Refraction — Set 4

Correct Answer: D. sound speed with temperature and wind

• **sound speed with temperature and wind** = the speed of sound in air increases with temperature (v ≈ 331 + 0.6T m/s) and varies with wind; vertical gradients in temperature and wind create gradients in sound speed, causing sound wavefronts to curve (refract). • **Day vs night** — during the day, ground air is hotter so sound refracts upward (bending away from ground); at night, ground air is cooler so sound refracts downward, allowing distant sounds to be heard more clearly. • 💡 Wrong-option analysis: sound amplitude with time: amplitude relates to loudness and decreases with distance due to spreading and absorption — it is not the cause of refraction; sound frequency with distance: frequency of a sound wave does not change as it propagates (no Doppler if source and observer are stationary) — frequency change is not the cause of refraction; sound wavelength becoming zero: wavelength becoming zero would require infinite frequency, a non-physical condition that does not describe atmospheric refraction.

Correct Answer: D. refract towards the ground

• **refract towards the ground** = at night, ground-level air is often cooler than air above; sound speed is lower near the ground, so wavefronts curve downward; sound is channeled along the ground and reaches distant listeners more efficiently. • **Temperature inversion** — the temperature inversion at night (warm air above, cool air below) creates a natural acoustic duct that bends sound downward, increasing its range. • 💡 Wrong-option analysis: travel in vacuum: sound cannot travel in vacuum — it requires a medium; this option is physically impossible; get absorbed less by air: absorption by air is slightly less at night (lower humidity sometimes), but this is a minor effect and not the main reason for increased audibility at night; stop reflecting from ground: sound still reflects from the ground at night; ground reflection does not explain the additional clarity — the key is downward refraction channeling more energy towards the listener.

Correct Answer: A. total internal reflection

• **total internal reflection** = the core of an optical fiber has higher n than the cladding; light hitting the core-cladding boundary at angles greater than the critical angle undergoes total internal reflection and is trapped inside the core. • **Step-index fiber** — n_core > n_cladding ensures a critical angle at the interface; each reflection loses virtually no energy, so light signals travel thousands of kilometres with low loss in modern single-mode fibers. • 💡 Wrong-option analysis: scattering only: scattering in the glass causes loss (attenuation) and is a problem to be minimised, not the guiding mechanism; photoelectric emission: the photoelectric effect involves electrons being ejected by photons — it is an effect at metal surfaces, not the mechanism that guides light inside fibers; diffraction only: diffraction spreads light and would cause energy to leak out of a fiber, not guide it — it is a loss mechanism, not a guiding one.

Correct Answer: D. endoscope

• **endoscope** = an endoscope uses coherent bundles of optical fibers to carry illuminating light to the inside of a body cavity and relay the image back to the doctor's eye or camera. • **Coherent bundle** — in an endoscope the fiber bundle is 'coherent' (fibers maintain their relative spatial arrangement) so each fiber carries one pixel of the image, faithfully reconstructing the picture at the output end. • 💡 Wrong-option analysis: battery: a battery converts chemical energy to electrical energy — it has no mechanism to transmit light or images; thermometer: a thermometer measures temperature using thermal expansion or electrical resistance — optical fibers are not its working principle; barometer: a barometer measures atmospheric pressure using liquid column height or mechanical deflection — it has nothing to do with light transmission.

Correct Answer: A. 1

• **1** = by definition, the absolute refractive index is n = c/v; in vacuum v = c, so n = c/c = 1; vacuum is the reference standard for all absolute refractive index measurements. • **Air ≈ 1.0003** — the refractive index of air at standard conditions is only slightly above 1 (≈ 1.0003); for most practical calculations air is treated as n = 1, like vacuum. • 💡 Wrong-option analysis: less than 1 always: no physical medium has n < 1 for real (non-evanescent) light at normal frequencies — vacuum has the minimum value of exactly 1; infinite: n = ∞ would mean v = 0, i.e. light stands still in vacuum — physically impossible; 0: n = 0 would mean v = c/0 = ∞, i.e. light travels infinitely fast — violating special relativity.

