Answers · Refraction and total internal reflection

Total internal reflection — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. NQ14 · Practice

Model answer. The critical angle is the angle of incidence, inside a medium such as glass, at which the angle of refraction is 90°, so that the refracted ray runs along the boundary. Total internal reflection happens at angles bigger than this.

A full answer includes

I chose: the angle of incidence, inside the medium, at which the angle of refraction is 90°.
I did not choose “all of the light is reflected”: that happens only above the critical angle.

Also accept

the angle of incidence inside the medium at which the angle of refraction is 90°

Watch for

“The critical angle is the angle at which all of the light is reflected.” — At the critical angle itself, the ray in the air runs along the boundary. All of the light is reflected only at angles bigger than the critical angle.
“The critical angle is an angle in the air, at which light stops getting into the medium.” — Light going from air into a medium always gets in. The critical angle is measured inside the medium, as light tries to get out.
“The critical angle is measured between the ray and the surface.” — Angles in ray diagrams are measured from the normal, not from the surface, and at the critical angle the ray in the air runs along the boundary.

2. NQ15 · Practice

Options as numbered on the pupil sheet: 1 — The ray in the air runs along the flat side.; 2 — No light leaves: all of it is reflected inside.; 3 — All of it leaves, and none is reflected.; 4 — Most leaves, bent away; a weak ray is reflected..

Model answer. Ray Q meets the flat side at 22°, less than the critical angle of 37°, so most of it refracts out into the air, bending away from the normal, and a weak ray is reflected back inside: internal reflection. Ray R meets it at exactly the critical angle, so its angle of refraction is 90° and the ray in the air runs along the flat side. Ray P meets it at 61°, more than the critical angle, so no light gets out: it is totally internally reflected, at 61° to the normal.

A full answer includes

I matched ray P to: no light leaves, and all of it is reflected inside.
I matched ray Q to: most of it leaves, bent away from the normal, and a weak ray is reflected.
I matched ray R to: the ray in the air runs along the flat side.
I did not match ray Q to “all of it leaves and none is reflected”: some light is always reflected inside.

Also accept

P: all reflected inside; Q: most leaves, a weak ray reflected; R: runs along the flat side

Watch for

“Ray Q: all of it leaves, and none is reflected.” — Below the critical angle most of the light leaves, but a weak ray is always reflected back inside the plastic.
“Ray R: all of it is reflected inside.” — At exactly the critical angle the ray in the air still runs along the flat side. All of the light is reflected only at angles bigger than 37°.
“Ray P: the ray in the air runs along the flat side.” — That happens only at the critical angle, 37°. At 61°, more than the critical angle, no light gets out.

3. NQ20 · Practice

Model answer. The light is going from air into water, into the medium in which it travels more slowly, so it bends towards the normal. Its angle of refraction is always less than 90°, so it always gets in; only part of it is reflected. Total internal reflection needs light inside the water trying to get out.

A full answer includes

I chose: no, going from air into water the light bends towards the normal and always gets in.
I used the fact that total internal reflection needs light inside the slower medium.
I did not choose “80° is bigger than the critical angle of water”.

Also accept

no: going into water the light always gets in

Watch for

“80° is bigger than the critical angle of water, so all of the light is reflected.” — The critical angle matters only for light inside the water trying to get out. Leon’s light is going into the water.
“At 80° the ray in the water runs along the surface.” — A ray runs along the surface only for light leaving the water at the critical angle. Going in, the ray bends towards the normal, so its angle in the water is less than 80°.
“At 80° the light slows down so much at the surface that it cannot get in.” — Light slows down as it enters water at every angle, and it still gets in. Slowing down makes it bend towards the normal; it does not stop it.

4. NQ16 · Practice

Model answer. Total internal reflection needs both: the light must be inside the medium in which it travels more slowly, such as glass or water, meeting a boundary with a medium in which it travels faster, such as air; and its angle of incidence must be bigger than the critical angle. Row C. At exactly the critical angle the ray runs along the boundary, below it most of the light gets out, and light going into a slower medium always gets in.

A full answer includes

I chose row C.
I checked the first condition: the light is in the slower medium, meeting a faster one.
I checked the second condition: the angle of incidence is bigger than the critical angle.
I did not choose row B: at the critical angle itself, light still leaves along the boundary.

