Physics uses basic models to simplify complex systems. A common model is the particle. Particles are point-like objects with no internal structure, just properties like mass and velocity. More complicated properties like rotation, and friction are often ignored. We've used the particle model to understand phenomena like: momentum, gravitation, Coulomb's law, and current.

Another commonly used model is the wave. Waves are different from the particle model in some interesting ways. They don't exist at a point. They spread out in every direction at a constant speed. They have zero mass, and they can overlap with each other without interacting.

A wave is produced when a medium is disrupted. The wave is the disruption spreading through the medium.

The nature of each medium gives waves different properties. If you throw a pebble into a pond you see a wave spread out along the surface of the water. When a tree falls in the forest you can hear the sound waves spread out through the air. When you look at the sky you see electro-magnetic waves made by stars.

wave	medium	examples
sound	matter (any phase)	music, speakers
light	electro-magnetic fields	sunlight, Wi-Fi, X-Rays
surface	interface between two media	ocean waves, ripples
string	strings, ropes	whip, guitar string
seismic	solids	S and P waves
electrical	conductors	cable TV, phone lines
gravitational	spacetime	LIGO, Virgo
traffic	cars on a road	ripples of slow traffic
audience	stadium of people	The Wave

Media and Wave Speed

When a wave propagates, what is moving is energy, not matter. The speed of propagation is determined by the medium. Properties like frequency, amplitude, or wavelength generally don't affect the speed.

A wave is a disturbance that propagates through a medium.

Click the underlined words for their definition.

Each type of wave has a different mechanism of propagation. The speed and possibility of a wave propagating through a medium is different for each wave type.

speed (m/s)	vacuum	air	water	glass
sound	N/A	340	1484	4540
light	299 792 458	299 700 000	225 000 000	200 000 000

Sound travels faster in dense media because the atoms are closer together. This means the atoms don't have to move as far to collide.

A light wave is slower in dense media because as the light wave propagates through a medium it produces ripples that interfere in a way that slows the group velocity of the light wave.

Example: Two students are holding a slinky while standing 3.6 m apart. The first student sends a pulse which travels all the way down and back again. It takes 2.4 s for the wave to return. What is the speed of the wave?

solution

$$v = \frac{\Delta x}{\Delta t}$$ $$v = \frac{3.6 \, \mathrm{m}\times 2}{2.4\, \mathrm{s}} $$ $$v = 3 \, \mathrm{\tfrac{m}{s}}$$

Investigation: Can you figure out what factors affect the speed of a string wave? Go experiment with a string, or slinky to find out.

results

Tension is probably the easiest way to control the speed of a string wave. The equation below shows that string mass and length are also factors. $$ v = \sqrt{\frac{T}{\tfrac{m}{L}}}$$

v = velocity (m/s)
T = tension (N)

m = mass (kg)
L = string length (m)

Changing wave speed is used to change the pitch of stringed instruments. Notes from a guitar or harp are changed through tension, length, or string mass.

How Waves Propagate

Propagate is the word we use to describe waves moving. We try not to say move, because waves are a disturbance transmitting through the medium. The medium doesn't move. Each medium has their own mechanism of propagation, but they share some common principles.

A medium at rest is in equilibrium; the forces are in balance. A disruption spreads through a medium bringing it out of equilibrium.

Sound waves are vibrations that propagate through matter. They are produced by changes in pressure and velocity. Your ear senses the amplitude and frequency of the vibrations.

Each time you click these simulations you will send a pulse of compressed particles, sound!

I've graphed the density and highlighted the particles with higher speeds. You can see the wave trading between kinetic energy and pressure.

Waves don't transmit matter, just energy. In the simulation above one particle is highlighted. Watch how it oscillates when it becomes part of the wave, but over time it stays in the same area. Sound waves oscillate parallel to propagation, but most waves oscillate perpendicular to propagation.

Longitudinal Wave = The medium oscillates parallel to the direction of wave propagation.
Examples: sound, slinky

Transverse Wave = The medium oscillates perpendicular to the direction of wave propagation.
Examples: light, string, slinky, sound (in solids), gravity

Click to Run

Periodic Waves

Waves that repeat are called periodic. Periodic waves have these measurable properties.

