HSC Exam Practice

Sample response. Intensity is the power of electromagnetic radiation delivered per unit area perpendicular to the direction of propagation; SI unit is watts per square metre (W/m²). For a point source of power P radiating uniformly in all directions, the intensity at distance r is given by I = P/(4πr²).

Marking notes. 1 mark for “power per unit area” (or equivalent correct definition); 1 mark for correct SI unit (W/m²); 1 mark for the correct inverse square law formula I = P/(4πr²).

Sample response. Electromagnetic waves are transverse because the oscillating electric field (E) and magnetic field (B) are both perpendicular to the direction of wave propagation. The two fields are also perpendicular to each other. Since the disturbance (the field oscillation) is at right angles to the direction of energy travel, the wave satisfies the definition of a transverse wave.

Marking notes. 1 mark for stating that the oscillations (fields) are perpendicular to the direction of propagation; 1 mark for noting that E and B are also perpendicular to each other; 1 mark for linking this explicitly to the definition of a transverse wave (oscillation at right angles to energy transfer direction).

Sample response. (i) Radio waves from a broadcast antenna: produced by accelerated (oscillating) electrons in the antenna circuit, which creates a time-varying electric field that propagates outward. (ii) Infrared from a heated metal plate: produced by the thermal vibration of atoms in the lattice; higher temperature means more energetic vibrations and more intense IR emission. (iii) Gamma rays from a radioactive nucleus: produced by transitions between nuclear energy levels during radioactive decay, releasing high-energy photons.

Marking notes. 1 mark per correctly identified mechanism (3 total). Accept “electron transitions in antenna” for (i), “molecular/atomic vibration/thermal motion” for (ii), “nuclear decay/nuclear transitions” for (iii).

Sample response. Reflection occurs when an EM wave strikes a boundary and bounces back into the original medium. The angle of incidence equals the angle of reflection; frequency, wavelength, and speed all remain constant. Refraction occurs when an EM wave crosses a boundary between two media with different refractive indices, changing direction. Frequency remains constant (determined by the source), but wavelength and wave speed change as the wave moves into the new medium.

Marking notes. Reflection: 1 mark for “bounces back, angle in = angle out, no property changes”; 1 mark for explicitly stating what does not change (frequency, wavelength, speed). Refraction: 1 mark for “direction changes at boundary into new medium”; 1 mark for correctly identifying what changes (wavelength, speed) vs stays constant (frequency).

Sample response. The student is incorrect because the inverse square law states that intensity is proportional to 1/r², not 1/r. Moving three times further away means the distance factor is 3, so the intensity factor is 1/3² = 1/9. The intensity is reduced to one-ninth of its original value, not one-third. The inverse square law: I⊂2;/I⊂1; = (r⊂1;/r⊂2;)² = (1/3)² = 1/9.

Marking notes. 1 mark for identifying the error (confusing 1/r with 1/r²); 1 mark for stating the correct factor (one-ninth / ×1/9); 1 mark for citing the inverse square law and showing or writing the correct relationship.

Sample response. Glass is transparent to visible light because the photon energies of visible wavelengths do not match the energy gaps of the electron transitions in glass (silicon dioxide), so they are not absorbed and pass through. Ultraviolet photons have higher frequencies and therefore higher energies (by E = hf). These energies match electronic transition energies in the glass structure, causing efficient absorption — the UV energy is transferred to the glass rather than transmitted. This is also why UV-blocking sunscreens work by absorbing high-energy UV photons in a similar way.

Marking notes. 1 mark for explaining transparency using the absence of matching electronic transitions for visible photons; 1 mark for explaining UV absorption using higher photon energy matching electronic transitions (must reference frequency/energy relationship); 1 mark for correctly applying the concept of absorption (energy transferred to medium).

Sample response (a). I = P/(4πr²) = 200/(4π × 2.0²) = 200/(4π × 4) = 200/50.27 ≈ 3.98 W/m² ≈ 4.0 W/m². The calculated value of ~4.0 W/m² agrees with the measured value to within rounding error (<0.5%), consistent with the inverse square law model for a point source in free space.

Marking notes (a). 1 mark for correct substitution; 1 mark for I ≈ 4.0 W/m² (accept 3.98–4.0); 1 mark for commenting that measured and calculated agree (within rounding).

