Refraction — Set 5

Correct Answer: C. refraction at the water-air surface

• **refraction at the water-air surface** = light rays from the submerged part of the pencil bend away from the normal as they exit water into air; the observer's eye traces these rays back in straight lines and perceives the pencil tip at a shallower, displaced position. • **Apparent shift** — the apparent depth effect raises the submerged end of the pencil to an apparently higher position, so the pencil looks kinked or bent at the water surface. • 💡 Wrong-option analysis: reflection from the pencil only: reflection creates an image above the surface; it cannot displace the image of the submerged part sideways to make the pencil look bent; polarization by water: polarization at the water surface (Brewster angle) affects the state of polarisation of reflected light — it does not shift the apparent position of a submerged object; diffraction around the pencil: diffraction occurs around sharp edges and through slits at very small scales — it does not produce the large-scale apparent bending of the pencil observed in everyday life.

Correct Answer: D. n = sin i / sin r

• **n = sin i / sin r** = from Snell's law for air (n = 1) to the medium: 1 × sin i = n × sin r; rearranging gives n = sin i / sin r. • **Direct measurement** — by measuring the angle of incidence and angle of refraction for a ray entering the medium from air, the refractive index can be calculated using this ratio. • 💡 Wrong-option analysis: n = cos i / cos r: using cosines has no basis in Snell's law — the law involves sines, not cosines; this ratio is not the refractive index; n = r/i: dividing angles in degrees rather than their sines is not Snell's law; it gives the wrong value except coincidentally near normal incidence; n = i/r: this inverts r/i but is still wrong for the same reason — Snell's law uses sin i / sin r, not i/r.

Correct Answer: C. 18.75 cm

• **18.75 cm** = use n = real depth / apparent depth; real depth = n × apparent depth = 1.25 × 15 = 18.75 cm. • **n = 1.25** — the coin appears shallower than it really is; 15 cm apparent × 1.25 = 18.75 cm real; the object is actually 3.75 cm deeper than it looks. • 💡 Wrong-option analysis: 12 cm: 12 = 15 / 1.25, which uses apparent = real/n — this would give apparent depth from real, not the other way round; 30 cm: 30 = 15 × 2, which uses n = 2.0 instead of 1.25; 15 cm: 15 cm is the apparent depth given in the question — the real depth must be greater than the apparent depth, so 15 cm cannot be the answer.

Correct Answer: B. 53 degrees

• **53 degrees** = use sin c = 1/n = 1/1.25 = 0.8; c = arcsin(0.8) ≈ 53°. • **sin c = 0.8** — a 3-4-5 triangle gives sin 53° = 0.8 exactly; this is a commonly used angle in optics problems; beyond 53°, total internal reflection occurs in this medium. • 💡 Wrong-option analysis: 30 degrees: sin 30° = 0.5 would require n = 1/0.5 = 2.0 — much denser than this medium (n = 1.25); 42 degrees: sin 42° ≈ 0.67, corresponding to n ≈ 1.5 (common glass) — not n = 1.25; 48 degrees: sin 48° ≈ 0.74, corresponding to n ≈ 1.35 — close to water but not 1.25.

Correct Answer: A. r1 = r2 = A/2

• **r1 = r2 = A/2** = at minimum deviation the ray passes symmetrically through the prism; because the path is symmetric, r1 = r2; and since r1 + r2 = A (geometry of the prism), each internal angle equals A/2. • **Minimum deviation formula** — with r1 = r2 = A/2 and i = i2 at minimum deviation, applying Snell's law gives n = sin((A + Dm)/2) / sin(A/2). • 💡 Wrong-option analysis: r1 + r2 = 0: r1 + r2 = 0 would require both internal angles to be zero, meaning no refraction inside the prism — physically impossible for a non-zero prism angle; Dm = 0 always: Dm = 0 would mean no deviation at all — this only happens for a 'prism' with A = 0 (a flat slab), not for a real prism; i = 0 always: i = 0 means normal incidence — for a real prism with A > 0, normal incidence does not in general produce minimum deviation; i at minimum deviation is a specific oblique angle.

