Waves, Sound, and Light

26 questions

All types26 Multiple choice13 Matching6 Ordering7

The speed at which light travels depends directly on the optical density and refractive index of the medium through which it passes. Arrange the following media in order of the speed of light within them, from slowest to fastest.

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Answer

Diamond, Crown glass, Liquid water, Vacuum (from slowest to fastest)

Light travels fastest through empty space (

c \approx 3.00 \times 10^8\text{ m/s}

) where the refractive index is

1.00

. When entering denser transparent media, the wave interacts with the electromagnetic fields of atoms, causing an apparent slowdown proportional to the medium's index of refraction (

v = \frac{c}{n}

). Diamond has the highest index of refraction among the options (

n \approx 2.42

), making it the slowest, followed by crown glass (

n \approx 1.52

), liquid water (

n \approx 1.33

), and vacuum (

n = 1.00

Step-by-Step Solution

Identify the relationship between the index of refraction and the speed of light.

The speed of light in a medium is inversely proportional to its refractive index, given by

v = \frac{c}{n}

, where

c

is the speed of light in a vacuum and

n

is the refractive index.

Higher refractive indices indicate greater optical density and slower wave propagation speeds.

Rank the media by their refractive indices from highest to lowest.

Diamond (

n \approx 2.42

) > Crown glass (

n \approx 1.52

) > Liquid water (

n \approx 1.33

) > Vacuum (

n = 1.00

Ranking from highest index to lowest index establishes the reverse order of speed.

Invert the refractive index ranking to order the media from slowest speed to fastest speed.

Diamond (slowest,

\approx 1.24 \times 10^8\text{ m/s}

)

\rightarrow

Crown glass (

\approx 1.97 \times 10^8\text{ m/s}

)

\rightarrow

Liquid water (

\approx 2.25 \times 10^8\text{ m/s}

)

\rightarrow

Vacuum (fastest,

\approx 3.00 \times 10^8\text{ m/s}

Matches the required sequence criteria from slowest to fastest.

Key Concept

Inverse relationship between index of refraction and wave speed in optical media (

v = \frac{c}{n}

)

Estimated Time:1m 0s

Question 2Question →

A sound wave traveling through air at a constant temperature has a frequency of $250\text{ Hz}$ and a wavelength of $1.36\text{ m}$ . If the frequency of the sound source increases to $500\text{ Hz}$ , what is the new wavelength of the sound wave?

2.72\text{ m}

1.36\text{ m}

0.68\text{ m}

0.34\text{ m}

Show answer & explanation

Answer: $0.68\text{ m}$

Answer

The new wavelength is

0.68\text{ m}

Wave speed in a given medium at constant temperature is constant (

v = f\lambda

). The initial speed is

250\text{ Hz} \times 1.36\text{ m} = 340\text{ m/s}

. When the frequency doubles to

500\text{ Hz}

, the wavelength must be halved:

340\text{ m/s} \div 500\text{ Hz} = 0.68\text{ m}

Step-by-Step Solution

Calculate the constant wave speed in air using the initial conditions

v = f_1 \times \lambda_1 = 250\text{ Hz} \times 1.36\text{ m} = 340\text{ m/s}

The speed of a mechanical wave depends entirely on the medium and remains constant at a fixed temperature.

Rearrange the wave equation to solve for the new wavelength

\lambda_2 = \frac{v}{f_2} = \frac{340\text{ m/s}}{500\text{ Hz}} = 0.68\text{ m}

Because wave speed is constant, an increase in frequency causes an inversely proportional decrease in wavelength.

Key Concept

Inverse relationship between frequency and wavelength in a constant medium (

v = f\lambda

)

Estimated Time:1m 0s

Question 3Question →

Wave interactions with boundaries, media, and observers produce distinct observable phenomena. Match each wave phenomenon on the left with its corresponding physical scenario on the right.

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Answer

Refraction correctly matches the submerged pencil appearing broken at the water boundary; Diffraction correctly matches hearing sound around a hallway corner; Doppler Effect correctly matches the siren pitch increasing during vehicle approach; Total Internal Reflection correctly matches light signal confinement in optical fibers.

