Quiz: Characterizing Wave Phenomena — 21 questions

1. Which quantity is measured in watts per square metre and represents the energy transported by a sound wave per unit time and area?

A. Sound intensity level

B. Acoustic frequency

C. Sound intensity

D. Propagation speed

Answer: (C)

Explanation

Sound intensity measures energy flow per unit area and time, with units of watts per square metre. Sound intensity level is instead a logarithmic quantity expressed in decibels.

2. Which pair correctly identifies the audibility threshold and the pain threshold for sound intensity?

A. $1.0 \times 10^{-12}\ \mathrm{W\,m^{-2}}$ and $1.0 \times 10^{-6}\ \mathrm{W\,m^{-2}}$

B. $1\ \mathrm{W\,m^{-2}}$ and $1.0 \times 10^{-12}\ \mathrm{W\,m^{-2}}$

C. $1.0 \times 10^{-6}\ \mathrm{W\,m^{-2}}$ and $1.0 \times 10^{-3}\ \mathrm{W\,m^{-2}}$

D. $1.0 \times 10^{-12}\ \mathrm{W\,m^{-2}}$ and $1\ \mathrm{W\,m^{-2}}$

Answer: (D)

Explanation

The audibility threshold is $I_0 = 1.0 \times 10^{-12}\ \mathrm{W\,m^{-2}}$, while the pain threshold is $1\ \mathrm{W\,m^{-2}}$. Reversing these values confuses the much weaker hearing threshold with the much stronger pain threshold.

3. A sound has intensity $I = 100I_0$. What is its sound intensity level?

A. 10 dB

B. 2 dB

C. 20 dB

D. 100 dB

Answer: (C)

Explanation

Using $L = 10\log\left(I/I_0\right)$ gives $L = 10\log(100) = 20\ \mathrm{dB}$. The intensity ratio itself is not the decibel value, because the level is logarithmic.

4. A listener moves farther from a sound source in open air, while no absorbing material is introduced between them. Which attenuation mechanism primarily explains the reduced intensity?

A. Absorption attenuation from conversion into thermal energy

B. Resonant attenuation from matching the listener’s frequency

C. Diffraction attenuation from passage through a narrow opening

D. Geometric attenuation from spreading over larger spheres

Answer: (D)

Explanation

Geometric attenuation results because the same acoustic power spreads over increasingly large spherical surfaces as distance grows. Absorption attenuation instead requires material between the source and receiver that removes sound energy.

5. When does a wave undergo pronounced diffraction at an opening?

A. When the opening dimension is close to or smaller than the wavelength

B. When the opening dimension is much larger than the wavelength

C. When the opening blocks the wave without transmitting it

D. When the wave has zero frequency at the opening

Answer: (A)

Explanation

Diffraction becomes pronounced when the opening size is comparable to or smaller than the wavelength, allowing the wave to spread significantly after passing through. A much larger opening generally produces less noticeable spreading.

6. For an opening of size $a$ and wavelength $\lambda$, how does the characteristic diffraction angle change if $\lambda$ increases while $a$ remains fixed?

A. It decreases because the wave bends less at longer wavelengths

B. It remains fixed because the opening determines the angle

C. It becomes zero because the wavelength exceeds the opening

D. It increases because $\theta = \frac{\lambda}{a}$

Answer: (D)

Explanation

The relation $\theta = \frac{\lambda}{a}$ shows that increasing the wavelength increases the diffraction angle when the opening is unchanged. The angle decreases instead when the opening size increases.

7. Which statement correctly defines diffraction?

A. It is the addition of two waves with a constant phase difference.

B. It is the absorption of wave energy by a material between source and receiver.

C. It is the spreading of a wave after it encounters an opening no larger than its wavelength.

D. It is the reflection of a wave from a surface at equal angles.

Answer: (C)

Explanation

Diffraction is the spreading of a wave after it passes an opening whose size is less than or comparable to its wavelength. Interference concerns superposition of waves, while absorption and reflection describe different wave phenomena.

8. What physical process describes interference?

A. The spreading of one wave after passage through a narrow opening

B. The conversion of wave energy into heat inside a material

C. The superposition of two waves producing reinforcement in some regions and cancellation in others

D. The change in wave speed caused by entering a new medium

Answer: (C)

Explanation

Interference occurs when two waves superpose, creating regions of reinforcement and cancellation. Spreading at an opening is diffraction, not the defining process of interference.

9. Two sound waves have the same frequency, but their phase difference changes continuously with time. What is the most likely result?

A. A stable observable interference pattern will not be maintained.

B. The waves will necessarily produce constant constructive interference.

C. The waves will necessarily produce constant destructive interference.

D. The waves will stop propagating because their frequencies match.

Answer: (A)

Explanation

Observable interference requires synchronous waves whose phase difference remains constant, not merely equal frequencies. If the phase difference continually changes, reinforcement and cancellation do not remain fixed.

10. Which condition describes constructive interference?

A. The waves are in phase and their amplitudes add.

B. The waves are in opposite phase and their amplitudes cancel.

C. The waves pass through an opening smaller than their wavelength.

D. The waves have different frequencies and lose their amplitudes.

Answer: (A)

Explanation

Constructive interference occurs when the waves arrive in phase, so their amplitudes reinforce one another. Opposite-phase waves produce destructive interference, while a small opening describes a diffraction condition.

