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A Level H2 Physics Waves Sound Light Quiz

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Questions

A-Level Physics H2 Quiz - Waves Sound Light

Name: __________________________
Class: __________________________
Date: __________________________
Score: _______ / 40

Duration: 45 minutes
Total Marks: 40

Instructions:

Answer all questions.
Write your answers in the spaces provided.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a scientific calculator.
Assume the speed of light in vacuum $c = 3.00 \times 10^8 \text{ m s}^{-1}$ and Planck constant $h = 6.63 \times 10^{-34} \text{ J s}$ unless otherwise stated.

Section A: Multiple Choice & Short Structured Questions (10 Marks)

1. Which of the following statements correctly describes the formation of a stationary wave? A. Two waves of different frequencies traveling in opposite directions superpose. B. Two waves of the same frequency and amplitude traveling in the same direction superpose. C. Two waves of the same frequency traveling in opposite directions superpose. D. A wave reflects off a boundary and interferes with the incident wave, resulting in zero net energy transfer along the medium.

[1]

2. In a double-slit experiment using monochromatic light, the fringe separation is $x$ . If the distance between the slits and the screen is doubled and the slit separation is halved, what is the new fringe separation? A. $0.5x$ B. $x$ C. $2x$ D. $4x$

[1]

3. Sound waves of frequency 500 Hz travel through air at a speed of 340 m s⁻¹. Calculate the wavelength of these sound waves.

Answer: __________________________ m

[1]

4. State the condition required for two light sources to be considered coherent.

[1]

5. A diffraction grating has 500 lines per mm. Calculate the grating spacing $d$ in meters.

Answer: __________________________ m

[1]

6. In the photoelectric effect, increasing the intensity of the incident radiation (while keeping frequency constant) increases: A. The maximum kinetic energy of the photoelectrons. B. The work function of the metal. C. The number of photoelectrons emitted per second. D. The threshold frequency.

[1]

7. Define the term threshold frequency in the context of the photoelectric effect.

[1]

8. A pipe closed at one end has a length of 0.25 m. Calculate the fundamental frequency of the sound produced in the pipe. (Speed of sound = 340 m s⁻¹)

Answer: __________________________ Hz

[1]

9. Explain why the sky appears blue, referring to the scattering of light.

[2]

10. A laser beam passes through a single slit. Describe the appearance of the diffraction pattern observed on a screen.

[2]

Section B: Structured Questions (18 Marks)

11. (a) State the Principle of Superposition.

[2]

(b) Two coherent sound sources, $S_1$ and $S_2$ , are placed 2.0 m apart. A detector is moved along a line parallel to the line joining the sources, at a distance of 5.0 m from them. The wavelength of the sound is 0.50 m.

(i) Calculate the path difference between the waves from $S_1$ and $S_2$ arriving at a point P where the first constructive interference maximum (other than the central maximum) is observed.

Answer: __________________________ m

[1]

(ii) Explain why a maximum is observed at this point.

[2]

(c) If the frequency of the sound sources is doubled, state and explain the effect on the separation between adjacent maxima observed by the detector.

[3]

12. (a) Monochromatic light of wavelength 550 nm is incident normally on a diffraction grating. The second-order maximum is observed at an angle of 30° to the normal.

(i) Calculate the grating spacing $d$ .

Answer: __________________________ m

[3]

(ii) Calculate the number of lines per mm on the grating.

Answer: __________________________ lines mm⁻¹

[1]

(b) Determine the highest order of maximum that can be observed with this grating and light source.

Answer: __________________________

[2]

(c) White light is now incident on the same grating. Explain why a spectrum is observed for each order, except the zero-order maximum.

[3]

13. (a) In an X-ray tube, electrons are accelerated through a potential difference of 30 kV before striking a metal target.

(i) Calculate the maximum kinetic energy of the electrons in Joules. ( $e = 1.60 \times 10^{-19} \text{ C}$ )

Answer: __________________________ J

[2]

(ii) Calculate the minimum wavelength of the X-rays produced.

Answer: __________________________ m

[2]

(b) The X-ray spectrum produced consists of a continuous distribution of wavelengths with sharp peaks superimposed on it. Explain the origin of the continuous distribution.

[3]

Section C: Data Analysis & Application (12 Marks)

14. A student investigates the photoelectric effect using a zinc plate and a source of ultraviolet (UV) radiation. The work function of zinc is $6.9 \times 10^{-19} \text{ J}$ .