Each wave phenomenon matches its exact governing physical principle: Refraction is the bending of light due to speed variation across media; Diffraction is wave spreading around physical obstacles; Doppler Effect is the frequency shift due to relative source motion; Total Internal Reflection is complete optical confinement based on critical angle reflection.

Step-by-Step Solution

Analyze the phenomenon of wave speed changes across differing media boundaries.

Refraction occurs when light enters water from air, decelerating and changing path, which visually displaces the submerged portion of a pencil.

Different medium densities alter the phase velocity of light, bending the transmitted ray.

Analyze how waves navigate physical edges and barrier corners.

Diffraction enables acoustic wavefronts to bend around hallway corners without requiring medium changes or reflection.

Audible sound wavelengths are large enough relative to corner dimensions to produce substantial diffraction.

Analyze the impact of relative source-observer velocity on wave frequency.

The Doppler Effect causes wavefront compression ahead of an approaching siren, increasing observed frequency and pitch.

Relative motion changes the rate at which wave crests reach the observer.

Analyze the optical boundary condition where refraction ceases and complete internal reflection occurs.

Total Internal Reflection retains light pulses entirely inside higher-index fiber glass cores when incident angles exceed the critical angle.

Transmission into lower-index cladding is prevented when the critical angle is exceeded.

Key Concept

Wave interactions and principles: Refraction, Diffraction, Doppler Shift, and Total Internal Reflection

Estimated Time:1m 30s

Question 4Question →

Electromagnetic radiation spans a continuous spectrum of wavelengths and frequencies, with all waves traveling at the speed of light in a vacuum. What is the correct sequence of the following electromagnetic radiation bands when arranged in order from lowest frequency to highest frequency?

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Answer

The correct order from lowest frequency to highest frequency is Microwave radiation, followed by Infrared radiation, then Ultraviolet radiation, and finally Gamma radiation.

Electromagnetic waves propagate at

c \approx 3 \times 10^8\text{ m/s}

in a vacuum according to the relationship

c = f\lambda

. As wavelength decreases, wave frequency and photon energy (

E = hf

) increase. Arranging the given bands from lowest frequency to highest frequency yields Microwave radiation (

10^{9}\text{ to }10^{11}\text{ Hz}

), Infrared radiation (

10^{12}\text{ to }4 \times 10^{14}\text{ Hz}

), Ultraviolet radiation (

8 \times 10^{14}\text{ to }3 \times 10^{16}\text{ Hz}

), and Gamma radiation (

> 10^{19}\text{ Hz}

Step-by-Step Solution

Recall the inverse relationship between wavelength and frequency in electromagnetic waves.

All electromagnetic waves travel at the speed of light in vacuum (

c = f\lambda

), meaning that frequency (

f

) increases as wavelength (

\lambda

) decreases.

Determining the correct frequency sequence requires ordering the wave bands from longest wavelength to shortest wavelength.

Identify the band with the longest wavelength and lowest frequency.

Microwaves exhibit wavelengths between

1\text{ mm}

and

1\text{ m}

(

f \approx 10^{9}\text{ to }10^{11}\text{ Hz}

), making it the first item in the sequence.

Microwaves sit just above radio waves at the low-frequency end of the spectrum.

Order the intermediate bands relative to visible light.

Infrared lies below visible red light (

f \approx 10^{12}\text{ to }4 \times 10^{14}\text{ Hz}

), and ultraviolet lies above visible violet light (

f \approx 8 \times 10^{14}\text{ to }3 \times 10^{16}\text{ Hz}

Infrared has a lower frequency than ultraviolet because its photon energy is lower.

Identify the band with the shortest wavelength and highest frequency.

Gamma rays possess extremely short wavelengths (

< 10\text{ pm}

) and the highest frequencies (

> 10^{19}\text{ Hz}

Gamma rays sit at the highest-energy boundary of the electromagnetic spectrum.

Key Concept

Electromagnetic Spectrum Frequency and Wavelength Relationships

Estimated Time:1m 15s

Question 5Question →

Arrange the following examples of electromagnetic radiation in order of DECREASING wavelength, from the longest wavelength (first) to the shortest wavelength (last).

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Answer

The correct order from longest to shortest wavelength is: AM radio broadcasts (Radio), air traffic control radar (Microwave), thermal imaging (Infrared), sunburn-causing radiation (Ultraviolet), and PET scan emissions (Gamma rays).

