal:

$v_1 t = \ell\sin\theta_1$ and $v_2 t = \ell\sin\theta_2$

Dividing one by the other removes $t$ and $\ell$:

$\frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{v_2}$, that is $v_1\sin\theta_2 = v_2\sin\theta_1$

With $v = c/n$ the same line reads $n_1\sin\theta_1 = n_2\sin\theta_2$, Snell's law: the two forms are one law.

Speed in the water: $v_2 = \frac{c}{n} = \frac{3.00\times 10^{8}}{1.33} = 2.26\times 10^{8}$ m/s.

$\sin\theta_2 = \frac{v_2}{v_1}\sin\theta_1 = \frac{\sin 30.0^{\circ}}{1.33} = 0.376$, so $\theta_2 = 22.1^{\circ}$.

Direction check: the slower side has the smaller angle, so the beam bends TOWARDS the normal; with $v_2 = v_1$ the formula gives $\theta_2 = \theta_1$, a straight line, as it must when nothing changes.

The glass wall: across its first face $\frac{\sin\theta_g}{v_g} = \frac{\sin\theta_1}{v_1}$, across its second $\frac{\sin\theta_2}{v_2} = \frac{\sin\theta_g}{v_g}$. The ratio $\frac{\sin\theta}{v}$ is the same in every layer, so the wall drops out and $\theta_2 = 22.1^{\circ}$ whatever the glass.

Pitfall: swapping the speeds, $v_1\sin\theta_1 = v_2\sin\theta_2$, gives $\sin\theta_2 = 0.665$ and $\theta_2 = 41.7^{\circ}$, a beam bending AWAY from the normal into the slower medium. The direction check catches it in two seconds.

The same figure explains exercise 4 c: the fronts are one period apart on both sides, so their spacing, the wavelength, drops in the ratio $\frac{v_2}{v_1} = \frac{1}{n}$.

The two rays leave $S_1$ and $S_2$ in the SAME direction, because the point they reach is far away compared with $d$. Drop the perpendicular from $S_1$ onto the ray leaving $S_2$: beyond that foot the two paths are strictly parallel and equal, so everything that matters is the piece between $S_2$ and the foot. In the right triangle $S_1S_2F$ the hypotenuse is $d$ and the angle at $S_2$ equals $\theta$, so the side opposite $\theta$ is $d\sin\theta$. The cosine would be the side along the slits, which is common to both paths, and the tangent belongs to the other triangle, the one that runs from the slits to the screen.

Bright means $d\sin\theta = m\lambda$. For $m = 1$, $\sin\theta = \lambda / d = 633\times 10^{-9} / 2.50\times 10^{-4} = 2.532\times 10^{-3}$, so $\theta = 0.1451^{\circ}$. Keep the full value $2.532\times 10^{-3}$ in the calculator: rounding it to $2.5\times 10^{-3}$ already changes the third digit of the angle, and the next exercise multiplies that angle by two metres.

Dark means $\delta = (m + \tfrac{1}{2})\lambda$, and the FIRST dark fringe is $m = 0$, not $m = 1$. So $\sin\theta = \lambda / (2d) = 1.266\times 10^{-3}$ and $\theta = 0.0725^{\circ}$, exactly half the angle of part b, which is the check to make. Counting the first dark fringe as $m = 1$ shifts every dark fringe by one slot and is worth a full mark.

$\delta = d\sin(0.200^{\circ}) = 2.50\times 10^{-4}\times 3.4907\times 10^{-3} = 8.73\times 10^{-7}$ m. Divide by the wavelength: $\delta / \lambda = 8.73\times 10^{-7} / 6.33\times 10^{-7} = 1.38$. That is neither a whole number nor a half integer, so the point is neither a maximum nor a zero. The useful habit is exactly this one: convert the path difference into WAVELENGTHS and read the decimal part. $1.38$ sits between $1.0$, bright, and $1.5$, dark, and closer to dark, so the screen is dim there.

A bright fringe needs $\sin\theta = m\lambda / d \le 1$, so $m \le d / \lambda = 2.50\times 10^{-4} / 633\times 10^{-9} = 394.9$. The largest whole number is $m = 394$, giving $394$ bright fringes on each side plus the central one. The sine that must stay below $1$ is the fastest sanity check of the whole chapter: any time an order comes out with $\sin\theta > 1$, that order simply does not exist and the answer is that it is not observed, not a rounding complaint.