Period, T = time for one complete wave cycle to pass a point [s]
Frequency, f = number of cycles that pass per second [Hz, 1/s]
Wavelength, λ = distance over which a wave's shape repeats [m]
Phase, ϕ = fraction of a repeating cycle for point on the wave [0,2π] or [0,360]

Waves can have different shapes. The waveform below is a wave.

Graphing: Let's see if we can find a relationship between frequency and period. Change the wavelength and record about five different pairs of frequency and period. Graph the pairs with period as the x-coordinate and frequency as the y-coordinate. What does the graph look like?

result

It's an inverse function.

$$f = \frac{1}{T}$$

$f$ = frequency [Hz, 1/s, hertz]
How often an event happens in a second.

$T$ = time period [s, seconds]
How many seconds are between each event.

Example: An air horn sounds at a frequency of 220 Hz. How many seconds pass between each wave crest?

solution

$$T = \frac{1}{f}$$ $$T = \frac{1}{220 \, \mathrm{Hz}}$$ $$T = 0.0045 \, \mathrm{s}$$

Example: The Los Angeles Metro Expo Line has a train arrive at 6:50am, 6:56am, and 7:02am. What is the period and frequency of the trains?

solution

$$6:56-6:50 = 6\, \mathrm{min}$$ $$T = 6\, \mathrm{min} \left(\frac{60\, \mathrm{s}}{1\, \mathrm{min}}\right) = 360\, \mathrm{s}$$
$$f = \frac{1}{T}$$ $$f = \frac{1}{360 \, \mathrm{s}}$$ $$f=0.0027 \, \mathrm{Hz}$$

Click the circle to start. Click again to pause if it gets annoying.

Question: Which slider controls period and which controls frequency?

answer

Period is controlled by the left slider and frequency by the right.

Example: Red light has a frequency of 450 THz. What is its period?

table of metric prefixes

Name	Symbol	Factor		Power
tera	T	1 000 000 000 000		10¹²
giga	G	1 000 000 000		10⁹
mega	M	1 000 000		10⁶
kilo	k	1 000		10³
centi	c		0.01	10^-2
milli	m		0.001	10^-3
micro	μ		0.000 001	10^-6
nano	n		0.000 000 001	10^-9
pico	p		0.000 000 000 001	10^-12

solution

$$T = \frac{1}{f}$$ $$T = \frac{1}{450 \times 10^{12}}$$ $$T = 2.22 \times 10^{-15}\, \mathrm{s}$$

The Wave Equation

For periodic waves, the product of the frequency and wavelength is equal to the wave's velocity.

$$v = f \lambda \quad \quad v = \frac{\lambda}{T} $$

$v$ = propagation speed [m/s]
$\lambda$ = wavelength [m, meters]
$f$ = frequency [Hz, 1/s, hertz]
$T$ = time period [s, seconds]

Example: If I triple the wavelength of a sound wave while keeping the wave speed the same, what happens to the frequency?

solution

$${ \atop v} {\atop =} { \Downarrow \atop f} { \Uparrow \atop \lambda} $$ $$v=\left(\tfrac{1}{3} f \right) (3 \lambda) $$

λ and f are inversely proportional, and v is constant.

Tripling λ will reduce f by a factor of one third.

frequency = Hz
Example: Click to hear a wave. Calculate the length of the sound wave at the frequency you hear for both air and water.

table of wave speeds

speed (m/s)	vacuum	air	water	glass
sound	N/A	340	1484	4540
light	299 792 458	299 700 000	220 000 000	200 000 000

solution

$$\text{air}$$ $$v = f \lambda$$ $$\lambda = \frac{v}{f}$$

$$\text{water}$$ $$v = f \lambda$$ $$\lambda = \frac{v}{f}$$

Example: A cell phone sends and receives light in the microwave range, at wavelengths around 1 cm. How many cycles pass through the phone in one second?

solution

Use the speed of light in a vacuum.

$$v = 3.0 \times 10^{8} \, \mathrm{\tfrac{m}{s}}$$ $$\lambda = 1\, \mathrm{cm} = 0.01\, \mathrm{m}$$
$$v = f \lambda$$ $$f = \frac{v}{\lambda}$$ $$f = \frac{(3.0 \times 10^{8})}{0.01}$$ $$f=3\times 10^{10}\, \mathrm{Hz}$$ $$f=30 \, \mathrm{GHz}$$

Example: As a wave enters a new medium its speed decreases. The frequency of the wave stays the same, but how does the wavelength change?

solution

$$v = f \lambda $$

Frequency is constant so we can set it to one.

$${\Downarrow \atop v} {\atop =} {\Downarrow \atop \lambda} $$

v and λ are directly proportional. This means decreasing v will decrease λ.