The correct sequence maps each application to its electromagnetic band and orders them strictly by decreasing wavelength. Radio waves (AM radio) have the longest wavelengths, followed by microwaves (radar), infrared (thermal imaging), ultraviolet (sunburns), and finally gamma rays (PET scans) with the shortest wavelengths.

Step-by-Step Solution

Identify the electromagnetic spectrum band corresponding to each real-world application.

AM radio = Radio waves; Radar = Microwaves; Thermal imaging = Infrared; Sunburn = Ultraviolet; PET scan = Gamma rays.

Mapping the practical examples to their scientific classification is necessary to compare their physical properties.

Recall the standard order of the electromagnetic spectrum bands.

The bands in sequence are: Radio, Microwave, Infrared, Visible light, Ultraviolet, X-ray, Gamma ray.

The spectrum is a continuous sequence organized by the fundamental properties of the waves.

Determine the direction of decreasing wavelength across the spectrum.

Wavelength is longest for Radio waves and continuously decreases towards Gamma rays, which have the shortest wavelength.

Wavelength and frequency are inversely proportional. Since energy and frequency increase from Radio to Gamma, wavelength must decrease in that same direction.

Arrange the identified applications from longest wavelength to shortest wavelength.

Radio (longest), then Microwave, Infrared, Ultraviolet, and Gamma rays (shortest).

This sequence perfectly matches the decreasing wavelength criteria requested in the stem.

Key Concept

Electromagnetic Spectrum and Wavelength Properties

Question 6Question →

Two audio speakers in an acoustics laboratory emit identical sound waves in phase with a wavelength of $0.5\text{ m}$ . At a specific observation point, the distance to the first speaker is $4.0\text{ m}$ and the distance to the second speaker is $4.75\text{ m}$ . Which of the following describes the type of interference that occurs at this location and its effect on the detected sound?

Constructive interference, producing a louder sound because the path difference equals an integer multiple of the wavelength.

Destructive interference, producing a softer sound because the two waves arrive exactly half a wavelength out of phase.

Constructive interference, producing a higher pitch because the frequencies of the two sound sources combine additively.

Destructive interference, producing a lower pitch because the speed of sound decreases over unequal path lengths.

Show answer & explanation

Answer: Destructive interference, producing a softer sound because the two waves arrive exactly half a wavelength out of phase.

Answer

Destructive interference occurs, producing a softer sound because the path difference is an odd multiple of half-wavelengths (

1.5\lambda

), causing crests to meet troughs.

The difference in distance from the two coherent speakers to the observer is

4.75\text{ m} - 4.0\text{ m} = 0.75\text{ m}

. Dividing by the wavelength (

0.5\text{ m}

) yields

1.5\lambda

, or three half-wavelengths. When two in-phase coherent waves travel distances differing by an odd number of half-wavelengths, they arrive exactly out of phase (

180^\circ

). The crests of one wave coincide with the troughs of the other, resulting in destructive interference and a reduction in sound intensity (loudness).

Step-by-Step Solution

Calculate the path length difference (

\Delta d

) between the two sound sources and the observation point.

\Delta d = |4.75\text{ m} - 4.0\text{ m}| = 0.75\text{ m}

Interference patterns depend on the relative distance traveled by waves from coherent sources.

Determine the path difference in terms of the wave's wavelength (

\lambda = 0.5\text{ m}

\frac{\Delta d}{\lambda} = \frac{0.75\text{ m}}{0.5\text{ m}} = 1.5 = \frac{3}{2}\lambda

Comparing the path difference to the wavelength reveals the phase relationship upon arrival.

Apply the physical criteria for wave interference and evaluate the effect on the acoustic signal.

Because

\Delta d = \left(m + \frac{1}{2}\right)\lambda

with

m = 1

, the waves arrive

180^\circ

(

\pi\text{ radians}

) out of phase, producing destructive interference that minimizes sound amplitude (volume).

Odd half-integer multiples of wavelength cause wave crests from one source to superimpose onto wave troughs from the other, reducing overall loudness without changing wave frequency.

Key Concept

Wave Superposition and Path Difference Interference

Estimated Time:1m 15s

Question 7Question →

Waves transmit energy through space and matter and are described by specific physical characteristics. How should each wave property on the left be paired with its correct definition on the right?

