Reflection & Refraction

Stand at the edge of a swimming pool and look at your own legs. They bend at the waterline. Drop a coin in and the bottom looks closer than it is. Look up from underwater and the whole sky is squeezed into a bright circle overhead, with darkness around it. None of this is your eyes failing. It is light changing speed.

The last lesson treated light as a wave with a frequency that shifts when the source moves. Here you mostly get to forget the wave and follow straight lines called rays. A ray is the direction energy travels. Wherever the things light meets are much bigger than its wavelength (a pool surface, a prism, a raindrop) rays get you the right answer with geometry instead of wave mechanics.

Light slows down inside matter

In vacuum light travels at $c = 3.00\times10^8$ m/s. Inside glass or water it goes slower, because the atoms absorb and re-radiate it and that takes time. Define the refractive index of a material as the ratio

$n = \frac{c}{v}$

where $v$ is the speed of light in that material. So $n$ is always at least 1, vacuum has $n = 1$ exactly, and air is $n = 1.0003$, near enough to 1 that you usually write 1. Crown glass has $n = 1.52$, so light inside it travels at $v = c/n = (3.00\times10^8)/1.52 = 1.97\times10^8$ m/s, about two thirds of its vacuum speed.

One thing does not change at a boundary: the frequency. The wave on the far side is driven by the wave on the near side, crest for crest, so they must arrive at the same rate. Since $v = f\lambda$ and $f$ is fixed while $v$ drops by a factor $n$, the wavelength must drop by the same factor:

$\lambda_n = \frac{\lambda_0}{n}$

Slower speed, shorter waves, same frequency. Hold on to that; it is what makes refraction happen at all.

Reflection, and what mirrors do to angles

Light hitting a smooth surface bounces with the angle of reflection equal to the angle of incidence, both measured from the normal: the line perpendicular to the surface. Angles from the surface itself will wreck every problem you ever do. Measure from the normal, always.

A flat mirror puts the image as far behind the mirror as the object is in front. Two consequences show up constantly. If you walk toward a mirror at $1.20$ m/s, your image walks toward the mirror at $1.20$ m/s from the other side, so the gap between you and your image closes at $2.40$ m/s. And if you rotate a mirror by an angle $\alpha$ while the incoming beam is held fixed, the normal turns by $\alpha$, so the angle of incidence changes by $\alpha$ and the angle of reflection changes by $\alpha$ as well. The reflected beam turns by $2\alpha$. Rotate a galvanometer mirror $8.00°$ and the spot swings $16.0°$.

Two mirrors hinged at an angle $\theta$ multiply images: light can bounce off one, then the other, then back. Counting the distinct images gives

$N = \frac{360°}{\theta} - 1$

when $360°/\theta$ is a whole number. Hinge them at $72.0°$ and you get $360/72 - 1 = 5 - 1 = 4$ images.

Snell's law, from wavefronts

Picture a wide flat wavefront in air heading down onto water at a slant. The left edge of the front touches the water first and immediately slows down. The right edge is still in air, still fast. For the short time it takes the right edge to reach the surface, the left edge has crept forward a smaller distance. The front pivots. That pivot is refraction.

Make it quantitative. In a time $\Delta t$ the part of the front still in air covers $v_1\Delta t$ and the part already in water covers $v_2\Delta t$. Both pieces share the same strip of surface, of length $L$. Geometry of the two right triangles built on that strip gives $v_1\Delta t = L\sin\theta_1$ and $v_2 \Delta t = L\sin\theta_2$. Divide one by the other and $L$ and $\Delta t$ both vanish:

$\frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{v_2} = \frac{n_2}{n_1}$

which is usually written in the symmetric form

$n_1\sin\theta_1 = n_2\sin\theta_2$

Read it in words: the quantity $n\sin\theta$ is the same on both sides of the boundary. Going into a denser medium, $n$ goes up so $\sin\theta$ must come down and the ray bends toward the normal. Coming out into a rarer medium it bends away. At normal incidence, $\theta_1 = 0$, both sides are zero and nothing bends at all.

Worked example 1: a laser into a pool, and how deep it looks

A laser in air strikes a pool at $30.0°$ from the normal; water has $n = 1.33$. Apply Snell's law directly:

$\sin\theta_2 = \frac{n_1\sin\theta_1}{n_2} = \frac{(1.00)(0.500)}{1.33} = 0.3759$

so $\theta_2 = \arcsin(0.3759) = 22.1°$. Bent toward the normal, as promised.

Now the harder half. The pool is $2.40$ m deep and you look straight down. Rays from the bottom leave the water at small angles, so use the small-angle form of Snell's law: for small $\theta$, $\sin\theta \approx \tan\theta \approx \theta$, and $n_{\text{w}}\theta_{\text{w}} = \theta_{\text{air}}$. A ray from a point on the floor at true depth $h$ that crosses the surface a horizontal distance $x$ from straight-above has $\theta_{\text{w}} \approx x/h$. In air it leaves at $\theta_{\text{air}} \approx x/h'$, where $h'$ is the depth the backward extension points to. Putting those together, $x/h' = n_{\text{w}}(x/h)$, and the $x$ cancels:

$h' = \frac{h}{n}$

So the floor appears at $h' = 2.40/1.33 = 1.80$ m. The apparent depth is the real depth divided by the index. Stack two liquids and you apply this to each layer separately and add: $4.00$ cm of glycerin under $6.00$ cm of water gives $4.00/1.47 + 6.00/1.33 = 2.72 + 4.51 = 7.23$ cm.