Example: AC (alternating current) electricity in the U.S. has a frequency of 60 Hz. In most other countries it is 50 Hz. An electrical signal propagates through a wire at about 2/3 the speed of light. What is the wavelength for an AC wave in the United States?

solution

$$v = f \lambda$$ $$\lambda = \frac{v}{f} $$ $$\lambda = \frac{(\tfrac{2}{3})(3.0 \times 10^8)}{60} $$ $$\lambda = \frac{2.0 \times 10^8}{60} $$ $$\lambda = 0.033 \times 10^8 \, \mathrm{m} $$ $$\lambda = 3300\, \mathrm{km} $$

Example: Gravity waves are ripples in spacetime that propagate at the speed of light. The first detected gravity wave was on 11 February 2016 when the LIGO and Virgo Scientific Collaboration observed gravitational waves originating from a pair of merging black holes. If you convert the gravity wave into a sound wave you can hear a chirp.

How big are gravity waves? Over the 0.2 second detection, the frequency increased from 35 Hz to 250 Hz. Calculate the starting and ending wavelength.

solution

$$ v = 3 \times 10^{8} \, \mathrm{\tfrac{m}{s} }$$ $$v = f \lambda$$ $$\lambda = \frac{v}{f}$$

$$\lambda = \frac{3 \times 10^{8}}{35}$$ $$\lambda = 8\,500\,000\, \mathrm{m}$$

$$\lambda = \frac{3 \times 10^{8}}{250}$$ $$\lambda = 1\,200\,000\, \mathrm{m}$$

The wavelength ranged from 8500 km to 1200 km. We can also find the length of the entire signal.

$$v = \frac{\Delta x}{\Delta t}$$ $$\Delta x = v\Delta t$$ $$\Delta x = (3 \times 10^{8}) (0.2)$$ $$\Delta x = 0.6 \times 10^{8} \, \mathrm{m}$$

The entire signal was 60 000 000 m long. That's about 5 Earths long.

Reflection and Refraction

Reflection and refraction occur when the wave speed changes as it enters a new medium. Part of the wave refracts into the new medium, and part of the wave reflects back into the old medium.

Refraction = After entering a new medium, a wave will change its direction of propagation. This occurs because different media have different speeds.

When slowing down, the wave will bend towards the normal.
When speeding up, the wave will bend away from the normal.

Reflection = When hitting a new medium, part of a wave stays in the original medium. The angle between the incident ray and the normal line equals the angle between the reflected ray and the normal. $$\theta_i = \theta_r$$

Most substances are bumpy at the molecular level. Because of the bumpiness, they scatter reflected light in many directions. Scattered light is called diffuse light. Materials like dirt, skin, and wallpaper reflect diffuse light.

Smooth materials produce a clear mirror image, called a specular reflection. Specular reflections are found in materials like metal, glass, and water. Mirrors are typically made by coating one side of a plate of glass with metal.

Example: How will the light ray bend as it enters the denser media? Complete the path of light in the diagram above.

solution

It will refract up, towards the normal.

Click to Run

Simulation: Can you figure out how to produce total internal reflection? This is when the light only reflects and doesn't refract.

answer

Total internal reflection occurs when the light starts in the slower medium and hits the faster medium at a small angle.

Simulation: Different frequencies refract different amounts. This phenomenon is called dispersion. How does refraction work differently for high vs. low frequency? Test it out in the simulation.

answer

The wavelength/color/frequency of the light changes the angle of refraction. A blue light has a shorter wavelength and it will refract more.

Light slows down in dense media. As a gas increases in temperature, it expands and becomes less dense. This allows light to speed up when it enters a hot pocket of gas. As light changes speed it will refract and bend. This phenomenon produces a mirage effect when you look down a hot road.

Here are some examples of reflection and refraction in a 2-D ripple tank:

Snell's Law

When a wave enters a new medium, several ratios have equal proportions.

$$\frac{sin \theta_1}{sin \theta_2} = \frac{v_1}{v_2} = \frac{ \lambda_1}{\lambda_2} =\frac{n_2}{n_1} $$

$ \theta_1 $ = angle of incidence
$ \theta_2 $ = angle of refraction
$v$ = propagation speed [m/s]
$\lambda$ = wavelength [m, meters]
$n$ = index of refraction [no units]

Question: Why can't we compare the ratio of the frequencies in each medium?

solution

Wavelength, speed, and angle change in each medium. Frequency is determined by the source of the wave, not the medium.