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Answer

Amplitude pairs with maximum displacement from rest position; Frequency pairs with number of cycles per unit time; Wavelength pairs with distance between successive crests or troughs; Period pairs with time required for one full wave cycle.

Each term is accurately matched to its standard physical definition: amplitude describes height from rest position, frequency measures cycles per second, wavelength measures physical distance between peaks, and period measures the duration of one cycle.

Step-by-Step Solution

Identify spatial wave dimensions

Amplitude corresponds to displacement height from equilibrium, while wavelength is the linear distance between identical points on consecutive waves.

Separating vertical displacement from horizontal spatial extent clarifies the two geometric dimensions of a wave.

Identify temporal wave characteristics

Frequency measures cycle count per second (rate), while period measures total time per cycle (duration).

Frequency (

f

) and period (

T

) are reciprocal properties related by

f = \frac{1}{T}

Key Concept

Fundamental wave parameters (amplitude, frequency, wavelength, and period) define the spatial and temporal behavior of waves.

Estimated Time:45s

Question 8Question →

A sensor hovering above a lake simultaneously emits a sound wave and a beam of monochromatic light downward into the water. As both waves transition from the air into the water, which of the following accurately describes the changes in their respective wave properties?

The sound wave's speed and wavelength increase, while the light wave's speed and wavelength decrease; the frequency of both waves remains constant.

Both waves experience a decrease in speed and a corresponding decrease in wavelength, while the frequency of both waves remains constant.

The sound wave's speed decreases and its frequency increases, while the light wave's speed increases and its frequency decreases.

The sound wave's speed and frequency increase, while the light wave's speed and frequency decrease; the wavelength of both waves remains constant.

Show answer & explanation

Answer: The sound wave's speed and wavelength increase, while the light wave's speed and wavelength decrease; the frequency of both waves remains constant.

Answer

The sound wave's speed and wavelength increase, while the light wave's speed and wavelength decrease; the frequency of both waves remains constant.

When any wave crosses a boundary between two media, its frequency remains constant because the number of cycles per second is locked to the original source. Sound is a mechanical wave and travels faster in water than in air because water is denser and more rigid. According to the wave equation (

v = f \lambda

), if speed (

v

) increases and frequency (

f

) is constant, wavelength (

\lambda

) must increase. Conversely, light is an electromagnetic wave that interacts with the atoms in the water, causing its propagation speed to decrease compared to its speed in air. Therefore, with a lower speed and constant frequency, the light wave's wavelength must decrease.

Step-by-Step Solution

Analyze the change in frequency for both waves.

The frequency of both the sound wave and the light wave remains constant.

Frequency is determined exclusively by the wave source and does not change when transitioning between different media.

Determine how the speed of each wave changes when moving from air to water.

The sound wave speeds up, while the light wave slows down.

Sound is a mechanical wave that requires a medium and travels faster in denser, less compressible materials (water > air). Light is an electromagnetic wave that is impeded by matter, so its speed decreases in media with higher optical density (water < air).

Apply the wave equation (

v = f \lambda

) to determine wavelength changes.

The sound wave's wavelength increases, and the light wave's wavelength decreases.

Because frequency (

f

) is constant, wavelength (

\lambda

) must change directly proportionally to wave speed (

v

). Higher speed means longer wavelength; lower speed means shorter wavelength.

Key Concept

Wave Propagation Across Boundaries

Question 9Question →

Which property of a sound wave directly determines how loud it sounds to a listener?

Frequency

Amplitude

Wavelength

Wave speed

Show answer & explanation

Answer: Amplitude

Answer

Amplitude

The loudness or volume of a sound is directly governed by its amplitude. A sound wave with a larger amplitude carries more mechanical energy and produces greater pressure fluctuations, which human ears perceive as a louder sound.

Step-by-Step Solution

Identify what physical characteristic of a sound wave corresponds to sound volume or intensity.

Loudness is determined by the energy carried by the wave, which is directly proportional to the square of its amplitude.

Greater amplitude means particles in the medium vibrate with greater displacement, creating larger pressure variations detected as louder sound.

Distinguish amplitude from frequency and wavelength.

Frequency and wavelength determine the pitch of the sound, whereas amplitude dictates loudness.

