13.3 Refraction

1. AA person sees their own face in a flat mirror on a bathroom wall.
   A student who uses refraction as another name for reflection picks this. The mirror sends light back into the air it came from; that is reflection. Refraction needs light to pass from one medium into another.
2. BLight spreads out after it passes through a very narrow slit in a card.
   A student who confuses refraction with diffraction picks this. The light is in air on both sides of the slit, so no boundary between two media is crossed; spreading through a narrow opening is diffraction, a wave effect.
3. CLight passes straight through a window that it strikes head-on.
   A student who thinks any light passing through a transparent material is refracted picks this. Light that strikes the glass head-on (along the normal) does not change direction, so it is transmitted without being refracted.
4. DA straw in a glass of water looks bent at the water's surface. Correct
   Refraction is a change in direction of light as it passes from one medium into another. Light from the part of the straw under water changes direction as it passes from the water into the air at the surface, so that part of the straw appears displaced and the straw looks bent where it meets the surface.

CED 13.3.A.1 · Read this in Fix

1. ALight travels more slowly in plastic P than in plastic Q. Correct
   Refraction is caused by the change in the speed of light at the boundary. With the same angle of incidence, the ray that changes direction more has the smaller angle of refraction, so P has the greater index of refraction, and n = c/v means light travels more slowly in P.
2. BLight travels faster in plastic P than in plastic Q.
   A student who thinks light travels faster in a medium with a greater index picks this. The greater bending shows that P has the greater index, and n = c/v means the speed of light in P is less, not greater.
3. CPlastic P has a larger mass density than plastic Q.
   A student who links refraction to mass density picks this. How much light bends depends on the speed of light in each material, which is not set by mass density; the observation says nothing about which plastic is denser.
4. DLight travels at the same speed in P as it does in Q.
   A student who thinks light travels at c in every material picks this. c is the speed in a vacuum; the different amounts of bending show that light's speed differs in P and Q, and it is slower in P, which bends the ray more.

CED 13.3.A.2 · Read this in Fix

1. ALight travels 1.5 times as fast in the plastic as in a vacuum.
   A student who treats the index as proportional to the speed picks this. The index is inversely proportional to the speed: n = c/v, so the light is slower in the plastic, and no material makes light faster than c.
2. BThe plastic's mass density is 1.5 times the mass density of water.
   A student who identifies the index with mass density picks this. The index compares speeds of light, n = c/v; it is not a ratio of densities. Olive oil, for example, is less dense than water but has a greater index.
3. CA ray entering from air is bent to 1/1.5 of its angle of incidence.
   A student who applies Snell's law to the angles themselves picks this. Snell's law relates the sines: sin θ₂ = (sin θ₁)/1.5 for a ray entering from air, and the angles are not in that ratio.
4. DLight travels 1.5 times as fast in a vacuum as it does in the plastic. Correct
   n = c/v, so n = 1.5 means c = 1.5v: light in a vacuum travels 1.5 times as fast as light in the plastic, and v = c/1.5 = 2.0 × 10⁸ m/s.

Working n = c/v → v = c/n = (3.0 × 10⁸ m/s)/1.5 = 2.0 × 10⁸ m/s; c/v = 1.5.

CED 13.3.A.3 · Read this in Fix

1. A25.4° Correct
   The 50.0° angle in the diagram is measured from the surface, so the angle of incidence (from the normal) is 90.0° − 50.0° = 40.0°. Snell's law: (1.00) sin 40.0° = 1.50 sin θ₂, so sin θ₂ = 0.643/1.50 = 0.429 and θ₂ = 25.4°.
2. B26.7°
   A student who applies Snell's law to the angles themselves picks this: 40.0°/1.50 = 26.7°. Snell's law relates the sines: sin θ₂ = sin 40.0°/1.50, which gives 25.4°.
3. C74.6°
   A student who pairs each index with the angle in the other medium picks this: sin θ₂ = 1.50 sin 40.0° = 0.964. That would bend the ray away from the normal on entering glass; the index 1.50 belongs with the angle in the glass.
4. D30.7°
   A student who uses the 50.0° angle measured from the surface as the angle of incidence picks this: sin θ₂ = sin 50.0°/1.50. Angles in Snell's law are measured from the normal, so θ₁ = 40.0°.