$\Delta y = \lambda L / d = (633\times 10^{-9}\times 2.00) / 2.50\times 10^{-4} = 5.064\times 10^{-3}$ m $= 5.06$ mm. Check the order of magnitude before going on: a fringe pattern on a lab bench is a few millimetres per fringe. A result in metres means $d$ was entered in millimetres, a result in micrometres means $L$ was.

With the approximation, $y_3 = 3\times 5.064 = 15.192$ mm. Exactly: $\sin\theta = 3\lambda / d = 7.596\times 10^{-3}$, so $\theta = 7.5961\times 10^{-3}$ rad and $y = 2.00\tan\theta = 15.1924$ mm. The two agree to five digits, which is far better than any ruler. That agreement is not luck: it holds because $\theta$ is under half a degree.

The dark fringes sit exactly halfway between the bright ones, so the one between $m = 2$ and $m = 3$ is at $y = 2.5\,\Delta y = 12.66$ mm. Writing it as $(m + \tfrac{1}{2})\lambda L / d$ with $m = 2$ gives the same thing; the half integer is the whole content of the word dark.

For $m = 50$: the approximation gives $50\times 5.064 = 253.2$ mm, while $\sin\theta = 0.1266$, $\theta = 0.12694$ rad and $y = 2.00\tan\theta = 255.3$ mm. The relative error is $(255.3 - 253.2) / 253.2 = 0.0081$, that is $0.81$ percent. The approximation does not fail suddenly, it degrades slowly, and its relative error grows as the SQUARE of the angle, about $\theta^{2}/2$. The rule to remember: below about $8^{\circ}$ the error stays under one percent, and above that you write $\tan$.

Measuring one fringe spacing of $5.06$ mm with a ruler good to $0.5$ mm is a ten percent measurement, which is useless. Measuring ten spacings gives $50.6$ mm with the SAME absolute error of $0.5$ mm, so one percent, and dividing by ten at the end does not bring the error back. From the reading, $\Delta y = 5.06$ mm and $\lambda = \Delta y\, d / L = (5.06\times 10^{-3}\times 2.50\times 10^{-4}) / 2.00 = 6.33\times 10^{-7}$ m $= 633$ nm, safely inside the visible range of $400$ to $700$ nm. A wavelength coming out at $6.33$ micrometres or $63$ nm is not a wavelength of visible light and the calculation has to be redone.

By ratio: $\Delta y$ is proportional to $\lambda$, so the new spacing is $5.064\times (450 / 633) = 3.600$ mm. The direct computation confirms it, $450\times 10^{-9}\times 2.00 / 2.50\times 10^{-4} = 3.600\times 10^{-3}$ m. Blue light gives a TIGHTER pattern than red, which is the opposite of what most students guess: the shorter the wavelength, the less the wave bends.

$\Delta y$ is inversely proportional to $d$, so doubling $d$ halves the spacing: $5.064 / 2 = 2.532$ mm. Spreading the slits apart squeezes the fringes together. The same inversion runs through the whole chapter and it is worth saying out loud once: the pattern is a magnified image of the INVERSE of the object.

In water $\lambda_n = \lambda / n = 633 / 1.33 = 476$ nm, so $\Delta y = 5.064 / 1.33 = 3.81$ mm. The frequency is set by the source and never changes when the light enters a new medium; it is the speed that drops to $c / n$, and with $\lambda = v / f$ the wavelength drops in the same ratio. Answering that nothing changes because the geometry did not move is the classic loss of a full mark here.

The centre is the one place where $\delta = 0$ for EVERY wavelength at once, so all colours are bright there and the central fringe is white; this is how you recognise the order zero in a photograph. On either side each colour has its own $m = 1$ position: $y = 400\times 10^{-9}\times 2.00 / 2.50\times 10^{-4} = 3.20$ mm for violet and $5.60$ mm for red, so the first order spectrum is a band $5.60 - 3.20 = 2.40$ mm wide with violet closest to the centre.

Order $m$ runs from $400m$ to $700m$ in units of $L / d$ nanometres. Orders $1$ and $2$ would overlap if $800 < 700$, which is false, so they stay separate. Orders $2$ and $3$ overlap if $3\times 400 < 2\times 700$, that is $1200 < 1400$, which is true: the violet end of the third order falls inside the red end of the second. In millimetres the second order red edge is at $2\times 5.60 = 11.2$ mm and the third order violet edge at $3\times 3.20 = 9.60$ mm, so the overlap runs from $9.60$ mm to $11.2$ mm. This is exactly why a spectrometer is used at low order, and it is the first real limit of the instrument.