Separating wave energy characteristics (amplitude) from wave oscillation rates (frequency) identifies amplitude as the correct property.

Key Concept

Relationship between sound wave amplitude and perceived loudness

Estimated Time:30s

Question 10Question →

Arrange the following media in order of the speed at which sound waves travel through them, from slowest to fastest.

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Answer

Air at sea level, followed by Liquid water, and finally Solid steel bar.

Sound is a mechanical longitudinal wave that relies on molecular collisions to transfer kinetic energy. In gases such as air, molecules are far apart, resulting in the slowest speed. In liquids such as water, molecules are closer together, increasing speed. In rigid solids such as steel, strong atomic bonds and high elasticity allow vibrations to transmit most rapidly.

Step-by-Step Solution

Identify the physical state of matter for each listed medium.

Air is a gas, water is a liquid, and steel is a solid.

Sound is a mechanical wave that requires a physical medium to propagate via particle collisions.

Apply the general rule for sound propagation speed across states of matter.

Sound travels slowest in gases, faster in liquids, and fastest in solids.

Closer particle spacing and higher elastic properties in condensed matter allow vibrational energy to transfer more rapidly.

Sequence the media from slowest transmission speed to fastest transmission speed.

Air at sea level (

343\text{ m/s}

) < Liquid water (

1{}480\text{ m/s}

) < Solid steel bar (

5{}000\text{ m/s}

Matches the requested ordering criterion from lowest to highest wave speed.

Key Concept

Sound Wave Propagation Speed in Different States of Matter

Estimated Time:45s

Question 11Question →

When a ray of light passes from air into a glass lens, it slows down and changes direction at the boundary. Which wave phenomenon describes this bending of light due to a change in speed?

Diffraction

Refraction

Reflection

Polarization

Show answer & explanation

Answer: Refraction

Answer

Refraction is the wave phenomenon that describes the bending of light caused by a change in speed when passing between different media.

When light passes from one medium to another with a different optical density (such as from air into glass), its speed changes. This change in speed causes the light ray to bend at the boundary, a phenomenon known as refraction.

Step-by-Step Solution

Analyze the physical scenario described in the question.

Light travels from air into glass, resulting in a change in speed and a change in propagation direction across the boundary.

Identifying the medium transition and speed variation is essential to classifying the wave behavior.

Identify the wave phenomenon associated with speed-induced directional changes at a boundary.

Refraction is defined as the bending of a wave when it passes from one medium to another due to a difference in wave speed.

This matches the exact physical principle occurring when light enters a glass lens.

Key Concept

Refraction of light waves at medium boundaries

Estimated Time:45s

Question 12Question →

Waves are classified based on the direction of particle vibration relative to energy transport and whether they require matter to travel. Match each wave category on the left with its defining physical property on the right.

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Answer

Transverse wave pairs with perpendicular particle oscillation; longitudinal wave pairs with parallel particle oscillation; electromagnetic wave pairs with propagation through a vacuum; mechanical wave pairs with the requirement of a material medium.

Each term accurately links to its fundamental physical definition: transverse waves have perpendicular displacement, longitudinal waves have parallel displacement, electromagnetic waves do not need matter, and mechanical waves require a medium.

Step-by-Step Solution

Classify wave types by directional motion.

Transverse motion means particles move perpendicular to wave travel; longitudinal motion means particles move parallel (along the line of travel).

Wave geometric behavior is determined by the orientation of the medium's displacement relative to energy propagation.

Classify wave types by medium requirements.

Electromagnetic waves do not require matter to travel (can travel in a vacuum), whereas mechanical waves strictly require a solid, liquid, or gas medium.

Wave transmission mechanism depends on whether energy is carried via particle collisions (mechanical) or synchronized field oscillations (electromagnetic).

Key Concept

Wave classification by vibration direction (transverse vs. longitudinal) and medium dependence (mechanical vs. electromagnetic)

Estimated Time:45s

Question 13Question →

A musician tightens a guitar string, which causes the string to vibrate at a higher frequency when plucked. How does this increase in frequency change the sound that is heard?

The loudness of the sound increases.

The pitch of the sound becomes higher.

The speed of the sound wave through the surrounding air increases significantly.

The wavelength of the sound wave in the air becomes longer.