(a) Calculate the threshold frequency for zinc.

Answer: __________________________ Hz

[2]

(b) The UV source emits radiation with a wavelength of 250 nm.

(i) Show that the energy of a single photon of this radiation is approximately $7.9 \times 10^{-19} \text{ J}$ .

[2]

(ii) Calculate the maximum kinetic energy of the emitted photoelectrons.

Answer: __________________________ J

[2]

(c) The intensity of the UV radiation is increased while keeping the wavelength constant.

(i) State the effect, if any, on the maximum kinetic energy of the photoelectrons.

[1]

(ii) State the effect, if any, on the photoelectric current.

[1]

(iii) Explain your answer to (c)(ii) in terms of the photon model of light.

[3]

(d) If the zinc plate is replaced with a metal having a higher work function, but the same UV source is used, explain whether photoelectric emission will still occur.

[3]

15. A stationary wave is formed on a string fixed at both ends. The length of the string is 1.2 m.

(a) Calculate the wavelength of the fundamental mode (first harmonic).

Answer: __________________________ m

[1]

(b) If the frequency of the fundamental mode is 50 Hz, calculate the speed of the wave on the string.

Answer: __________________________ m s⁻¹

[1]

(c) Sketch the shape of the string for the second harmonic (first overtone). Label the nodes (N) and antinodes (A).

[2]

16. Light travels from glass (refractive index $n = 1.5$ ) into air.

(a) Calculate the critical angle for the glass-air boundary.

Answer: __________________________ °

[2]

(b) State what happens to a ray of light incident on the boundary at an angle of incidence greater than the critical angle.

[1]

(c) Explain why optical fibers use total internal reflection to transmit data.

[2]

17. Two speakers emit sound waves of the same frequency and amplitude. They are placed 3.0 m apart.

(a) Define what is meant by "constructive interference".

[2]

(b) A student walks along a line parallel to the speakers. Explain why they hear alternating loud and quiet regions.

[2]

(c) If the frequency of the sound is increased, state and explain what happens to the distance between adjacent loud regions.

[2]

18. A diffraction grating is used to analyze the light from a sodium lamp.

(a) State one advantage of using a diffraction grating over a double slit for spectral analysis.

[1]

(b) The first order maximum for a specific yellow line is observed at 15°. If the grating has 600 lines per mm, calculate the wavelength of this light.

Answer: __________________________ m

[3]

(c) Explain why the second order spectrum is wider than the first order spectrum.

[2]

19. Consider the photoelectric effect equation: $hf = \Phi + KE_{max}$ .

(a) Identify the physical quantity represented by $\Phi$ .

[1]

(b) Sketch a graph of $KE_{max}$ (y-axis) against frequency $f$ (x-axis). Label the intercepts on both axes.

[3]

(c) Explain the significance of the gradient of this graph.

[1]

20. Ultrasound waves are used in medical imaging.

(a) State the typical frequency range of ultrasound waves.

[1]

(b) Explain why ultrasound is preferred over X-rays for imaging a fetus.

[2]

(c) Calculate the wavelength of an ultrasound wave with frequency 2.0 MHz traveling through soft tissue where the speed of sound is 1540 m s⁻¹.

Answer: __________________________ m

[2]

Answers

A-Level Physics H2 Quiz - Waves Sound Light (Answer Key)

1. C
[1]
Reasoning: Stationary waves are formed by the superposition of two progressive waves of the same frequency traveling in opposite directions.

2. D
[1]
Reasoning: Fringe separation $x = \frac{\lambda D}{a}$ .
New $D' = 2D$ , new $a' = 0.5a$ .
$x' = \frac{\lambda (2D)}{0.5a} = 4 \frac{\lambda D}{a} = 4x$ .

3. 0.68 m
[1]
Reasoning: $\lambda = \frac{v}{f} = \frac{340}{500} = 0.68 \text{ m}$ .

4. They must have a constant phase difference.
[1]
Note: "Constant phase difference" is the key phrase. Same frequency is usually implied or accepted as part of the condition.

5. $2.0 \times 10^{-6}$ m
[1]
Reasoning: $d = \frac{1}{N} = \frac{1}{500 \times 10^3 \text{ lines/m}} = \frac{1}{5 \times 10^5} = 2.0 \times 10^{-6} \text{ m}$ .

