A Level H2 Physics Waves Sound Light Quiz

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]