Also accept

Row C

Watch for

“Row A.” — Light going into a slower medium bends towards the normal and always gets in, whatever its angle of incidence.
“Row B.” — At exactly the critical angle the ray in the air runs along the boundary, so some light still leaves. The angle must be bigger.
“Row D.” — Below the critical angle most of the light gets out; only a weak ray is reflected inside.

5. NQ17 · Practice

Model answer. Light leaving the plastic follows the same path as light entering it, backwards, so the angle in the air goes on top: n = sin(angle in air) ÷ sin(angle in plastic). So sin(angle in air) = n × sin 28° = 1.45 × 0.4695 = 0.681, and the angle in the air is sin⁻¹ 0.681 = 42.9°. It is bigger than 28°: leaving the plastic, the ray speeds up and bends away from the normal.

Checked automatically

Angle between the ray in the air and the normal (°): 42.9 (within ±0.5)
- Common wrong answer 18.9 : You divided by n. Leaving the plastic, the ray bends away from the normal, so the angle in the air is bigger: sin(angle in air) = 1.45 × sin 28°.
- Common wrong answer 40.6 : You multiplied the angle by 1.45. Multiply its sine instead, then take sin⁻¹.
- Common wrong answer 0.681 : That is the sine. Use sin⁻¹ to turn it into an angle.
- Common wrong answer 28 : The ray changes speed at the surface, so it changes direction.

A full answer includes

Method: I put the sine of the angle in the air on top of the equation.
Method: I found sin(angle in air) = 1.45 × sin 28° = 1.45 × 0.4695 = 0.681.
Method: I used sin⁻¹ to turn 0.681 into an angle.
Answer: I wrote 42.9° (or 43°).
I did not divide sin 28° by 1.45 (that gives 18.9°, closer to the normal).

Also accept

42.9°

Watch for

“sin(angle in air) = sin 28° ÷ 1.45.” — That would put the ray in the air closer to the normal. Leaving the plastic it speeds up and bends away, so multiply by n.
“The angle in the air is 1.45 × 28° = 40.6°.” — Multiply the sine of the angle by n, then take sin⁻¹.
“The ray goes straight on at 28°, because it is leaving the plastic, not entering it.” — It changes speed at the surface, so unless it is along the normal it changes direction.

6. NQ18 · Practice

Model answer. (a) The surface is at 90° to the normal, so a ray at 32° to the surface is 90° − 32° = 58° from the normal: the angle of incidence is 58°. This is bigger than the critical angle, 41°, so the ray is totally internally reflected. (b) Total internal reflection obeys the law of reflection: the angle of reflection equals the angle of incidence, 58°. (c) The ray in the air runs along the surface when the angle of incidence is the critical angle, 41° from the normal. From the surface, that is 90° − 41° = 49°. At more than 49° to the surface, light gets out.

Checked automatically

(a) Angle of incidence (°): 58 (within ±0.5)
- Common wrong answer 32 : That is the angle to the surface. Measure from the normal: 90° − 32°.
(b) Angle of reflection (°): 58 (within ±0.5)
- Common wrong answer 41 : That is the critical angle. The angle of reflection equals the angle of incidence.
- Common wrong answer 32 : That is the angle to the surface; the angle of reflection is measured from the normal.
(c) Angle between the ray in the glass and the surface (°): 49 (within ±0.5)
- Common wrong answer 41 : That is the critical angle, measured from the normal. The question asks for the angle to the surface: 90° − 41°.

A full answer includes

(a) Method: I took the angle to the surface away from 90°: 90° − 32°.
(a) Answer: I wrote 58°.
(b) Answer: I wrote 58°.
(b) I gave the reason: the angle of reflection equals the angle of incidence.
(c) Method: I used the critical angle, 41°, as the angle of incidence.
(c) Method: I took 41° away from 90°.
(c) Answer: I wrote 49°.
I did not give 41°, the critical angle, as the angle of reflection.

Also accept

(a) 58° (b) 58° (c) 49°

Watch for

“The angle of reflection is the critical angle, 41°.” — The reflected ray obeys the law of reflection: it leaves at the angle of incidence, 58°.
“The ray in the air runs along the surface when the ray in the glass is at 41° to the surface.” — The critical angle, 41°, is measured from the normal. From the surface, that is 90° − 41° = 49°.