The pairing argument needs two strips separated by half the slit, and their path difference must be half a wavelength. With $m = 0$ the difference is zero, which is the opposite of cancellation: every strip arrives in step. So $\theta = 0$ is the CENTRAL MAXIMUM, the brightest point of the whole pattern, and writing $m = 0$ into the minimum formula claims the middle of the beam is dark. That single line has cost more marks on this chapter than any calculation.

$\sin\theta = \lambda / a = 633\times 10^{-9} / 1.00\times 10^{-4} = 6.33\times 10^{-3}$, so $\theta = 0.3627^{\circ}$ and $y_1 = L\tan\theta = 2.00\times 6.3301\times 10^{-3} = 1.266\times 10^{-2}$ m $= 12.66$ mm. The angle is small, so the approximation $y_1 = \lambda L / a$ gives $12.660$ mm, the same to four digits.

The central band runs from $-y_1$ to $+y_1$, so its width is $2\lambda L / a = 25.3$ mm. Compare with the two slit spacing of $5.06$ mm: the single bright blob of one slit is FIVE times wider than the distance between two neighbouring fringes of the double slit. The ratio is $2d / a$, and it is the reason a real double slit pattern shows a set of fringes sitting inside a broad hump, the hump being the diffraction of each individual slit.

$2\lambda L / a$ with $a$ halved doubles: $50.6$ mm. This is the result students refuse to believe, so state it as a slogan: the narrower the slit, the wider the pattern. It follows from the same ratio $\lambda / a$ as exercise 1, and it is what makes a pinhole camera blurry rather than sharp beyond a certain smallness.

In $d\sin\theta = m\lambda$ the letter $d$ is the distance BETWEEN two slits, and the equation locates the bright fringes. In $a\sin\theta = m\lambda$ the letter $a$ is the WIDTH of a single slit, and the same equation locates the dark ones. No amount of algebra tells them apart, because the algebra is identical: you decide by reading the sentence. Two openings, or a grating, or a wire in a beam, means separation and maxima. One opening, an obstacle of a given size, a beam clipped by an aperture, means width and minima. Write the letter with its meaning on the exam paper before substituting any number, and the mistake becomes impossible.

In the water the wavelength is $\frac{\lambda}{n}$ and nothing else in the pairing argument changes, so $\lambda$ becomes $\frac{\lambda}{n}$ in part c:

Dimensions: $\lambda$, $L$ and $a$ are lengths and $n$ is a pure number, so $W$ is a length times a length over a length, a length, as a width must be.

Limit $n = 1$: $W = \frac{2\lambda L}{a}$, the result of part c in air.

Direction: a larger $n$ or a wider slit narrows the band, a longer tank widens it, each in proportion.

$W = \frac{2\times 633\times 10^{-9}\times 0.600}{1.33\times 1.00\times 10^{-4}} = 5.71\times 10^{-3}$ m $= 5.71$ mm

The exact form $2L\tan\theta_1$ with $\sin\theta_1 = \frac{\lambda}{na} = 4.76\times 10^{-3}$ gives the same $5.71$ mm, since the angle is about a quarter of a degree.

Limit $a \to \frac{\lambda}{n} = 476$ nm: the exact condition gives $\sin\theta_1 \to 1$, the first zero goes to $90^{\circ}$, and the central band spreads over the whole far wall; there is no dark direction left at all.

The small angle formula, at that same limit, gives $W = 2L = 1.20$ m, a finite band. The limit check is what tells you the formula has left its domain, which happens as soon as $\frac{\lambda}{na}$ stops being small.

The opposite limit, $a$ very large, gives $W \to 0$: the slit casts a sharp shadow and the ray model of exercise 1 comes back.

Pitfall: multiplying by $n$ instead of dividing, $W = \frac{2n\lambda L}{a}$, because slower light is believed to spread more. The dimensions pass and so does $n = 1$; only the direction check catches it, since a shorter wavelength must give a narrower band.

Six hundred lines per millimetre means the lines are $1 / 600$ of a millimetre apart: $d = 1.00\times 10^{-3} / 600 = 1.667\times 10^{-6}$ m, that is $1.67$ micrometres. Notice the order of magnitude, about three wavelengths of visible light. A grating works precisely because it is built at the scale of the wavelength, which is exercise 1 turned into an instrument.

$\sin\theta = \lambda / d$. For $589.0$ nm, $\sin\theta = 0.35340$ and $\theta = 20.695^{\circ}$. For $589.6$ nm, $\sin\theta = 0.35376$ and $\theta = 20.717^{\circ}$. The separation is $\Delta\theta = 0.0221^{\circ} = 3.85\times 10^{-4}$ rad. This is where the small angle habit must be dropped: $20^{\circ}$ is not small, $\sin\theta$ and $\theta$ differ by two percent, and writing $\theta = \lambda / d$ here gives $20.25^{\circ}$, wrong in the second digit.