Working Angle between the ray and the surface = 50.0°, so θ₁ = 90.0° − 50.0° = 40.0° from the normal. n_air sin θ₁ = n_glass sin θ₂: sin θ₂ = (1.00)(sin 40.0°)/1.50 = 0.6428/1.50 = 0.4285, θ₂ = 25.4°. (Angles: 40.0/1.50 = 26.7°; swapped: sin θ₂ = 1.50 × 0.6428 = 0.964, 74.6°; surface angle: sin θ₂ = 0.7660/1.50 = 0.511, 30.7°.)

CED 13.3.A.4 · Read this in Fix

1. AA refracted ray travels along the boundary, at 90° to the normal. Correct
   At the critical angle Snell's law gives n_water sin θ_critical = n_air sin θ₂ = n_air, so sin θ₂ = 1 and θ₂ = 90°: the refracted ray leaves at 90° to the normal and travels along the surface.
2. BA refracted ray leaves along the normal, at 90° to the boundary.
   A student who measures angles from the surface picks this, reading 'refracts at 90°' as 90° to the boundary. The angle of refraction is measured from the normal, so 90° means along the boundary.
3. CAll of the light reflects back, and no refracted ray exists at all.
   A student who thinks total reflection starts at the critical angle itself picks this. At exactly the critical angle a refracted ray still exists and grazes the surface; all of the light is reflected only beyond the critical angle.
4. DThe ray continues into the air in the direction it had in the water.
   A student who thinks light changes direction only on entering a material picks this. Light leaving water for air refracts away from the normal; at the critical angle it is bent all the way to 90°.

Working n_water sin θ_critical = n_air sin θ₂ with sin θ_critical = n_air/n_water gives sin θ₂ = 1, θ₂ = 90° from the normal: the refracted ray grazes the boundary.

CED 13.3.A.5.ii · Read this in Fix

1. Av_oil > v_ethanol > v_water
   A student who thinks light travels faster in a medium with a greater index picks this. Since v = c/n, the liquid with the greatest index, olive oil, has the smallest speed of light, not the largest.
2. Bv_water > v_ethanol > v_oil Correct
   The index of refraction is inversely proportional to the speed of light in the medium, v = c/n. Ranking the indices from smallest to largest (water 1.33, ethanol 1.36, olive oil 1.47) ranks the speeds from largest to smallest. The mass densities are irrelevant.
3. Cv_ethanol > v_oil > v_water
   A student who ranks by mass density, expecting light to be slowest in the densest liquid, picks this. The speed of light depends on the index of refraction, v = c/n: olive oil is less dense than water but has the greater index, so light is slower in it.
4. Dv_water = v_ethanol = v_oil
   A student who thinks light travels at c in every material picks this. c is the speed in a vacuum; in each liquid v = c/n, and the three indices differ, so the three speeds differ.

CED 13.3.A.3 · Read this in Fix

1. A4.01 × 10⁸ m/s
   A student who treats the index as proportional to the speed picks this, multiplying c by n = 1.34. The index is inversely proportional to the speed, v = c/n; no material slows light to more than c.
2. B2.10 × 10⁸ m/s
   A student who applies Snell's law to the angles instead of their sines picks this: n = 50.0/35.0 = 1.43, so v = 2.10 × 10⁸ m/s. Snell's law relates sin θ₁ and sin θ₂, which gives n = 1.34.
3. C2.25 × 10⁸ m/s Correct
   Snell's law gives the index of the liquid: n = (1.00)(sin 50.0°)/sin 35.0° = 0.766/0.574 = 1.34. Then v = c/n = (3.00 × 10⁸ m/s)/1.34 = 2.25 × 10⁸ m/s, less than c as it must be in a material.
4. D3.00 × 10⁸ m/s
   A student who thinks light travels at c in every material picks this. c is the speed in a vacuum; the refraction shows that light's speed changes at the surface, and v = c/n = 2.25 × 10⁸ m/s in the liquid.