7. NQ19 · Practice

Model answer. (a) The light is inside the water, the medium in which it travels more slowly, and the air outside is a medium in which it travels faster. Along the smooth jet, the light meets the water surface at angles of incidence bigger than the critical angle of water, so it is totally internally reflected, again and again, and follows the jet round its curve, like light in an optical fibre. (b) A drop is small and round, so its surface curves sharply. Light inside a drop meets the surface at angles of incidence smaller than the critical angle, so much of it refracts out into the air, and the drops sparkle.

A full answer includes

(a) I wrote that the light is inside the water, where it travels more slowly than in the air outside.
(a) I wrote that it meets the surface of the jet at more than the critical angle.
(a) I wrote that it is totally internally reflected, again and again, along the jet.
(b) I wrote that the surface of a small drop curves sharply.
(b) I wrote that the light then meets the surface at less than the critical angle.
(b) I wrote that this light refracts out into the air.
I did not write that the water surface reflects like a silvered mirror.

Also accept

(a) total internal reflection, again and again, at more than the critical angle (b) the light meets the curved surface of a drop at less than the critical angle

Watch for

“The light stays in because the water surface is shiny, like a mirror.” — There is no mirror. The light is totally internally reflected at the water–air boundary, because it meets it at more than the critical angle.
“Light comes out of the drops because the light is weaker there.” — Weak or bright, light that meets the surface at more than the critical angle stays in. Round a small drop the surface curves sharply, so the light meets it at less than the critical angle and gets out.

8. NQ27 · Practice

Model answer. The angle of incidence at each short face is fixed at 45°. Total internal reflection needs the angle of incidence to be bigger than the critical angle, so the material needs a critical angle less than 45°. Only the glass with a critical angle of 40° works: 45° is bigger than 40°, so all of the light is reflected at both faces. With a critical angle of 45°, the ray in the air runs along the face; with 47° or 52°, 45° is smaller than the critical angle and most of the light gets out, so the image is dim.

A full answer includes

I chose the glass with a critical angle of 40°.
I used the rule: all of the light is reflected only when 45° is bigger than the critical angle.
I concluded that the critical angle must be less than 45°.
I did not choose the plastic with a critical angle of 45°: at the critical angle itself, light still runs along the face.

Also accept

the glass with a critical angle of 40°

Watch for

“The plastic with a critical angle of 45°.” — At exactly the critical angle the ray in the air runs along the face, so not all of the light is reflected. The angle of incidence must be bigger than the critical angle.
“The plastic with a critical angle of 47°.” — 45° is smaller than 47°, so most of the light refracts out of each short face, and only a weak ray is reflected.
“The resin with a critical angle of 52°, because a bigger critical angle reflects more.” — A bigger critical angle makes total internal reflection harder: at 45°, smaller than 52°, most of the light gets out.

9. NQ26 · Practice

Model answer. (a) Only a ray aimed at O, the centre of the curved side, travels along a radius, and only a radius meets the curved side at 90°, along the normal. A ray aimed 1 cm to the side of O meets the curved side at an angle to the normal there, so it bends as it enters the glass. Its direction inside the glass is no longer the direction set on the scale, and it meets the flat side away from O, where the scale’s normal is not drawn, so the angle read from the scale is not its angle of incidence. (b) The angle of incidence at which the ray in the air runs along the flat side is the critical angle, so c = 38°. n = 1 ÷ sin c = 1 ÷ sin 38° = 1 ÷ 0.616 = 1.62.

Checked automatically

(b) Supplement: Refractive index of the glass: 1.62 (within ±0.03)
- Common wrong answer 0.616 : That is sin c. The refractive index is 1 ÷ sin c.
- Common wrong answer 0.026 : That is 1 ÷ 38. Take the sine of the angle first.

A full answer includes

(a) I wrote that the ray no longer travels along a radius of the curved side.
(a) I wrote that it meets the curved side at an angle to the normal there.
(a) I wrote that it bends as it enters the curved side.
(a) I wrote that its direction inside the glass is then not the direction set on the scale.
(b) Method: I used n = 1 ÷ sin c = 1 ÷ sin 38° = 1 ÷ 0.616.
(b) Answer: I wrote 1.62 (or 1.6).
I did not write that a ray enters the curved side without bending wherever it is aimed.