It helps to draw a rough diagram for each problem.
Label which medium is 1 and which is 2.
Focus on two ratios, and set them equal to each other.

Example: 7.0 mm light travels from air to water. Use the index of refraction to find the wavelength of the light in water.

solution

$$\frac{n_2}{n_1} = \frac{\lambda_1}{\lambda_2}$$ $$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}$$ $$\lambda_2 = \frac{\lambda_1 n_1}{n_2} $$ $$\lambda_2 = \frac{(0.007) (1)}{1.33} $$ $$\lambda_2 = 0.0053\, \mathrm{m}$$ $$\lambda_2 = 5.3\, \mathrm{mm}$$

Example: Light enters air from glass at 36.0º to the normal. This glass has an index of 1.52. What angle does the light leave relative to the normal?

solution

$$\frac{\mathrm{sin} \theta_1}{\mathrm{sin} \theta_2} = \frac{n_2}{n_1} $$ $$n_1 \mathrm{sin} (\theta_1) = n_2 \mathrm{sin} (\theta_2) $$ $$(1.52) \mathrm{sin} (36.0) = \mathrm{sin} (\theta_2) $$ $$ 0.893 = \mathrm{sin} (\theta_2)$$ $$63.3^{\circ} = \theta_2 $$

Example: Diamonds have a very high index of refraction. Its high index is why they seem sparkly. How fast does light move through a diamond?

strategy

Setup a situation where the light goes from air to diamonds.
Build an equation with the index of refractions and the velocities.

solution

$$\frac{v_1}{v_2} = \frac{n_2}{n_1} $$ $$v_1 = \frac{n_2 v_2}{n_1}$$ $$v_1 = \frac{(1)(3\times 10^8)}{2.42}$$ $$v_1 = 1.24\times 10^8 \mathrm{\tfrac{m}{s}}$$

As the light wave enters a new medium some of the light is reflected, and some is refracted. If the angle of refraction is 90º or higher, the light will only reflect. This is called total internal reflection. The angle of incidence that produces a 90º angle of refraction is called the critical angle.

You can see total internal reflection if you swim under some water and look upwards.

Example: A 600 nm wave of red light starts in water and enters air. What is the critical angle for total internal reflection?

strategy

Look up the index of refraction for air and water.
Set the angle of refraction to 90º.
Solve for the angle of incidence.

solution

$$\frac{\mathrm{sin} \theta_1}{\mathrm{sin}\theta_2} = \frac{n_2}{n_1} $$ $$n_1\mathrm{sin} \theta_1 = n_2\mathrm{sin} \theta_2$$ $$(1.33)\mathrm{sin} \theta_1 = (1)\mathrm{sin}(90)$$ $$\mathrm{sin} \theta_1 = \frac{1}{1.33}$$ $$\mathrm{sin}^{-1}\mathrm{sin} \theta_1 = \mathrm{sin}^{-1}(0.751)$$ $$\theta_1 = 48.6 \degree$$

Click to Run

Simulation: Find the index of refraction for substances A and B. Use the simulated protractor to measure the angles or use the speed tool.

solution A

$$\text{1 = air} \quad \quad \text{ 2 = mystery A}$$ $$\theta_1 = 39.0^{\circ} \quad \quad \quad n_1 = 1.0$$ $$\theta_2 = 15.0^{\circ} \quad \quad \quad n_2 = \ ?$$ $$\frac{\mathrm{sin} \theta_1}{\mathrm{sin} \theta_2} =\frac{n_2}{n_1} $$ $$n_2 = \frac{n_1 \mathrm{sin} \theta_1}{\mathrm{sin} \theta_2} $$ $$n_2 = \frac{(1.0) \mathrm{sin} (39.0)}{\mathrm{sin} (15.0)} $$ $$ n_2 = 2.43 $$

solution B

$$\text{1 = air} \quad \quad \text{ 2 = mystery B}$$ $$\theta_1 = 39.0^{\circ} \quad \quad \quad n_1 = 1.0$$ $$\theta_2 = 26.5^{\circ} \quad \quad \quad n_2 = \ ?$$ $$\frac{\mathrm{sin} \theta_1}{\mathrm{sin} \theta_2} =\frac{n_2}{n_1} $$ $$n_2 = \frac{n_1 \mathrm{sin} \theta_1}{\mathrm{sin} \theta_2} $$ $$n_2 = \frac{(1.0) \mathrm{sin} (39.0)}{\mathrm{sin} (26.5)} $$ $$ n_2 = 1.41 $$