When there is nowhere to refract to

Run Snell's law the other way, from dense to rare, with $n_1 > n_2$:

$\sin\theta_2 = \frac{n_1}{n_2}\sin\theta_1$

The prefactor is bigger than 1, so $\sin\theta_2 > \sin\theta_1$ and the ray tips away from the normal. Keep raising $\theta_1$ and at some point the formula demands $\sin\theta_2 > 1$. No angle has a sine above 1. There is no refracted ray. All of the energy goes into the reflected one. This is total internal reflection, and the threshold is where $\theta_2$ would be exactly $90°$:

$\sin\theta_c = \frac{n_2}{n_1}$

Water to air gives $\sin\theta_c = 1/1.33 = 0.7519$, so $\theta_c = 48.8°$. Glass to air is about $42°$, which is why a $45°$-$45°$-$90°$ prism turns a beam through a right angle with no silver coating at all: the internal angle at the hypotenuse is $45°$, safely past $42°$. Run that backwards and you get the smallest index a periscope prism can have: you need $\theta_c \le 45°$, so $n \ge 1/\sin 45° = 1.41$.

Note the direction. TIR only exists going from high index to low. There is no critical angle on the way into glass.

That bright circle is called Snell's window. A lamp at depth $h$ lights a patch of surface of radius $R = h\tan\theta_c$. At $h = 1.50$ m in water, $\tan(48.8°) = 1.14$, so $R = 1.71$ m.

Worked example 2: escaping through a layer of water

A glass block, $n = 1.60$, is covered by water, $n = 1.33$, with air above. Light inside the glass hits the glass-water boundary at $\theta$. Above what $\theta$ does none of it reach the air?

You could do two Snell's laws in a row. Do it once and you will see why you never need to again. At the first boundary $1.60\sin\theta = 1.33\sin\theta_w$. At the second, $1.33\sin\theta_w = 1.00\sin\theta_a$. The middle term is identical in both, so it cancels straight out:

$1.60\sin\theta = 1.00\sin\theta_a$

The intermediate layer is invisible to the bookkeeping. Whenever your boundaries are all parallel, $n\sin\theta$ is a constant all the way through, and only the first and last media matter. Light stops escaping when $\theta_a$ would exceed $90°$:

$\sin\theta = \frac{1.00}{1.60} = 0.625 \quad\Rightarrow\quad \theta = 38.7°$

The water changes nothing. The same invariant runs the mirage over a hot road, where air near the tarmac is hot and thin, $n = 1.000230$, while air at eye level is cooler, $n = 1.000280$. There the ray is nearly horizontal, so work with the angle $\phi$ from the horizontal, for which $\sin\theta = \cos\phi$:

$n_{\text{eye}}\cos\phi = n_{\text{road}}\cos 0°$ $\cos\phi = \frac{1.000230}{1.000280} = 0.99995 \quad\Rightarrow\quad \phi = 0.573°$

A ray leaving your eye barely more than half a degree below horizontal curves back up. You see sky on the road.

Prisms: two refractions and a total deviation

A prism has two flat faces meeting at an apex angle $A$. Send a ray in at $\theta_1$, let it refract to $\theta_2$ inside, cross to the second face where it hits at $\theta_3$ inside, and refract out at $\theta_4$. The geometry of the triangle formed by the ray and the two normals gives one clean relation:

$\theta_2 + \theta_3 = A$

and the total turning of the ray, its deviation, is the sum of what each face did:

$D = (\theta_1 - \theta_2) + (\theta_4 - \theta_3) = \theta_1 + \theta_4 - A$

Worked example 3: through an equilateral prism, and the minimum

Take $A = 60.0°$, $n = 1.52$, and a ray in at $\theta_1 = 40.0°$.

First face: $\sin\theta_2 = \sin 40.0°/1.52 = 0.6428/1.52 = 0.4229$, so $\theta_2 = 25.0°$.

Inside the glass: $\theta_3 = A - \theta_2 = 60.0° - 25.0° = 35.0°$.

Second face, going glass to air: $\sin\theta_4 = 1.52\sin 35.0° = 1.52 \times 0.5736 = 0.8719$, so $\theta_4 = 60.6°$. That is the answer, and the deviation is $40.0 + 60.6 - 60.0 = 40.6°$.

Now the thing the figure shows. Sweep $\theta_1$ and $D$ falls, bottoms out, and climbs. At the bottom the path is symmetric, $\theta_1 = \theta_4$ and $\theta_2 = \theta_3 = A/2$. Call the deviation there $D_m$. Then $\theta_1 = (A + D_m)/2$, and Snell at the first face reads

$n = \frac{\sin\!\big((A + D_m)/2\big)}{\sin(A/2)}$

For $A = 60.0°$ and $n = 1.50$: $\sin((60 + D_m)/2) = 1.50\sin 30.0° = 0.750$, so $(60 + D_m)/2 = 48.6°$ and $D_m = 37.2°$. Run it backwards to measure an index: a measured $D_m = 40.0°$ gives $n = \sin 50.0°/\sin 30.0° = 0.766/0.500 = 1.53$.