Show answer & explanation

Answer: The pitch of the sound becomes higher.

Answer

The pitch of the sound becomes higher.

The pitch of a sound is directly determined by the frequency of the sound wave. When the frequency of vibration increases, the ear perceives the sound at a higher pitch.

Step-by-Step Solution

Identify the acoustic property controlled by the frequency of a sound wave.

Frequency measures the number of wave cycles per second (

Hz

) and directly determines perceived pitch.

Pitch is the psychological perception of wave frequency in human hearing.

Determine the physical consequence of increasing the frequency of vibration.

An increase in frequency produces a higher-pitched tone.

Faster vibrations create compressions that reach the ear more frequently, resulting in a higher pitch.

Key Concept

Relationship between wave frequency and sound pitch

Estimated Time:45s

Question 14Question →

When white light passes through a prism, it disperses into different colors of the visible spectrum. What is the correct order of the following colors of visible light when arranged from lowest wave frequency to highest wave frequency?

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Answer

The correct sequence from lowest frequency to highest frequency is Red light, Yellow light, Green light, Violet light.

In the visible light spectrum, colors progress in order of decreasing wavelength and increasing frequency according to the sequence Red, Orange, Yellow, Green, Blue, Indigo, and Violet (ROYGBIV). Red light has the longest wavelength and lowest frequency, followed by yellow light, green light, and finally violet light, which has the shortest wavelength and highest frequency.

Step-by-Step Solution

Recall the order of the visible light spectrum from longest wavelength to shortest wavelength.

The standard sequence of visible light colors is Red, Orange, Yellow, Green, Blue, Indigo, and Violet (ROYGBIV).

Visible light wavelengths decrease progressively from red light (~700 nm) to violet light (~400 nm).

Determine the relationship between wavelength and frequency using the wave speed equation

c = \lambda f

Frequency is inversely proportional to wavelength (

f = \frac{c}{\lambda}

Because electromagnetic waves travel at the speed of light in a vacuum, longer wavelengths correspond to lower frequencies and shorter wavelengths correspond to higher frequencies.

Sequence the given colors in order of increasing frequency.

Red light (lowest frequency) -> Yellow light -> Green light -> Violet light (highest frequency).

Following the ROYGBIV sequence, wave frequency increases continuously from red to violet.

Key Concept

Relationship between wavelength, frequency, and color sequence in the visible light spectrum

Estimated Time:45s

Question 15Question →

When a person shouts inside a large empty gymnasium, they hear their voice repeated a moment later as an echo. Which wave behavior is directly responsible for producing an echo?

Diffraction

Reflection

Refraction

Absorption

Show answer & explanation

Answer: Reflection

Answer

Reflection is the wave behavior responsible for producing an echo.

An echo is heard when sound waves strike a rigid barrier and bounce back toward the listener. This bouncing back of a wave from a surface is defined as reflection.

Step-by-Step Solution

Analyze the physical scenario of an echo

Sound waves travel outward from the source, strike a hard obstacle such as a gymnasium wall, and bounce back to the listener.

Identifying the motion of the wave as returning after hitting a boundary isolates the specific physical mechanism.

Associate the wave behavior with standard physical terminology

The bouncing back of waves upon encountering an impenetrable surface is defined as reflection.

An echo is a direct real-world example of sound wave reflection.

Key Concept

Sound wave reflection and echo formation

Estimated Time:45s

Question 16Question →

Wave interactions and behaviors produce distinct physical effects in sound and light. Match each wave phenomenon to its corresponding real-world observation.

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Answer

Doppler effect matches the shift in pitch of a moving siren; Resonance matches the shattering of a glass at its natural frequency; Total internal reflection matches light confinement in fiber optics; Diffraction matches radio waves bending around hills and structures.

Each wave phenomenon maps directly to its underlying principle: relative motion alters perceived pitch via the Doppler effect; matching natural frequencies drives maximum vibrational amplitude via resonance; optical containment in fiber cores relies on total internal reflection; wave propagation around physical obstacles occurs through diffraction.

Step-by-Step Solution

Analyze the effect of source motion on perceived sound pitch.

Compression and expansion of wavefronts due to motion produces the Doppler effect, matching the approaching and receding siren.

Perceived frequency changes result directly from relative velocity between the source and observer.