An order exists as long as $\sin\theta = m\lambda / d \le 1$, so $m \le d / \lambda = 1.667\times 10^{-6} / 589\times 10^{-9} = 2.83$. The highest whole number is $m = 2$. Asking for the third order of sodium with this grating is asking for $\sin\theta = 1.06$, which has no answer: the correct exam sentence is that the third order does not exist, not that the angle is large.

At a maximum, neighbouring slits differ by exactly $m\lambda$, so all $N$ contributions add. Move slightly off: neighbours now differ by $m\lambda + \varepsilon$, and the slit number $N/2$ further along differs from the first by $N\varepsilon / 2$. As soon as that reaches half a wavelength, the first half of the grating cancels the second half exactly, and the intensity is zero. With $N$ large, a very small $\varepsilon$ is enough, so the maximum is narrow; with $N = 2$ there is no third slit to do the cancelling and the fringe is broad. Width falls as $1/N$, which is the entire design principle of a spectrometer.

The illuminated width is $3.00$ mm, so $N = 600\times 3.00 = 1800$ lines. In the first order $R = mN = 1800$. The doublet needs $R = \lambda / \Delta\lambda = 589.0 / 0.6 = 982$. Since $1800 > 982$, the two lines are separated, and comfortably so. Two consequences worth remembering: stopping the beam down to $1.00$ mm would give $R = 600$ and merge the doublet, and going to the second order doubles $R$ for free, which is why spectroscopists work at the highest order the geometry allows.

Top face: air to soap, $1.00$ to $1.33$, the index goes up, so there is a flip. Bottom face: soap to air, the index goes down, no flip. Net: one flip, worth $\lambda / 2$. A bright reflection therefore needs the geometric difference to be a HALF integer: $2nt = (m + \tfrac{1}{2})\lambda$. The thinnest is $m = 0$: $t = \lambda / (4n) = 550 / (4\times 1.33) = 103$ nm. A bubble film is a tenth of a micrometre thick, which is why it bursts.

As $t$ goes to zero the term $2nt$ goes to zero, but the flip does not go away. The two reflected rays are then exactly half a wavelength out of step at every wavelength at once, so they cancel for the whole spectrum and the top of the film looks BLACK against the light. Seeing that black band is the experimental proof that the flip is real, and it is the standard exam question on the phase change.

Top face: air to coating, $1.00$ to $1.38$, index up, flip. Bottom face: coating to glass, $1.38$ to $1.52$, index up again, flip. Two flips, which together are a whole wavelength and therefore count for nothing. Killing the reflection now needs the geometric part to be the half integer: $2nt = (m + \tfrac{1}{2})\lambda$, so $t = \lambda / (4n) = 550 / (4\times 1.38) = 99.6$ nm. The arithmetic is almost the same as in part a, but the physical CONCLUSION is the opposite, bright there and dark here, which is exactly why the flips must be counted before anything is written down.

Top: air to oil, $1.00$ to $1.20$, flip. Bottom: oil to water, $1.20$ to $1.33$, index up again, flip. Two flips cancel, so bright needs $2nt = m\lambda$, and the thinnest non zero solution is $m = 1$: $t = \lambda / (2n) = 550 / (2\times 1.20) = 229$ nm. Note that $m = 0$ would be $t = 0$, no film at all, so it is rejected; saying so is part of the answer. And an oil film over glass instead of water would flip only once and give the other rule, which is how the same puddle can look different on a wet road and on a glass plate.

The film here is AIR between two glass plates. Top face: glass to air, index down, no flip. Bottom face: air to glass, index up, flip. One flip, and the film has $n = 1$, so DARK is $2t = m\lambda$. The wedge grows linearly: at distance $x$ from the contact edge the gap is $t = xD / L_w$ with $D = 5.0\times 10^{-5}$ m and $L_w = 0.120$ m. Two neighbouring dark fringes differ by $\lambda / 2$ in thickness, so $\Delta x = \lambda L_w / (2D) = (589\times 10^{-9}\times 0.120) / (2\times 5.0\times 10^{-5}) = 7.07\times 10^{-4}$ m $= 0.707$ mm. At the contact edge $t = 0$ and the condition $2t = m\lambda$ holds with $m = 0$, so the edge is DARK. That dark edge is the free check on every air wedge problem: if your formula makes the contact line bright, you dropped the flip.