Working n_liquid = n_air sin θ₁ / sin θ₂ = (1.00)(sin 50.0°)/(sin 35.0°) = 0.7660/0.5736 = 1.336. v = c/n = (3.00 × 10⁸ m/s)/1.336 = 2.25 × 10⁸ m/s. (v = cn = 4.01 × 10⁸ m/s; n = 50.0/35.0 = 1.429 gives 2.10 × 10⁸ m/s; v = c gives 3.00 × 10⁸ m/s.)

CED 13.3.A.3 · Read this in Fix

1. Asin A ÷ n
   A student who pairs each index with the angle in the other medium picks this: n sin θ = sin A. That would make the ray bend toward the normal as it leaves the glass; in fact n goes with the angle A inside the glass.
2. Bn × sin A Correct
   At the first face the angle of incidence is 0, so the ray is not refracted. The normals to the two faces make the same angle A as the faces, so the ray meets the second face at an angle of incidence A. Snell's law there: n sin A = (1.00) sin θ, so sin θ = n sin A. Passing into air, the ray bends away from the normal.
3. Csin (n×A)
   A student who applies Snell's law to the angles themselves picks this, taking θ = nA. Snell's law relates the sines: n sin A = sin θ, and sin (nA) is not equal to n sin A.
4. Dn × cos A
   A student who measures the angle of incidence from the surface picks this: the ray makes an angle of 90° − A with the second face, and sin (90° − A) = cos A. The angle of incidence is measured from the normal, and it equals A.

Working Face 1: θ₁ = 0, so sin θ₂ = 0 and the ray continues undeviated. Inside, the ray is perpendicular to face 1; the normal to face 2 is rotated by the angle A between the faces, so the angle of incidence at face 2 is A. Snell's law at face 2: n sin A = n_air sin θ = sin θ. So sin θ = n sin A (> sin A: away from the normal). Errors: n sin θ = sin A gives sin A ÷ n; θ = nA gives sin (n×A); taking the angle from the surface, 90° − A, gives n cos A.

CED 13.3.A.4.i · Read this in Fix

1. AMedium 2 has a lower index of refraction than medium 1.
   A student who has the rule reversed, linking bending toward the normal with entering a lower index, picks this. In Snell's law the larger index goes with the smaller angle, and the angle is smaller in medium 2.
2. BLight travels faster in medium 2 than it does in medium 1.
   A student who correctly finds that medium 2 has the greater index but thinks light is faster in such a medium picks this. n = c/v, so the greater index in medium 2 means light travels more slowly there.
3. CLight travels at the same speed in medium 2 as in medium 1.
   A student who thinks light travels at c in every material picks this. The ray bends toward the normal, so medium 2 has the greater index, and v = c/n means light is slower in medium 2.
4. DMedium 2 has a greater index of refraction than medium 1. Correct
   In medium 2 the ray makes a smaller angle with the normal than in medium 1: it has refracted toward the normal. From n₁ sin θ₁ = n₂ sin θ₂, a smaller angle goes with a larger index, so n₂ > n₁.

Working In medium 2 the ray is closer to the normal: θ₂ < θ₁. Snell's law n₁ sin θ₁ = n₂ sin θ₂ gives n₂/n₁ = sin θ₁/sin θ₂ > 1, so n₂ > n₁, and v = c/n gives v₂ < v₁.