More Practice

For wave practice, slow down enough to name what is moving, what is oscillating, and what is only carrying energy. Many mistakes come from treating a wave like a single object instead of a pattern that propagates through a medium. Use each problem to connect the words crest, trough, amplitude, wavelength, period, frequency, reflection, and refraction to a physical situation.

Question: A pebble lands in a quiet pond and circular ripples spread outward. Why is the ripple better described with a wave model than a particle model?

solution

The water is not traveling outward with the ripple. A small part of the water mainly moves up and down near its starting location.

The wave pattern propagates outward and carries energy away from the splash. That is the key wave idea: energy moves through the medium, but the medium does not need to travel with the wave.

Example: Two students are 4.2 m apart and stretch a slinky between them. One student sends a pulse to the other end, and the pulse reflects back to the start in 2.8 s. What is the wave speed in the slinky?

solution

The pulse travels to the far student and back, so the total distance is twice the separation.

$$d = 2(4.2\,\mathrm{m})$$ $$d = 8.4\,\mathrm{m}$$ $$v = \frac{d}{t}$$ $$v = \frac{8.4\,\mathrm{m}}{2.8\,\mathrm{s}}$$ $$v = 3.0\,\mathrm{m/s}$$

The wave speed is 3.0 m/s.

Question: Identify the medium for each wave: a sound wave from a speaker, a light wave from the Sun, a wave on a guitar string, and a traffic wave on a highway.

solution

The medium for the sound wave is air, or whatever matter the sound is passing through.

The light wave uses changing electric and magnetic fields, so it does not need matter as a medium.

The medium for the guitar wave is the string. The medium for the traffic wave is the pattern of cars slowing down and speeding up.

Example: A small speaker plays a tone at 250 Hz. Each air molecule near the speaker oscillates back and forth many times each second. What is the period of the sound wave?

solution

Frequency tells how many cycles happen each second. Period is the time for one cycle.

$$T = \frac{1}{f}$$ $$T = \frac{1}{250\,\mathrm{Hz}}$$ $$T = 0.0040\,\mathrm{s}$$

The period is 0.0040 s, or 4.0 ms.

Example: A bus line says a bus reaches the stop every 12 min. If the arrival pattern is treated like a repeating wave in time, what are the period and frequency of the arrivals?

solution

The period is the time for one repeat. First convert minutes to seconds.

$$T = 12\,\mathrm{min}\left(\frac{60\,\mathrm{s}}{1\,\mathrm{min}}\right)$$ $$T = 720\,\mathrm{s}$$ $$f = \frac{1}{T}$$ $$f = \frac{1}{720\,\mathrm{s}}$$ $$f = 0.00139\,\mathrm{Hz}$$

The period is 720 s, and the frequency is about 0.00139 Hz.

Example: A floating marker on a lake has an equilibrium height of 1.20 m above the lake bottom. As a wave passes, the marker reaches a crest at 1.70 m and a trough at 0.70 m. What is the amplitude of the wave, and what is the vertical distance from crest to trough?

solution

Amplitude is measured from equilibrium to a crest or trough, not from crest to trough.

$$A = 1.70\,\mathrm{m} - 1.20\,\mathrm{m}$$ $$A = 0.50\,\mathrm{m}$$

The trough gives the same amplitude check.

$$A = 1.20\,\mathrm{m} - 0.70\,\mathrm{m}$$ $$A = 0.50\,\mathrm{m}$$

The full vertical distance from crest to trough is twice the amplitude.

$$\Delta y = 2A$$ $$\Delta y = 1.00\,\mathrm{m}$$

The amplitude is 0.50 m, and the crest-to-trough distance is 1.00 m.

Question: A sound wave moves through air toward the right. A wave pulse on a rope also moves toward the right. How are the oscillations different?

solution

Sound in air is longitudinal. The air molecules oscillate back and forth parallel to the direction the sound travels.

A typical rope wave is transverse. The rope moves up and down while the wave travels sideways along the rope.