Also accept

(a) the ray is not along a radius, so it bends at the curved side and its angle at the flat side is not the scale reading (b) 1.62

Watch for

“The ray enters the curved side without bending wherever it is aimed.” — Only a ray along a radius meets the curved side along the normal. Aimed away from O, the ray meets it at an angle to the normal and bends.
“The reading is still right, because the flat side is straight.” — The trouble starts at the curved side: the ray bends there, so it no longer reaches the flat side at the angle set on the scale.
“n = sin 38° = 0.62.” — That is sin c. The refractive index is 1 ÷ sin c, bigger than 1.

10. NQ21 · Practice

Model answer. n = 1 ÷ sin c = 1 ÷ sin 34.4° = 1 ÷ 0.565 = 1.77. A small critical angle goes with a large refractive index.

Checked automatically

Refractive index of sapphire: 1.77 (within ±0.03)
- Common wrong answer 0.565 : That is sin c. The refractive index is 1 ÷ sin c.
- Common wrong answer 0.029 : That is 1 ÷ 34.4. Take the sine of the angle first.

A full answer includes

Method: I used n = 1 ÷ sin c = 1 ÷ sin 34.4° = 1 ÷ 0.565.
Answer: I wrote 1.77 (or 1.8).
I did not stop at sin c, 0.565.

Also accept

1.77

Watch for

“n = sin 34.4° = 0.565.” — That is sin c. The refractive index is 1 divided by it.
“n = 1 ÷ 34.4.” — Take the sine of the angle first.

11. NQ22 · Practice

Model answer. sin c = 1 ÷ n = 1 ÷ 2.15 = 0.465, so c = sin⁻¹ 0.465 = 27.7°. This is much smaller than the critical angle of ordinary glass, so light inside the stone is easily trapped by total internal reflection, and the stone sparkles almost like a diamond.

Checked automatically

Critical angle of cubic zirconia (°): 27.7 (within ±0.5)
- Common wrong answer 0.465 : That is sin c. Use sin⁻¹ to find the angle.

A full answer includes

Method: I found sin c = 1 ÷ 2.15 = 0.465.
Method: I used sin⁻¹ to turn 0.465 into an angle.
Answer: I wrote 27.7° (or 28°).
I did not give 0.465, the sine, as the angle.

Also accept

27.7°

Watch for

“c = 0.465.” — That is sin c. Use sin⁻¹ to find the angle.
“A bigger refractive index gives a bigger critical angle.” — The bigger the refractive index, the smaller the critical angle: sin c = 1 ÷ n.

12. NQ23 · Practice

Model answer. Optical fibres carry far more information than copper wires of the same size, the signal travels long distances with little loss, and the light inside is not affected by electrical interference from nearby cables or machines. They carry light, not electric current, and they keep it in by total internal reflection, not by a mirror coating. Joining a broken fibre needs special tools to line up the tiny glass cores, so repairs are not cheaper.

A full answer includes

I chose: far more information.
I chose: long distances with little loss.
I chose: not affected by electrical interference.
I did not choose “electric current”: optical fibres carry light.

Also accept

more information; little loss over long distances; no electrical interference

Watch for

“Optical fibres carry electric current with less resistance.” — They carry pulses of light, not electric current.
“The light is kept in by a mirror coating.” — It is kept in by total internal reflection at the boundary between the core and the cladding.
“Optical fibres are cheaper to repair than copper wires.” — Joining a broken glass fibre needs special tools to line up its tiny core exactly, so this is not an advantage.

13. NQ24 · Practice

Model answer. (a) Light in the core meets the boundary between the core and the cladding with an angle of incidence bigger than the critical angle, so it is totally internally reflected. This happens again and again along the fibre, so the light stays in the core round gentle bends. (b) Light travels more slowly in a material of higher refractive index, so the light would be going from the core into a slower medium. It bends towards the normal and gets in at every angle, so there is no total internal reflection: the light passes into the cladding and leaks out of the fibre, and the signal is lost.