11. Which path difference produces destructive interference for wavelength $\lambda$?

A. $\delta = k\lambda$

B. $\delta = 2k\lambda + \frac{\lambda}{4}$

C. $\delta = \left(k+\frac{1}{2}\right)\lambda$

D. $\delta = \frac{\lambda}{k+1}$

Answer: (C)

Explanation

Destructive interference occurs when the path difference is an odd half-wavelength, expressed as $\delta = \left(k+\frac{1}{2}\right)\lambda$. The condition $\delta = k\lambda$ instead corresponds to constructive interference.

12. In Young’s double-slit experiment, why can the two slits produce a stable interference pattern?

A. They act as synchronous sources derived from the same laser

B. They emit unrelated waves with independently changing phases

C. They reflect light from two mirrors with different temperatures

D. They create sound waves that remain synchronized in air

Answer: (A)

Explanation

The two slits are illuminated by the same laser and therefore act as synchronous sources, maintaining a stable phase relationship. Unrelated sources generally do not preserve the phase coherence required for stable fringes.

13. For a fixed slit separation and screen distance, how does the optical path difference change as a point moves farther from the central axis?

A. It reverses sign because the screen distance becomes larger

B. It increases because the position coordinate x becomes larger

C. It decreases because the two paths become more similar

D. It remains constant because the wavelength does not change

Answer: (B)

Explanation

The path difference is given by $\delta = \frac{e x}{D}$, so increasing x increases the value of $\delta$ when e and D remain fixed. The wavelength does not determine this particular geometric dependence.

14. What does the interfringe represent on the screen in a Young double-slit experiment?

A. The distance from a slit to the nearest screen edge

B. The smallest distance between consecutive constructive points

C. The total width occupied by the bright interference pattern

D. The separation between the two physical slits

Answer: (B)

Explanation

The interfringe is defined as the smallest distance separating two consecutive points of constructive interference. It is therefore a spacing on the screen, not the slit separation or the full pattern width.

15. If the wavelength and slit separation remain fixed while the screen is moved farther away, what happens to the Young double-slit interfringe?

A. It stays unchanged because the slits are unchanged

B. It becomes independent of wavelength and slit separation

C. It increases in proportion to the screen distance

D. It decreases in proportion to the screen distance

Answer: (C)

Explanation

The interfringe is $i = \frac{\lambda D}{e}$, so increasing D increases i when λ and e are fixed. The slit geometry still matters because e remains in the denominator.

16. What physical change defines the Doppler effect for an observer?

A. A change in wavelength caused by the material’s chemical composition

B. A change in the source’s emitted frequency while it remains at rest

C. A change in the perceived frequency caused by relative motion

D. A change in wave amplitude caused by the observer’s distance

Answer: (C)

Explanation

The Doppler effect is a change in the frequency perceived by an observer when the source and observer move relative to one another. The source’s emitted frequency at rest is not itself changed merely because the observer detects a shifted frequency.

17. A siren approaches a stationary listener and then moves away; how does its perceived pitch change?

A. It is higher during approach and lower during recession

B. It remains unchanged because the siren’s emitted tone is fixed

C. It becomes louder during approach and quieter during recession without a pitch change

D. It is lower during approach and higher during recession

Answer: (A)

Explanation

Approach causes the perceived frequency to increase, producing a higher pitch, while recession decreases the perceived frequency and pitch. A change in loudness is not the defining frequency shift in this situation.

18. For a source approaching an observer at speed v, which expression gives the received frequency?

A. $f_R = f_e\frac{c-v}{c}$

B. $f_R = f_e\frac{c+v}{c}$

C. $f_R = f_e\frac{c}{c-v}$

D. $f_R = f_e\frac{c}{c+v}$

Answer: (C)

Explanation

For an approaching source, the received frequency is $f_R = f_e\frac{c}{c-v}$, whose denominator is smaller than c and therefore raises the received frequency. The expression with $c+v$ applies to a receding source.

19. When a source speed v is much smaller than the wave speed c, which approximation describes the Doppler frequency shift for an approaching source?

A. $\Delta f \approx f_e\frac{v}{c}$

B. $\Delta f \approx f_e\frac{c-v}{f_e}$

C. $\Delta f \approx f_e\frac{v}{c-v}$

D. $\Delta f \approx f_e\frac{c}{v}$

Answer: (A)

Explanation

For $v \ll c$, the exact shift $\Delta f = f_e\frac{v}{c-v}$ is approximated by $\Delta f \approx f_e\frac{v}{c}$. The exact expression remains valid, but the approximation simplifies the calculation when v is small compared with c.

20. If the frequency of light decreases while its speed remains c, what happens to its wavelength?

A. It remains fixed because the wave speed remains c

B. It becomes zero because the frequency has decreased

C. It decreases because wavelength and frequency vary directly

D. It increases because wavelength and frequency vary inversely

Answer: (D)

Explanation

The relationship $\lambda = \frac{c}{f}$ shows that wavelength increases when frequency decreases at fixed wave speed. A lower frequency therefore corresponds to a longer wavelength.

21. What does a redshift in a star’s spectral lines indicate?

A. The star is stationary and its lines shift because of higher intensity

B. The star is approaching and its lines shift toward shorter wavelengths

C. The star is rotating and its lines shift because its temperature falls

D. The star is moving away and its lines shift toward longer wavelengths

Answer: (D)

Explanation

Redshift is the displacement of stellar spectral lines toward longer wavelengths and indicates that the star is receding from Earth. A shift toward shorter wavelengths is associated with approach and is called blueshift.