Identify the behavior where matching frequencies produce maximum vibration amplitude.

Energy transfer at an object's natural frequency defines resonance, matching the shattering crystal glass.

When forcing frequency matches natural frequency, energy accumulates constructively to cause large mechanical stress.

Examine the mechanism allowing light transmission without leakage in fiber optics.

Complete reflection at the core-cladding boundary beyond the critical angle is total internal reflection, matching fiber-optic light pulses.

Light cannot refract out when traveling from high to low refractive index at incident angles exceeding the critical threshold.

Determine how long-wavelength waves navigate physical terrain barriers.

The bending of waves around large structures without a change in medium is diffraction, matching obstructed radio wave reception.

Waves naturally spread into geometric shadow zones behind obstacles comparable to their wavelength.

Key Concept

Wave behaviors and boundary interactions

Estimated Time:1m 15s

Question 17Question →

A communications antenna emits radio waves that travel through uniform atmospheric air. If the transmitter is adjusted to emit radio waves at a higher frequency, how do the speed and wavelength of the transmitted waves change?

The speed increases, and the wavelength remains unchanged.

The speed increases, and the wavelength increases.

The speed remains constant, and the wavelength decreases.

The speed remains constant, and the wavelength increases.

Show answer & explanation

Answer: The speed remains constant, and the wavelength decreases.

Answer

The speed remains constant, and the wavelength decreases.

In a uniform medium such as air, the speed of an electromagnetic wave is constant. According to the wave equation

v = f \lambda

, when the speed

v

is constant, frequency

f

and wavelength

\lambda

are inversely proportional. Therefore, increasing the frequency causes the wavelength to decrease while the speed remains unchanged.

Step-by-Step Solution

Identify the factor that determines wave propagation speed.

The speed of electromagnetic radio waves is constant in a uniform medium (

v = c

in air).

Wave speed is an intrinsic property of the medium through which the wave travels, not the frequency of the source.

Apply the wave equation relating speed, frequency, and wavelength.

v = f \lambda

, which rearranges to

\lambda = \frac{v}{f}

This relationship mathematically couples frequency and wavelength for a given constant speed.

Determine the effect of increasing frequency on wavelength.

As frequency (

f

) increases while speed (

v

) remains fixed, wavelength (

\lambda

) must decrease proportionally.

Frequency and wavelength are inversely proportional when wave speed is held constant.

Key Concept

The speed of a wave is determined by the transmission medium, creating an inverse relationship between frequency and wavelength (

v = f \lambda

Estimated Time:1m 0s

Question 18Question →

Electromagnetic radiation travels through space as oscillating electric and magnetic fields, with each type characterized by its specific frequency, wavelength, and energy. Arrange the following regions of the electromagnetic spectrum in order of increasing photon energy, from the lowest energy to the highest energy.

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Answer

The correct sequence from lowest to highest photon energy is Radio waves, Microwaves, Visible light, Ultraviolet radiation, and X-rays.

Photon energy is governed by the relation

E = hf = \frac{hc}{\lambda}

, where

h

is Planck's constant and

c

is the speed of light. Because energy increases directly with frequency and inversely with wavelength, radiation with the longest wavelengths (radio waves) carries the least energy per photon, followed in ascending order by microwaves, visible light, ultraviolet radiation, and X-rays.

Step-by-Step Solution

Recall the fundamental relationship between wave frequency and photon energy.

Photon energy is directly proportional to frequency as defined by Planck's equation,

E = hf

, and inversely proportional to wavelength,

E = \frac{hc}{\lambda}

Establishing that higher frequency and shorter wavelength correspond directly to higher photon energy.

Identify the positions of the given wave types across the continuous electromagnetic spectrum.

In order of increasing frequency: Radio waves (

< 10^9\text{ Hz}

)

\rightarrow

Microwaves (

10^9 - 10^{12}\text{ Hz}

)

\rightarrow

Visible light (

4 \times 10^{14} - 7.5 \times 10^{14}\text{ Hz}

)

\rightarrow

Ultraviolet radiation (

10^{15} - 10^{16}\text{ Hz}

)

\rightarrow

X-rays (

10^{16} - 10^{19}\text{ Hz}

Classifying each radiation type by its frequency band determines its relative energy.