A full answer includes

(a) I wrote that the light meets the core–cladding boundary at more than the critical angle.
(a) I wrote that it is totally internally reflected.
(a) I wrote that this happens again and again along the fibre.
(b) I wrote that the light is not totally internally reflected: it passes into the cladding.
(b) I gave the reason: light is slower in this cladding than in the core.
(b) I wrote that light going into a slower medium bends towards the normal and always gets in.
I did not write that the cladding works like a mirror.

Also accept

(a) total internal reflection at more than the critical angle, again and again (b) it passes into the cladding: going into a slower medium there is no total internal reflection

Watch for

“The cladding is a mirror.” — The cladding is glass of lower refractive index; the reflection is total internal reflection, not a mirror.
“A cladding of higher refractive index holds the light in more strongly.” — Light is totally internally reflected only at a boundary with a medium of lower refractive index, in which it travels faster.

14. NQ25 · Practice

Options as numbered on the pupil sheet: 1 — At the far end, the pulses are detected and changed back into an electrical signal.; 2 — The electrical signal of the call is coded into very fast pulses of infrared light.; 3 — So the pulses travel along the whole fibre, reflected again and again, round gentle bends.; 4 — The pulses are sent into one end of the fibre’s core.; 5 — Each pulse meets the boundary between the core and the cladding at more than the critical angle and is totally internally reflected..

Model answer. The electrical signal of the call is coded digitally into very fast pulses of light, usually infrared. The pulses are sent into one end of the fibre’s core. Each pulse meets the boundary between the core and the cladding at more than the critical angle and is totally internally reflected. This happens again and again, so the pulses travel along the fibre, round gentle bends, with little loss. At the far end, the pulses are detected and changed back into an electrical signal.

A full answer includes

I put first: the call is coded into pulses of infrared light.
I put second: the pulses are sent into the core.
I put third: each pulse is totally internally reflected at the core–cladding boundary.
I put fourth: the pulses travel along the whole fibre, reflected again and again.
I put last: the pulses are changed back into an electrical signal.
I did not put the reflections before the pulses are sent into the core.

Also accept

coded into pulses; into the core; total internal reflection; again and again along the fibre; changed back at the far end

Watch for

“The light travels in a straight line down the middle of the fibre.” — It meets the sides again and again and is totally internally reflected each time.

15. NT2 · Apply

Model answer. (a) sin c = 1 ÷ n = 1 ÷ 1.34 = 0.746, so c = sin⁻¹ 0.746 = 48.3°. (b) The beam is inside the water, the slower medium, and meets the surface at 70°, which is bigger than the critical angle. So no light gets out into the air: all of it is totally internally reflected back down into the sea, at 70° to the normal, and the friend sees none of it. (c) Yes, the light gets into the water. The friend’s beam is going from air into water, the slower medium, so it bends towards the normal, and its angle of refraction is always less than 90°: it always gets in, with only a little reflected. Total internal reflection needs light inside the slower medium, trying to get out, so the critical angle does not matter for this beam.

Checked automatically

(a) Supplement: Critical angle of seawater (°): 48.3 (within ±0.5)
- Common wrong answer 0.746 : That is sin c. Use sin⁻¹ to find the angle.

A full answer includes

(a) Method: I found sin c = 1 ÷ 1.34 = 0.746.
(a) Method: I used sin⁻¹ to turn 0.746 into an angle.
(a) Answer: I wrote 48.3° (or 48°).
(b) I wrote that 70° is bigger than the critical angle.
(b) I wrote that no light gets out into the air.
(b) I wrote that all of the beam is totally internally reflected back into the sea.
(c) I predicted that the light gets into the water.
(c) I gave the reason: going into water, the slower medium, light bends towards the normal.
(c) I wrote that total internal reflection needs light inside the slower medium.
I did not write that the friend’s beam at 70° is totally internally reflected.

Also accept

(a) 48.3° (b) 70° is bigger than the critical angle, so all of it is totally internally reflected (c) yes: going into water it bends towards the normal and gets in

Watch for

“The friend’s beam at 70° is totally internally reflected, because 70° is bigger than the critical angle.” — The critical angle matters only for light inside the water trying to get out. Light going from air into water always gets in.
“At 70° the diver’s beam leaves along the surface.” — That happens only at the critical angle. At 70°, bigger than the critical angle, no light gets out.
“c = 0.746.” — That is sin c. Use sin⁻¹ to find the angle.