6. C
[1]
Reasoning: Intensity is proportional to the number of photons. More photons mean more photoelectrons (current). Kinetic energy depends on frequency ( $hf$ ).

7. The minimum frequency of incident radiation required to eject electrons from the surface of a metal.
[1]
Note: Must mention "minimum frequency" and "eject electrons/emission".

8. 340 Hz
[1]
Reasoning: For a pipe closed at one end, fundamental $\lambda = 4L$ .
$\lambda = 4 \times 0.25 = 1.0 \text{ m}$ .
$f = \frac{v}{\lambda} = \frac{340}{1.0} = 340 \text{ Hz}$ .

9. Shorter wavelengths (blue/violet) are scattered more strongly by atmospheric particles (Rayleigh scattering) than longer wavelengths (red).
[2]
Note: 1 mark for "blue scatters more/shorter wavelength scatters more", 1 mark for reference to atmospheric particles/scattering.

10. A central bright fringe (maximum) which is wider and brighter than the secondary fringes. Secondary fringes are less intense and decrease in intensity further from the center.
[2]
Note: 1 mark for central maximum description (wider/brighter), 1 mark for secondary fringes (dimmer/narrower).

11.
(a) When two or more waves meet at a point, the resultant displacement is the vector sum of the individual displacements.
[2]
Note: 1 mark for "meet/superpose", 1 mark for "vector sum of displacements".

(b)
(i) $\lambda = 0.50 \text{ m}$ . For the first maximum (order $n=1$ ), path difference $= n\lambda = 1 \times 0.50 = 0.50 \text{ m}$ .
[1]

(ii) Constructive interference occurs when the path difference is a whole number of wavelengths ( $n\lambda$ ). The waves arrive in phase.
[2]
Note: 1 mark for "path difference is integer wavelength/in phase", 1 mark for "constructive interference".

(c) Frequency doubles $\rightarrow$ wavelength halves ( $\lambda' = \lambda/2$ ).
Fringe separation $x \propto \lambda$ .
Therefore, the separation between maxima decreases (halves).
[3]
Note: 1 mark for linking freq to wavelength, 1 mark for linking wavelength to separation, 1 mark for correct conclusion (decreases).

12.
(a)
(i) Grating equation: $d \sin \theta = n \lambda$ .
$n=2, \lambda = 550 \times 10^{-9} \text{ m}, \theta = 30^\circ$ .
$d \sin 30^\circ = 2 \times 550 \times 10^{-9}$
$d \times 0.5 = 1100 \times 10^{-9}$
$d = 2200 \times 10^{-9} = 2.2 \times 10^{-6} \text{ m}$ .
[3]
Note: 1 mark for formula, 1 mark for substitution, 1 mark for answer.

(ii) Lines per mm $= \frac{1}{d \text{ (in mm)}} = \frac{1}{2.2 \times 10^{-3} \text{ mm}} \approx 455$ lines mm⁻¹.
[1]

(b) Max order when $\sin \theta \le 1$ .
$n = \frac{d}{\lambda} = \frac{2.2 \times 10^{-6}}{550 \times 10^{-9}} = 4$ .
Answer: 4
[2]
Note: 1 mark for calculation, 1 mark for correct integer.

(c) White light contains a range of wavelengths.
From $d \sin \theta = n \lambda$ , $\sin \theta \propto \lambda$ .
Different wavelengths are diffracted at different angles.
Zero order ( $n=0$ ) has $\theta=0$ for all $\lambda$ , so it remains white.
Higher orders spread out into spectra.
[3]
Note: 1 mark for range of wavelengths, 1 mark for angle depends on wavelength, 1 mark for zero order exception.

13.
(a)
(i) $KE = eV = 1.60 \times 10^{-19} \times 30 \times 10^3 = 4.8 \times 10^{-15} \text{ J}$ .
[2]

(ii) $KE_{max} = \frac{hc}{\lambda_{min}}$ .
$\lambda_{min} = \frac{hc}{KE} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{4.8 \times 10^{-15}}$
$\lambda_{min} \approx 4.14 \times 10^{-11} \text{ m}$ .
[2]

(b) Electrons collide with target atoms and are decelerated.
The loss in kinetic energy is converted into photon energy.
Electrons lose varying amounts of energy (from zero to max KE), producing photons of varying energies/wavelengths.
[3]
Note: 1 mark for deceleration/collision, 1 mark for KE to photon energy, 1 mark for range of energy loss.