Example: A stadium wave takes 8.0 s for one bright jacket in the crowd to go from sitting, to standing, and back to sitting again. Treat that as one cycle. What is the frequency of the stadium wave at that seat?

solution

The time for one complete cycle is the period.

$$T = 8.0\,\mathrm{s}$$ $$f = \frac{1}{T}$$ $$f = \frac{1}{8.0\,\mathrm{s}}$$ $$f = 0.125\,\mathrm{Hz}$$

The frequency is 0.125 Hz.

Example: A wave on a long spring has amplitude 0.30 m and period 0.50 s. The wave has been moving for 12 s. What is its wavelength?

solution

This problem looks like it should use wave equations, but it is missing information.

$$f = \frac{1}{T}$$ $$f = \frac{1}{0.50\,\mathrm{s}}$$ $$f = 2.0\,\mathrm{Hz}$$

To find wavelength, you also need the wave speed or a distance measurement.

$$v = f\lambda$$

Amplitude does not determine wavelength, and the 12 s only tells how long the wave has been moving. The wavelength cannot be determined from the given information.

Question: A student shakes a rope harder and makes a larger amplitude wave. The rope tension and the rope itself stay the same. Should the wave travel faster just because the amplitude is larger?

solution

No. In this course, wave speed is determined by the medium and its physical conditions.

For a rope, the rope tension and mass per length matter. A larger amplitude means a larger displacement from equilibrium, but it does not by itself make the wave speed larger.

Example: A wave machine in a science museum makes water ripples with frequency 2.5 Hz. The crests are 0.80 m apart. What is the speed of the wave?

solution

The distance from one crest to the next crest is the wavelength.

$$\lambda = 0.80\,\mathrm{m}$$ $$v = f\lambda$$ $$v = (2.5\,\mathrm{Hz})(0.80\,\mathrm{m})$$ $$v = 2.0\,\mathrm{m/s}$$

The wave speed is 2.0 m/s.

Example: A low note from a large organ pipe has frequency 170 Hz. Use 340 m/s for the speed of sound in air. What is the wavelength of the sound in the room?

solution

Use the wave equation and solve for wavelength.

$$v = f\lambda$$ $$\lambda = \frac{v}{f}$$ $$\lambda = \frac{340\,\mathrm{m/s}}{170\,\mathrm{Hz}}$$ $$\lambda = 2.0\,\mathrm{m}$$

The wavelength is 2.0 m.

Question: Two sound waves travel through the same room. One is a low note and the other is a high note. Which one has the shorter period, and why?

solution

The high note has the shorter period.

High pitch means high frequency. Since period and frequency are reciprocals, a larger frequency means a smaller period.

Example: A microwave oven uses electromagnetic waves with wavelength 12 cm inside the oven cavity. Use 3.0 × 10⁸ m/s for the wave speed. What is the frequency of the microwaves?

solution

Convert centimeters to meters before using the wave equation.

$$\lambda = 12\,\mathrm{cm}\left(\frac{1\,\mathrm{m}}{100\,\mathrm{cm}}\right)$$ $$\lambda = 0.12\,\mathrm{m}$$ $$v = f\lambda$$ $$f = \frac{v}{\lambda}$$ $$f = \frac{3.0 \times 10^8\,\mathrm{m/s}}{0.12\,\mathrm{m}}$$ $$f = 2.5 \times 10^9\,\mathrm{Hz}$$

The frequency is 2.5 × 10⁹ Hz.

Example: A guitar string is tightened to 45 N. The vibrating part of the string is 0.80 m long and has mass 0.012 kg. What is the speed of waves on the string?

solution

For a string, use the relationship from the string wave investigation.

$$v = \sqrt{\frac{T}{m/L}}$$

First find the mass per length.

$$\frac{m}{L} = \frac{0.012\,\mathrm{kg}}{0.80\,\mathrm{m}}$$ $$\frac{m}{L} = 0.015\,\mathrm{kg/m}$$

Now find the wave speed.

$$v = \sqrt{\frac{45\,\mathrm{N}}{0.015\,\mathrm{kg/m}}}$$ $$v = \sqrt{3000\,\mathrm{m^2/s^2}}$$ $$v = 54.8\,\mathrm{m/s}$$

The wave speed is about 55 m/s.