CED 13.3.A.4.ii · Read this in Fix

1. AThe claim is wrong: every ray entering glass is bent, including a ray along the normal.
   A student who thinks every ray entering a new medium changes direction picks this. For θ₁ = 0, n₁ sin 0 = n₂ sin θ₂ gives θ₂ = 0, so a ray along the normal is not bent. The claim is wrong for a different reason: the speed does change.
2. BThe claim is right: light changes speed at a boundary just when it changes direction.
   A student who thinks light changes speed only when it is bent picks this. The speed in the glass is c/n for every ray; along the normal the speed still drops, but the ray has no angle to change.
3. CThe claim is wrong: the light slows down in the glass even though the ray is not bent. Correct
   The speed of light depends only on the medium: v = c/n in the glass whatever the angle. At normal incidence θ₁ = 0, so Snell's law gives θ₂ = 0: the change of speed produces no change of direction. Not being bent is not evidence that the speed is unchanged.
4. DThe claim is right: light has the same speed, c, in glass as it does in air.
   A student who thinks light travels at c in every material picks this. c is the speed in a vacuum. In glass, v = c/n; with n = 1.5, light travels at about 2 × 10⁸ m/s.

CED 13.3.A.4.iii · Read this in Fix

1. AIt can occur for light in the glass heading toward the water. Correct
   Total internal reflection may occur when light travels toward a medium with a lower index of refraction. Here that means light inside the glass (1.50) meeting the water (1.33), at angles of incidence beyond θ_critical = sin⁻¹(1.33/1.50) = 62.5°.
2. BIt can occur for light that is in the water and heading toward the glass.
   A student who thinks light is turned back from the optically denser medium picks this. Going from water into glass, sin θ₂ = (1.33/1.50) sin θ₁ is always less than 1, so there is always a refracted ray.
3. CIt can occur for light going either way, if the angle is large enough.
   A student who thinks total internal reflection depends only on the size of the angle picks this. It also needs the light to be heading toward the lower index; light going from water into glass is refracted at every angle.
4. DIt needs air beyond the boundary, so it does not occur at this one.
   A student who thinks total internal reflection happens only where a material meets air picks this. Any boundary where n₂ < n₁ has a critical angle; for glass to water it is sin⁻¹(1.33/1.50) = 62.5°.

CED 13.3.A.5 · Read this in Fix

1. Av₁ sin θ_c
   A student who treats the index as proportional to the speed picks this, writing sin θ_c = n₂/n₁ = v₂/v₁. The index is inversely proportional to the speed, so n₂/n₁ = v₁/v₂.
2. Bv₁/cos θ_c
   A student who takes the critical angle to be measured from the surface picks this, using 90° − θ_c as the angle from the normal. The critical angle is an angle of incidence, measured from the normal.
3. Cv₁·90°/θ_c
   A student who applies Snell's law to the angles instead of their sines picks this: n₁ θ_c = n₂ (90°). Snell's law relates sines: n₁ sin θ_c = n₂ sin 90°.
4. Dv₁/sin θ_c Correct
   At the critical angle the refracted ray is at 90°: n₁ sin θ_c = n₂ sin 90° = n₂. With n = c/v, sin θ_c = n₂/n₁ = (c/v₂)/(c/v₁) = v₁/v₂, so v₂ = v₁/sin θ_c, which is greater than v₁, as it must be for total internal reflection.

Working Snell's law at the critical angle: n₁ sin θ_c = n₂ sin 90° → sin θ_c = n₂/n₁. n = c/v: n₂/n₁ = (c/v₂)/(c/v₁) = v₁/v₂. So v₂ = v₁/sin θ_c. (n ∝ v gives v₁ sin θ_c; angle from the surface gives v₁/cos θ_c; angles instead of sines give n₂/n₁ = θ_c/90°, so v₂ = v₁·90°/θ_c.)