Correct Answer: C. n = sin((A+Dm)/2)/sin(A/2)

• **n = sin((A+Dm)/2)/sin(A/2)** = derived from Snell's law at both faces of the prism under the minimum deviation condition (i1 = i2, r1 = r2 = A/2); the incidence angle at minimum deviation is i = (A + Dm)/2. • **Derivation** — at minimum deviation: i = (A + Dm)/2 and r = A/2; applying Snell's law: sin i / sin r = n, which gives n = sin((A + Dm)/2) / sin(A/2). • 💡 Wrong-option analysis: n = (A+Dm)/A: this uses angle values directly instead of their sines — angles in degrees cannot be used directly in the refractive index formula; n = sin(A/2)/sin((A+Dm)/2): this inverts the correct formula — numerator and denominator are swapped, giving n < 1 for most prisms, which is incorrect; n = A/Dm: a simple ratio of prism angle to deviation angle has no basis in Snell's law — it would be dimensionally correct but numerically meaningless.

Correct Answer: C. 1.53

• **1.53** = use n = sin((A+Dm)/2)/sin(A/2) = sin((60+40)/2)/sin(60/2) = sin 50°/sin 30° = 0.766/0.500 ≈ 1.53. • **sin 50°/sin 30°** — (60+40)/2 = 50°, A/2 = 30°; sin 50° ≈ 0.766 and sin 30° = 0.500; the ratio ≈ 1.53 is typical for dense flint glass. • 💡 Wrong-option analysis: 2.00: sin((A+Dm)/2)/sin(A/2) = 2.00 would require sin i = 2 sin r, giving sin 50° = 2 sin 30° = 1.0 — but sin 50° ≈ 0.766 ≠ 1.0; 1.33: n = 1.33 corresponds approximately to water; glass prisms are denser than water; 1.20: n = 1.20 would give sin 50° / sin r = 1.20, so sin r = 0.766/1.20 = 0.638, r = 39.7° — but r must equal A/2 = 30° at minimum deviation, so 1.20 is inconsistent.

Correct Answer: B. angle of incidence is 0 degrees

• **angle of incidence is 0 degrees** = at normal incidence (i = 0°), the ray enters perpendicular to both faces with no bending; there is no refraction angle difference, so (i − r) = 0 and the lateral displacement formula d = t sin(i−r)/cos r gives d = 0. • **Zero lateral shift** — when a ray hits the slab at 90° to the surface it passes straight through; the emergent ray coincides with the incident ray, giving zero lateral displacement. • 💡 Wrong-option analysis: refractive index is very large: very large n makes r very small, so (i − r) ≈ i; the displacement d = t sin(i−r)/cos r ≈ t sin i/cos 0 = t sin i — it does not go to zero; faces are rough: rough faces increase scattering and diffuse reflection but do not make lateral displacement zero — they just make the slab non-transparent; angle of incidence is 90 degrees: at i = 90° the ray just grazes the surface and does not enter the slab — no transmitted ray, so the concept of lateral displacement does not apply.

Correct Answer: B. away from the normal

• **away from the normal** = water is denser than air (n_water > n_air); light speeds up on leaving water, and by Snell's law the refracted ray bends away from the normal (r > i). • **r > i** — n_water sin i = n_air sin r; since n_water > n_air, sin r > sin i and r > i; the ray bends away from the normal towards the surface; if i exceeds the critical angle (~49° for water), TIR occurs. • 💡 Wrong-option analysis: only reflects and never refracts: total internal reflection only occurs if i > critical angle; for i below the critical angle, refraction does occur and the ray passes into air; along the normal always: traveling along the normal means r = 0°, which only happens at i = 0° (normal incidence), not generally; towards the normal: bending towards the normal occurs when entering a denser medium — water to air is going from denser to rarer, so bending is away.

Correct Answer: C. Apparent depth is less than real depth

• **Apparent depth is less than real depth** = when an observer in air looks into water, refraction bends rays away from the normal on exit; the backward extension of these rays meets at a shallower point, so the object appears closer to the surface (apparent depth = real depth/n < real depth). • **Formula** — apparent depth = real depth / n; for water (n = 1.33), a 2 m deep object appears at 1.5 m — 0.5 m shallower than reality. • 💡 Wrong-option analysis: Apparent depth equals real depth always: equal depths would require n = 1 (no refraction), meaning the medium is the same as air — not true for water; Apparent depth is greater than real depth: apparent depth is greater than real depth only when looking from the denser medium into the rarer one (e.g. a fish looking up at an object above water); Apparent depth is independent of refractive index: the formula apparent depth = real depth/n shows direct dependence on n — doubling n halves the apparent depth.