Electromagnetic spectrum

1. Use λ = c/f: the lower frequency has wavelength 6.0×10⁻⁷ m and the higher 3.0×10⁻⁷ m.
2. Ephoton = hf gives 3.313×10⁻¹⁹ and 6.626×10⁻¹⁹ J. Dividing equal total pulse energy by these values gives 2.0×10¹³ and 1.0×10¹³ photons.
3. 600 nm is visible light and 300 nm is ultraviolet. Both propagate at c in vacuum; the higher-frequency pulse contains fewer photons for the same total energy.
4. Spatial interference patterns follow phase and superposition, which a simple stream of classical particles does not explain.
5. Localized detector events with frequency-dependent energy are described by photons. These are complementary descriptions of quantum light rather than literal tiny balls riding classical waves.
6. Absorption, penetration, dose and photon energy depend on wavelength and material. Ultraviolet can cause photochemical damage; exposure assessment cannot be inferred from vacuum speed alone.

Answer (a) 600 nm, 3.313×10⁻¹⁹ J, 2.0×10¹³ photons; 300 nm, 6.626×10⁻¹⁹ J, 1.0×10¹³ photons. (b) Different models describe different observations; common vacuum speed does not imply identical interaction or hazard.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. Snell's law gives sin r = sin30°/n. The blue/violet value is asin(0.5/1.54)≈18.9° and the red value is asin(0.5/1.50)≈19.5°.
2. Speeds are c/n, about 1.95×10⁸ and 2.0×10⁸ m s⁻¹. Higher index gives greater bending toward the normal on entry.
3. Frequency is unchanged at entry, so λglass = λvacuum/n, giving about 260 and 467 nm. Dispersion separates directions because n depends on wavelength.
4. A second appropriately oriented prism or lens arrangement can recombine the separated spectral components into the original white-light mixture.
5. The first prism separates pre-existing frequencies; it does not manufacture colours or convert red photons into blue ones.
6. Coloured pigments absorb and scatter different bands, so pigment mixing is subtractive and does not follow the same colour rules as adding light beams.

Answer (a) 400 nm: r≈18.9°, v≈1.95×10⁸ m s⁻¹, λglass≈260 nm; 700 nm: r≈19.5°, v=2.0×10⁸ m s⁻¹, λglass≈467 nm. (b) Light addition combines spectral components; paints remove selected components through absorption.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. An ideal polariser transmits half the average intensity of unpolarised light, giving 80/2 = 40 W m⁻².
2. The middle axis at 45° transmits 40cos²45° = 20 W m⁻² and defines a new output polarisation direction.
3. The last axis is another 45° from the middle, so it transmits 20cos²45° = 10 W m⁻². Without the middle element, the crossed pair transmits zero in the ideal model.
4. A transverse oscillation has a selectable direction perpendicular to propagation. A simple longitudinal wave has no equivalent independent transverse axis to filter.
5. The evidence supports the transverse nature of light, not every microscopic detail of a classical-wave picture.
6. Reflected glare may be only partially polarised, and real filters leak light and absorb other components. Measured minima therefore need not be exactly zero.

Answer (a) First output = 40 W m⁻²; with the middle polariser, final output = 10 W m⁻². (b) Polarisation supports transverse electric-field oscillations; real glare and real filters need not reach complete extinction.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. The half-value thickness is 0.01 m, so μ = ln2/0.01 ≈69.3 m⁻¹. Four such thicknesses transmit (1/2)⁴ = 1/16.
2. The greatest photon energy is the full electron energy eV = (1.60×10⁻¹⁹)(60000) = 9.60×10⁻¹⁵ J.
3. Using E = hc/λ gives λmin = hc/(eV) ≈2.07×10⁻¹¹ m. Most emitted photons need not have this maximum energy.
4. Different tissue thicknesses and attenuation coefficients give different transmitted intensities at the detector, creating contrast.
5. If detected photons are independently counted, relative random counting uncertainty scales approximately as 1/√N; increasing N can improve statistical image quality.
6. X-rays are ionising, so increased photon fluence can increase patient dose. The exposure should be justified by the diagnostic task, with collimation, shielding and detector efficiency considered; ultrasound may answer some questions without ionising radiation.

Answer (a) μ≈69.3 m⁻¹; transmission at 4 cm = 1/16 = 6.25%; λmin≈2.07×10⁻¹¹ m. (b) Different attenuation gives contrast; more incident photons can reduce relative counting noise but also increases exposure.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.