CED 13.3.A.5.i · Read this in Fix

1. A41.8°
   A student who thinks the critical angle belongs to the glass alone picks this. θ_critical = sin⁻¹(n₂/n₁) depends on the medium beyond the boundary too; with water outside it is larger.
2. B62.4° Correct
   sin θ_critical = n₂/n₁. With the glass (n₁) unchanged, sin θ_critical is proportional to the index of the outer medium, which increases from 1.00 to 1.33. So sin θ_critical = 1.33 × sin 41.8° = 0.886, and θ_critical = 62.4°.
3. C30.1°
   A student who puts the outer medium's index in the denominator picks this: sin θ = sin 41.8°/1.33. From Snell's law, sin θ_critical = n₂/n₁, so a larger outer index makes sin θ_critical larger, not smaller.
4. D55.6°
   A student who scales the angle itself by 1.33 picks this: 41.8° × 1.33 = 55.6°. It is the sine of the critical angle that is proportional to n₂, not the angle.

Working Glass–air: sin 41.8° = 1.00/n_g, so n_g = 1/0.6665 = 1.50. Glass–water: sin θ_c = 1.33/n_g = 1.33 × sin 41.8° = 1.33 × 0.6665 = 0.8865, θ_c = 62.4°. (Unchanged: 41.8°; sin θ = 0.6665/1.33 = 0.501, 30.1°; 41.8° × 1.33 = 55.6°.)

CED 13.3.A.5.i · Read this in Fix

1. A2.23 × 10⁸ m/s
   A student who takes the critical angle to be the 48.0° measured from the face picks this: v = c sin 48.0°. The critical angle is measured from the normal: 42.0°.
2. B1.40 × 10⁸ m/s
   A student who applies Snell's law to the angles picks this: n (42.0°) = (1.00)(90°) gives n = 2.14. With sines, n = 1/sin 42.0° = 1.49.
3. C2.01 × 10⁸ m/s Correct
   A refracted ray that travels along the surface means the ray meets the flat face at the critical angle. The 48.0° in the diagram is measured from the face, so θ_critical = 90.0° − 48.0° = 42.0°. Then n sin 42.0° = (1.00) sin 90°, so n = 1/sin 42.0° = 1.49, and v = c/n = c sin 42.0° = 2.01 × 10⁸ m/s.
4. D4.48 × 10⁸ m/s
   A student who inverts the ratio in the critical-angle equation picks this, taking sin 42.0° = n/1.00, so n = 0.669 and v = c/n. The correct relation is sin θ_critical = n_air/n, so n = 1.49 and v is less than c.

Working The ray enters along a radius, so it is not refracted at the curved surface. At C the refracted ray grazes the flat face, so the angle of incidence is the critical angle. Angle from the face = 48.0°, so θ_c = 90.0° − 48.0° = 42.0°. n sin θ_c = n_air sin 90°: n = 1/sin 42.0° = 1/0.6691 = 1.494. v = c/n = (3.00 × 10⁸ m/s)(0.6691) = 2.01 × 10⁸ m/s. (c sin 48.0° = 2.23 × 10⁸ m/s; n = 90/42.0 = 2.14 gives 1.40 × 10⁸ m/s; n = sin 42.0° = 0.669 gives 4.48 × 10⁸ m/s.)

CED 13.3.A.5.ii · Read this in Fix

1. AMost of it is reflected, but a weak ray still passes out into the air.
   A student who thinks some light is always transmitted picks this. Beyond the critical angle Snell's law would need sin θ₂ = 1.50 × sin 45° = 1.06, which is impossible, so no light is transmitted.
2. BAll of it is reflected back into the prism; none is transmitted. Correct
   From the prism's angles, the ray meets the longest face at an angle of incidence of 45°. The critical angle for glass to air is sin⁻¹(1.00/1.50) = 41.8°. Since 45° is beyond the critical angle, there is no refracted ray: all of the light is reflected back into the glass.
3. CIt passes out into the air, bending away from the normal.
   A student who thinks light always gets out through a transparent surface picks this. That happens below the critical angle; here the angle of incidence, 45°, exceeds the critical angle of 41.8°, so the light is totally reflected.
4. DIt leaves the prism and travels along the prism's longest face.
   A student who expects a grazing ray beyond the critical angle picks this. A ray travels along the surface only at exactly the critical angle (41.8° here); at 45° there is no refracted ray at all.