Sample response (b). Using the ratio method: I⊂1;/I⊂2; = r⊂2;²/r⊂1;². Let I⊂1; = 4.0 W/m² at r⊂1; = 2.0 m, and I⊂2; = 50 W/m² (we seek r⊂2;). Then r⊂2;² = r⊂1;² × I⊂1;/I⊂2; = 4.0 × (4.0/50) = 4.0 × 0.08 = 0.32 m². So r⊂2; = √0.32 ≈ 0.57 m. The technician must remain more than approximately 0.57 m from the transmitter to stay below 50 W/m².

Marking notes (b). 1 mark for correctly rearranging the ratio formula to solve for r⊂2;; 1 mark for correct substitution; 1 mark for r⊂2; ≈ 0.57 m with units.

Sample response (c). Assumption: the transmitter radiates uniformly in all directions (isotropically) in free space with no obstacles between source and measurement point. On a real rooftop, metal structures and surrounding walls reflect microwaves, creating multipath propagation that could increase or decrease the local intensity compared with the free-space prediction; in some locations the reflected waves constructively interfere, raising intensity above the calculated value; in others they destructively interfere, lowering it.

Marking notes (c). 1 mark for a valid stated assumption (isotropic / free space / point source); 1 mark for explaining that reflections from rooftop structures create multipath, leading to higher or lower actual intensity than the model predicts.

Sample response. The claim that all electromagnetic waves behave identically when interacting with matter is incorrect. While all EM waves share fundamental properties — they are all transverse, travel at c in vacuum, and obey the inverse square law from a point source — their interactions with matter are strongly frequency-dependent. The three key interaction processes are absorption, reflection, and refraction, and each varies dramatically across the spectrum. Absorption depends on whether the photon energy matches an energy transition in the material. Visible light photons have energies that do not match electronic transitions in glass, so glass transmits visible light but absorbs UV (higher energy, matching electron transitions in the Si–O bonds). This is exploited in Australian building standards: window glass in homes blocks UV to reduce skin cancer risk, a significant public health consideration given Australia’s high UV index. In contrast, glass is transparent to radio waves, which is why mobile phone signals penetrate buildings. Reflection also varies with frequency. Metals reflect radio waves and microwaves very efficiently (used in dish antennas and parabolic radar reflectors), but the same metals absorb visible light (appearing metallic grey) rather than reflecting it perfectly at optical frequencies. The Parkes Radio Telescope (“The Dish”) uses a metal surface that is highly reflective at centimetre radio wavelengths to focus weak astronomical signals. Refraction is also frequency-dependent: glass has a higher refractive index for blue/violet light than for red (dispersion), causing rainbows and chromatic aberration in lenses. Radio waves refract in the ionosphere at certain frequencies, enabling long-distance AM radio broadcasting in Australia across the outback, while higher-frequency FM signals travel in straight lines and cannot be received beyond the horizon. In summary, the claim is demonstrably incorrect. Frequency determines photon energy (E = hf), and this in turn governs which energy levels in matter can be excited, making absorption highly wavelength-selective. Reflection and refraction coefficients at interfaces are also frequency-dependent. These differences are not incidental but are the physical basis for technologies ranging from UV-blocking glass in Australian homes to radio astronomy at Parkes and ionospheric radio propagation across the continent.

Marking criteria (7 marks).

1. [1] Identifies that all EM waves share common properties (transverse, travel at c, obey inverse square law) but correctly states the claim is still wrong because interactions with matter are frequency-dependent.
2. [1] Explains frequency-dependent absorption with a specific mechanism (photon energy matching energy transitions) and a named example.
3. [1] Explains frequency-dependent reflection with a named example or application.
4. [1] Explains frequency-dependent refraction (dispersion or wavelength dependence of refractive index) with a named example.
5. [1] First named Australian application correctly described with the specific EM wave property (absorption, reflection, or refraction) that is being exploited.
6. [1] Second named Australian application correctly described with the specific EM wave property.
7. [1] Explicit evaluative conclusion that clearly rejects the claim and synthesises the argument: differences in frequency-dependent interactions are the physical basis for distinct behaviour, not merely quantitative variations on the same phenomenon.