Question: A light wave enters glass from air and slows down. The frequency stays the same. What happens to the wavelength?

solution

The wavelength gets shorter.

The wave equation is still true. If frequency stays the same and speed decreases, wavelength must decrease too.

Example: A 500 Hz sound from a small underwater speaker travels first through air and then through water. Use 340 m/s for sound in air and 1484 m/s for sound in water. What is the wavelength in each medium?

solution

The frequency of the source stays the same, but the sound speed changes with the medium.

$$\lambda = \frac{v}{f}$$

In air:

$$\lambda_a = \frac{340\,\mathrm{m/s}}{500\,\mathrm{Hz}}$$ $$\lambda_a = 0.68\,\mathrm{m}$$

In water:

$$\lambda_w = \frac{1484\,\mathrm{m/s}}{500\,\mathrm{Hz}}$$ $$\lambda_w = 2.97\,\mathrm{m}$$

The wavelength is 0.68 m in air and 2.97 m in water.

Question: A laser beam hits a glass block. Some light bounces off the front surface and some light bends as it enters the glass. Which part is reflection, and which part is refraction?

solution

The light that bounces off the front surface is reflection.

The light that enters the glass and changes direction because its speed changes is refraction.

Example: A narrow light beam hits a flat mirror at an angle of incidence of 37° measured from the normal line. What is the angle of reflection?

solution

For reflection from a flat surface, the angle of reflection equals the angle of incidence.

$$\theta_r = \theta_i$$ $$\theta_r = 37^\circ$$

The angle of reflection is 37°, measured from the normal.

Question: A calm lake can show a clear image of the Moon, but a windy lake makes the Moon look broken into glittering patches. Which situation is more specular, and which is more diffuse?

solution

The calm lake is more specular because the smooth surface reflects light in a more organized way.

The windy lake is more diffuse because many small tilted surfaces reflect light in many directions.

Example: A red laser has wavelength 6.0 mm in air in a scaled classroom model of refraction. The model uses index of refraction 1.00 for air and 1.33 for water. What wavelength should the model use for the laser in water?

solution

Frequency stays the same when the wave crosses a boundary, so wavelength changes in the same ratio as speed. Index of refraction is inversely related to speed.

$$\frac{\lambda_1}{\lambda_2} = \frac{n_2}{n_1}$$ $$\lambda_2 = \lambda_1\frac{n_1}{n_2}$$ $$\lambda_2 = (6.0\,\mathrm{mm})\frac{1.00}{1.33}$$ $$\lambda_2 = 4.5\,\mathrm{mm}$$

The wavelength in the water part of the model is 4.5 mm.

Example: A ray of light travels from air into glass. The angle in air is 30° from the normal. Use index of refraction 1.00 for air and 1.50 for glass. What is the angle in the glass?

solution

Use Snell's law with angles measured from the normal.

$$n_1\sin\theta_1 = n_2\sin\theta_2$$ $$\sin\theta_2 = \frac{n_1\sin\theta_1}{n_2}$$ $$\sin\theta_2 = \frac{(1.00)\sin(30^\circ)}{1.50}$$ $$\sin\theta_2 = 0.333$$ $$\theta_2 = \sin^{-1}(0.333)$$ $$\theta_2 = 19.5^\circ$$

The angle in the glass is about 20° from the normal. The ray bends toward the normal because it slows down.

Question: In Snell's Law problems, why can you compare speed, wavelength, and index of refraction across two media, but not frequency?

solution

The frequency is set by the source of the wave, so it stays the same when the wave crosses the boundary.

Speed changes because the medium changes. Wavelength changes because the same frequency must fit the new speed.

Example: A swimmer shines a waterproof flashlight upward from underwater toward the flat surface. Use index of refraction 1.33 for water and 1.00 for air. What is the critical angle for total internal reflection?

solution

At the critical angle, the refracted ray would travel along the surface, so the refracted angle is 90°.

$$n_1\sin\theta_1 = n_2\sin\theta_2$$ $$n_1\sin\theta_c = n_2\sin(90^\circ)$$ $$\sin\theta_c = \frac{n_2}{n_1}$$ $$\sin\theta_c = \frac{1.00}{1.33}$$ $$\sin\theta_c = 0.752$$ $$\theta_c = \sin^{-1}(0.752)$$ $$\theta_c = 48.8^\circ$$

The critical angle is about 49° from the normal. Larger angles in the water would cause total internal reflection.