Working Angle of incidence at the longest face = 45° (the ray is perpendicular to the first face, which makes 45° with the longest face). θ_c = sin⁻¹(n_air/n_glass) = sin⁻¹(1.00/1.50) = 41.8°. 45° > 41.8°, so total internal reflection: 1.50 sin 45° = 1.06 > 1, no refracted ray.

CED 13.3.A.5.iii · Read this in Fix

1. Ad√(n² − 1)
   A student who takes the critical angle to be measured from the surface picks this: the ray's angle from the vertical becomes 90° − θ_critical, and r = d tan(90° − θ_critical) = d√(n² − 1). The critical angle is an angle of incidence, measured from the normal, which here is vertical.
2. Bd/√(n² − 1) Correct
   At the edge of the region, rays meet the surface at the critical angle, sin θ_critical = 1/n; rays at larger angles are totally reflected. The normal is vertical, so r = d tan θ_critical. With sin θ_critical = 1/n, cos θ_critical = √(n² − 1)/n and tan θ_critical = 1/√(n² − 1), so r = d/√(n² − 1).
3. Cd tan(90°/n)
   A student who applies Snell's law to the angles themselves picks this: n θ_critical = (1.00)(90°), so θ_critical = 90°/n. Snell's law relates the sines: n sin θ_critical = 1.00 × sin 90°, so sin θ_critical = 1/n.
4. Dd/n
   A student who finds sin θ_critical = 1/n correctly but then takes r = d sin θ_critical picks this. That would be right only if d were the length of the slanted ray; d is the vertical side adjacent to the angle, so r = d tan θ_critical.

Working Rays that reach the surface at an angle of incidence greater than θ_critical are totally reflected, so the edge of the region is where the angle of incidence equals θ_critical: sin θ_critical = n₂/n₁ = 1.00/n = 1/n. The normal is vertical, so a ray from the lamp at angle θ_critical from the vertical meets the surface a horizontal distance r = d tan θ_critical from the point above the lamp. Right triangle with opposite 1, hypotenuse n: adjacent √(n² − 1), so tan θ_critical = 1/√(n² − 1) and r = d/√(n² − 1). (n = 1.33: r = 1.14d.) Errors: θ_critical taken from the surface → r = d tan(90° − θ_critical) = d√(n² − 1); angles instead of sines, n θ_critical = 90° → r = d tan(90°/n); r = d sin θ_critical → d/n.

CED 13.3.A.5.i · Read this in Fix

1. A1.24t
   A student who thinks light travels faster in a medium with a greater index picks this: time ∝ thickness/n gives 2 × (1.50/2.42) = 1.24. Since n = c/v, light is slower in diamond than in glass, and the time is (2.42/1.50) × 2 = 3.23 times as long.
2. B2.00t
   A student who thinks light keeps its speed when it crosses a boundary along the normal, since it is not bent, picks this: only the doubled thickness then matters. Light slows down in each plate whether or not it is bent; in diamond its speed is c/2.42, less than in glass, so the time is 3.23t.
3. C1.61t
   A student who allows for the change of index but not the change of thickness picks this: 2.42/1.50 = 1.61. The pulse also has twice as far to travel in the diamond plate, which doubles the time again: 3.23t.
4. D3.23t Correct
   The crossing time is the thickness divided by the speed, and v = c/n, so the time is proportional to n × thickness. Both increase: (2.42/1.50) × 2 = 3.23, so the diamond plate takes 3.23t.

Working Speed in a medium: v = c/n. Crossing time = thickness/v = n × thickness/c, so the time is proportional to n × thickness. Factor = (2.42/1.50) × 2 = 3.23, so the diamond plate takes 3.23t. Errors: speed proportional to n (time ∝ thickness/n): 2 × 1.50/2.42 = 1.24; speed unchanged along the normal (time ∝ thickness): 2.00; thickness left out: 2.42/1.50 = 1.61.

CED 13.3.A.3 · Read this in Fix