Correct Answer: B. n2/n1

• **n2/n1** = relative refractive index n21 = n2/n1; it equals v1/v2 (ratio of speeds), which is also the ratio of absolute indices of the two media. • **Direction matters** — n21 (medium 2 w.r.t. medium 1) = n2/n1; the reciprocal n12 = n1/n2 gives medium 1 w.r.t. medium 2; these two are always reciprocals of each other. • 💡 Wrong-option analysis: n1×n2: the product of two absolute indices has no standard physical meaning as a relative index — it is not used in Snell's law; n1 - n2: the difference of indices also has no physical meaning as a relative index — Snell's law uses their ratio; n1 + n2: adding absolute indices is dimensionally consistent but physically meaningless — again, the ratio n2/n1 is what appears in Snell's law.

Correct Answer: B. 0.75

• **0.75** = from n = c/v, we get v/c = 1/n = 1/1.33 ≈ 0.75; light travels at about 75% of its vacuum speed in this medium (approximately water, n = 1.33). • **75% of c** — in water, light speed ≈ 2.26×10^8 m/s, which is 75% of 3×10^8 m/s; this is why light slows noticeably in water. • 💡 Wrong-option analysis: 1.33: v/c = 1.33 would mean v > c, which is impossible for light in a medium; 1.33 is the refractive index, not the speed ratio; 1.00: v/c = 1.00 would mean light travels at c in the medium, implying n = 1 (vacuum) — but n = 1.33 ≠ 1; 0.33: v/c = 0.33 would require n = 1/0.33 ≈ 3.0 — far denser than water; 0.33 is not 1/1.33.

Correct Answer: C. 90 degrees with the normal

• **90 degrees with the normal** = at the critical angle, Snell's law gives n_denser sin c = n_rarer sin 90° = n_rarer; the refracted ray grazes the boundary, making 90° with the normal (i.e., it travels along the surface). • **Grazing refraction** — the refracted ray lies exactly in the plane of the boundary; for angles of incidence even slightly larger, TIR occurs; this is the limiting case between refraction and total internal reflection. • 💡 Wrong-option analysis: 0 degrees with the boundary: 0° with the boundary is the same as 90° with the normal — this is actually the correct physical situation; but option C states '90 degrees with the normal' which is the standard way to express it; 180 degrees with the normal: 180° with the normal means the ray reverses direction — that is reflection, not refraction; 45 degrees with the normal: 45° with the normal would give a specific critical angle only for n = 1/sin45° = √2 ≈ 1.41; it is not a general answer.

Correct Answer: A. total internal reflection occurs

• **total internal reflection occurs** = when i > critical angle, Snell's law requires sin r = (n_denser/n_rarer) sin i > 1, which has no real solution; the light cannot pass into the rarer medium and is entirely reflected back inside the denser medium. • **100% reflection** — unlike partial reflection at normal incidence (~4% for glass-air), total internal reflection reflects 100% of the light energy with no loss; this is why optical fibers can transmit signals over great distances. • 💡 Wrong-option analysis: refraction increases strongly: refraction cannot increase beyond the grazing (90°) condition — once i exceeds the critical angle, refraction ceases entirely and reflection takes over; refraction becomes zero and the ray is absorbed: refraction does become zero (no transmitted ray), but the light is not absorbed — it is reflected; only highly absorbing media absorb light; the refractive index becomes 1 instantly: the refractive index is a property of the material and does not change with the angle of incidence — it remains constant regardless of the incident angle.

Correct Answer: C. towards the normal

• **towards the normal** = water (n ≈ 1.33) is optically denser than air (n ≈ 1.00); light slows on entering water and the refracted ray bends towards the normal (r < i). • **Snell's law** — n_air sin i = n_water sin r; since n_water > n_air, sin r < sin i and hence r < i; the bend is always towards the normal when entering a denser medium. • 💡 Wrong-option analysis: away from the normal: bending away from the normal occurs when going from denser to rarer medium (e.g. water to air) — the opposite direction to this question; without any change: no bending only occurs at i = 0° (normal incidence) or when both media have the same n; for oblique incidence from air to water there is always bending; along the surface: the refracted ray running along the surface corresponds to r = 90°, which only happens at the critical angle — and the critical angle condition only applies for denser-to-rarer travel, not air-to-water.