Two more things fall out of the prism relations. Because $\theta_3 = A - \theta_2$ and $\theta_2$ can never exceed $\arcsin(1/n)$, a large enough apex angle traps every ray: light emerges only if $A \le 2\arcsin(1/n)$. And because $n$ is slightly larger for violet light than for red (dispersion) each colour gets its own deviation, which is what spreads white light into a band.

Curved surfaces, and why rainbows sit at 42 degrees

Refraction at a curved surface is still just Snell's law, but with a normal that points along the radius. For paraxial rays (rays close to the axis and nearly parallel to it) the algebra collapses to

$\frac{n_1}{s} + \frac{n_2}{s'} = \frac{n_2 - n_1}{R}$

with $s$ the object distance, $s'$ the image distance, and $R$ positive when the surface bulges toward the incoming light. Try it on a solid glass sphere of radius $2.00$ cm, $n = 1.50$, lit from infinity. Front surface: $s = \infty$ kills the first term, so $1.50/s' = 0.50/2.00 = 0.250$ and $s' = 6.00$ cm past the front. That is $6.00 - 4.00 = 2.00$ cm past the back, so for the back surface the light is already converging and the object distance is $-2.00$ cm, with $R = -2.00$ cm as the surface now curves away:

$\frac{1.50}{-2.00} + \frac{1.00}{s'} = \frac{1.00 - 1.50}{-2.00} = 0.250$

$\frac{1.00}{s'} = 0.250 + 0.750 = 1.00 \quad\Rightarrow\quad s' = 1.00 \text{ cm}$

The sphere focuses one centimetre beyond its own back surface.

The rainbow needs no such approximation, only careful angle counting. A ray enters a spherical drop at incidence $\theta_1$, refracts to $\theta_2$, reflects once off the back, and refracts out. Each refraction turns it by $\theta_1 - \theta_2$ and the internal reflection turns it by $180° - 2\theta_2$. Total:

$D = 2(\theta_1 - \theta_2) + 180° - 2\theta_2 = 180° + 2\theta_1 - 4\theta_2$

Rays arrive across the whole face of the drop, so every $\theta_1$ is represented. Light piles up where $D$ barely changes as $\theta_1$ varies: that is why there is a bright arc rather than a smear. Set $dD/d\theta_1 = 0$. With $\sin\theta_1 = n\sin\theta_2$, differentiating gives $\cos\theta_1 = n\cos\theta_2 \, d\theta_2/d\theta_1$, and the stationary condition $d\theta_2/d\theta_1 = 1/2$ turns into

$\cos\theta_1 = \sqrt{\frac{n^2 - 1}{3}}$

For water, $n = 1.333$: $\cos\theta_1 = \sqrt{(1.777 - 1)/3} = \sqrt{0.2590} = 0.5089$, so $\theta_1 = 59.4°$. Then $\sin\theta_2 = 0.8609/1.333 = 0.6458$ and $\theta_2 = 40.2°$. Put them in:

$D = 180° + 2(59.4°) - 4(40.2°) = 180° + 118.8° - 160.8° = 138.0°$

The light comes back $138°$ from its original direction, which is $180° - 138° = 42.0°$ away from the direction pointing straight back at the sun. That is the radius of the rainbow. The secondary bow is the same calculation with two internal reflections, $D = 360° + 2\theta_1 - 6\theta_2$, and lands at $50.9°$.

Recognising which tool the problem wants

• The word "angle" plus two materials. Write $n_1\sin\theta_1 = n_2\sin\theta_2$ and check that you measured from the normal.
• Stacked parallel layers. Do not chain Snell's law layer by layer. Use $n\sin\theta$ constant and ignore everything in the middle.
• "Above what angle", "just barely escapes", "is guided", "trapped", a fiber, a periscope prism. That is a critical angle: $\sin\theta_c = n_2/n_1$ with $n_1$ the medium the light is in.
• "Looks like", "appears to be", straight down into water. Apparent depth, $h' = h/n$, one layer at a time.
• Two flat faces at an angle. Prism relations: $\theta_2 + \theta_3 = A$ and $D = \theta_1 + \theta_4 - A$. If the word "minimum" appears, jump straight to the symmetric path.
• Two indices quoted for one piece of glass. Dispersion: run the whole calculation twice and subtract the two answers.
• "Brightest", "stationary", "the rainbow angle". Write the deviation as a function of the entry angle and set its derivative to zero.
• A mirror that moves or turns. Angles double. A rotation $\alpha$ swings the beam $2\alpha$; approaching at $v$ closes the gap at $2v$.

Every one of these is the same law used with care about which angle is which. Draw the normal before you draw the ray, and most of the difficulty goes away.

Next you take the curved-surface result and let the two surfaces of a piece of glass work together, which is how a lens forms an image you can put on a screen: Thin Lenses and Mirrors.