Sequence the items from the lowest photon energy to the highest photon energy.

Radio waves

\rightarrow

Microwaves

\rightarrow

Visible light

\rightarrow

Ultraviolet radiation

\rightarrow

X-rays.

Arranging the items in ascending order of frequency matches the required ascending order of energy.

Key Concept

Photon energy across the electromagnetic spectrum is directly proportional to wave frequency (

E = hf

) and inversely proportional to wavelength (

E = hc/\lambda

Estimated Time:1m 0s

Question 19Question →

An acoustic engineer tests a noise-reduction device that emits a secondary sound wave exactly half a wavelength ( $180^\circ$ ) out of phase with an incoming noise wave of identical frequency and amplitude. What occurs when these two sound waves overlap in the air?

The waves undergo constructive interference, resulting in a sound wave with twice the original amplitude.

The waves undergo destructive interference, resulting in the cancellation of sound with zero net amplitude.

The waves undergo refraction, resulting in a permanent doubling of the wave's frequency.

The waves undergo dispersion, resulting in the sound separating into different component frequencies.

Show answer & explanation

Answer: The waves undergo destructive interference, resulting in the cancellation of sound with zero net amplitude.

Answer

The waves undergo destructive interference, resulting in the cancellation of sound with zero net amplitude.

When two waves with equal frequency and amplitude meet half a wavelength (

180^\circ

) out of phase, the crests of the first wave precisely coincide with the troughs of the second wave. According to the principle of superposition, their individual displacements subtract from each other, resulting in complete destructive interference and zero resultant amplitude.

Step-by-Step Solution

Analyze the phase relationship of the two sound waves.

The two waves have identical frequency and amplitude but are half a wavelength (

180^\circ

) out of phase.

Determining phase alignment indicates whether wave crests and troughs align constructively or destructively.

Apply the principle of wave superposition to calculate the net displacement.

Every positive peak (crest) of the primary wave perfectly aligns with an equal negative peak (trough) of the secondary wave, summing to zero net displacement (

A - A = 0

The principle of superposition states that overlapping waves combine algebraically at every point in space.

Identify the corresponding wave phenomenon.

Destructive interference.

Complete destructive interference occurs when opposing wave displacements of equal magnitude cancel each other out.

Key Concept

Destructive Wave Interference and Superposition

Estimated Time:1m 0s

Question 20Question →

Wave properties describe the spatial and temporal behavior of mechanical and electromagnetic oscillations. Which of the following correctly matches each wave parameter on the left with its standard physical definition on the right?

Click a left item, then click its matching right item

Show answer & explanation

Answer

Wavelength pairs with the spatial distance between two consecutive identical points on adjacent wave cycles; Frequency pairs with the number of complete wave cycles that pass a fixed reference point per second; Amplitude pairs with the maximum displacement of an oscillation from its central equilibrium position; and Wave speed pairs with the rate at which a wave disturbance propagates through a given medium.

Each fundamental wave term corresponds directly to its standard physical definition: wavelength is the distance between consecutive identical points on adjacent cycles; frequency is the number of cycles passing a reference point per second; amplitude is the maximum displacement from equilibrium; and wave speed is the propagation rate of the wave disturbance through a medium.

Step-by-Step Solution

Identify the spatial parameter describing the length of a full cycle

Wavelength (

\lambda

) is the physical distance between consecutive corresponding points, such as crest to crest or compression to compression.

This establishes the spatial periodicity of the wave.

Identify the temporal rate parameter measuring wave oscillations per unit time

Frequency (

f

) is the count of complete wave cycles passing a stationary point per unit time, measured in Hertz (

\text{Hz}

This establishes the temporal oscillation rate of the wave.

Identify the parameter measuring wave height or disturbance magnitude

Amplitude (

A

) is the maximum displacement of the medium or field from its central baseline equilibrium position.

This parameter indicates the energy carried by the wave, relating to loudness in sound and brightness in light.

Identify the propagation rate of the wave disturbance

Wave speed (

v

) is the distance a wave crest travels per unit time through a medium, determined by

v = f \lambda

This connects spatial wavelength and temporal frequency to linear propagation speed.

Key Concept

Fundamental wave parameters: wavelength, frequency, amplitude, and propagation speed

Estimated Time:1m 0s

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