14.
(a) $f_0 = \frac{\Phi}{h} = \frac{6.9 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 1.04 \times 10^{15} \text{ Hz}$ .
[2]

(b)
(i) $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{250 \times 10^{-9}} \approx 7.96 \times 10^{-19} \text{ J}$ .
This is approximately $7.9 \times 10^{-19} \text{ J}$ .
[2]

(ii) $KE_{max} = E - \Phi = 7.96 \times 10^{-19} - 6.9 \times 10^{-19} = 1.06 \times 10^{-19} \text{ J}$ .
Answer: $1.1 \times 10^{-19} \text{ J}$ .
[2]

(c)
(i) No change.
[1]

(ii) Increases.
[1]

(iii) Intensity is proportional to the number of photons incident per unit area per second.
More photons mean more interactions with electrons, so more photoelectrons are emitted per second.
Current is charge per second, so current increases.
[3]

(d) It depends on the magnitude of the new work function.
If the new work function is greater than the photon energy ( $7.96 \times 10^{-19} \text{ J}$ ), no emission occurs.
If the new work function is less than the photon energy, emission occurs.
Since the work function is "higher", it is possible it exceeds the photon energy, in which case emission stops.
[3]
Note: 1 mark for comparing photon energy and work function, 1 mark for condition of no emission, 1 mark for condition of emission.

15.
(a) For fundamental mode, $L = \lambda/2$ .
$\lambda = 2L = 2 \times 1.2 = 2.4 \text{ m}$ .
[1]

(b) $v = f \lambda = 50 \times 2.4 = 120 \text{ m s}^{-1}$ .
[1]

(c) Sketch should show two loops.
Nodes at both ends and in the center. Antinodes in the middle of each loop.
[2]
Note: 1 mark for correct shape (2 loops), 1 mark for correct labeling of N and A.

16.
(a) $\sin C = \frac{1}{n} = \frac{1}{1.5}$ .
$C = \sin^{-1}(0.666...) \approx 41.8^\circ$ .
[2]

(b) Total internal reflection occurs. The light is reflected back into the glass.
[1]

(c) Total internal reflection allows light to travel long distances with minimal loss of intensity (no refraction out of the core).
[2]
Note: 1 mark for minimal loss/no leakage, 1 mark for confinement within the fiber.

17.
(a) Constructive interference occurs when two waves meet in phase, resulting in a resultant displacement of maximum amplitude.
[2]

(b) Loud regions correspond to constructive interference (path difference = $n\lambda$ ).
Quiet regions correspond to destructive interference (path difference = $(n+0.5)\lambda$ ).
As the student moves, the path difference changes, alternating between these conditions.
[2]
Note: 1 mark for identifying constructive/destructive causes, 1 mark for changing path difference.

(c) Distance between loud regions decreases.
Higher frequency means shorter wavelength ( $\lambda = v/f$ ).
Fringe separation is proportional to wavelength.
[2]

18.
(a) Sharper/maxima are narrower / Better resolution / Higher dispersion.
[1]

(b) $d = \frac{1}{600 \times 10^3} = 1.67 \times 10^{-6} \text{ m}$ .
$d \sin \theta = n \lambda \Rightarrow \lambda = d \sin 15^\circ / 1$ .
$\lambda = 1.67 \times 10^{-6} \times 0.2588 \approx 4.32 \times 10^{-7} \text{ m}$ .
[3]

(c) Angular separation $\theta$ increases with order $n$ for a given $\Delta \lambda$ .
Or, dispersion is greater at higher orders.
[2]

19.
(a) Work function.
[1]

(b) Straight line with positive gradient.
X-intercept at $f_0$ (threshold frequency).
Y-intercept at $-\Phi$ .
[3]
Note: 1 mark for straight line, 1 mark for correct x-intercept, 1 mark for correct y-intercept.

(c) The gradient is equal to Planck's constant, $h$ .
[1]

20.
(a) Greater than 20 kHz.
[1]

(b) Ultrasound is non-ionizing and does not damage living tissue/DNA. X-rays are ionizing and can cause cell damage/mutation.
[2]

(c) $\lambda = \frac{v}{f} = \frac{1540}{2.0 \times 10^6} = 7.7 \times 10^{-4} \text{ m}$ .
[2]