Answers · Refraction and total internal reflection

Refraction of light — answer key

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

1. NQ1 · Practice

Options as numbered on the pupil sheet: 1 — The reflected ray; 2 — The incident ray; 3 — The refracted ray; 4 — The normal; 5 — The boundary.

Model answer. Line 1 is the boundary, the surface of the glass, with the air above it and the glass below it. Line 2 is the normal: it is dashed, because it is a construction line, and it is at 90° to the boundary where the ray meets it. Line 3 is the incident ray, with its arrow pointing towards the boundary, and line 4 is the refracted ray, the ray in the glass, with its arrow pointing away from the boundary.

A full answer includes

I matched line 1 to the boundary.
I matched line 2 to the normal.
I matched line 3 to the incident ray.
I matched line 4 to the refracted ray.
I did not match line 4 to a reflected ray.

Also accept

1 boundary; 2 normal; 3 incident ray; 4 refracted ray

Watch for

“Line 4 is the reflected ray.” — Line 4 goes on into the glass, so it is the refracted ray. A reflected ray would come back into the air.
“Line 1 is the normal, because it is at 90° to line 2.” — The normal is the dashed construction line. Line 1, with the glass on one side of it, is the boundary.
“Line 3 is the reflected ray, because it is in the air.” — Line 3 points towards the boundary, so it is the light arriving: the incident ray. A reflected ray would point away from the boundary, back into the air.

2. NQ2 · Practice

Model answer. The angle of refraction is the angle between the refracted ray and the normal. The refracted ray is line 4, in the glass, and the normal is the dashed line 2, so the angle of refraction is y. Angle x, between the incident ray and the normal, is the angle of incidence; w and z are angles to the surface of the glass.

A full answer includes

I chose y.
I checked that y is between the refracted ray, line 4, and the normal, line 2.
I did not choose z, the angle between the refracted ray and the surface.

Also accept

Watch for

“z, because it is the angle between the ray in the glass and the surface.” — The angles of incidence and refraction are measured from the normal, not from the surface.
“x, because it is next to the normal.” — x is between the normal and the incident ray: it is the angle of incidence.
“w, because it is the angle the light makes with the glass.” — w is between the incident ray and the surface, in the air. Both angles are measured from the normal, and the angle of refraction is in the glass.

3. NQ4 · Practice

Model answer. (a) The normal is at 90° to the surface, so the incident ray, 22° from the surface, is 90° − 22° = 68° from the normal: the angle of incidence is 68°. (b) The refracted ray is 51° from the surface, so it is 90° − 51° = 39° from the normal: the angle of refraction is 39°. It is smaller than 68°, as it should be for light going into glass.

Checked automatically

(a) Angle of incidence (°): 68 (within ±0.5)
- Common wrong answer 22 : That is the angle to the surface. Measure from the normal: 90° − 22°.
(b) Angle of refraction (°): 39 (within ±0.5)
- Common wrong answer 51 : That is the angle to the surface. Measure from the normal: 90° − 51°.

A full answer includes

(a) Method: I took the angle to the surface away from 90°: 90° − 22°.
(a) Answer: I wrote 68°.
(b) Method: I did the same for the refracted ray: 90° − 51°.
(b) Answer: I wrote 39°.
I did not give 22° or 51° as an answer: they are angles to the surface.

Also accept

(a) 68° (b) 39°

Watch for

“The angle of incidence is 22°.” — The 22° is measured from the surface. The angle of incidence is measured from the normal: 90° − 22° = 68°.
“The angle of refraction is bigger than the angle of incidence, 51° against 22°.” — Those are angles to the surface. From the normal, the ray in the glass is at 39°, smaller than 68°.

4. NQ5 · Practice

Model answer. Glass is the slower medium here. (1) From water into glass, light slows down, so it bends towards the normal. (2) From glass into water, it speeds up, so it bends away from the normal. (3) Along the normal, the ray crosses the boundary without changing direction, (4) but its speed still changes: it slows down going into the glass and speeds up coming out.

A full answer includes

(1) I wrote towards.
(2) I wrote away from.
(3) I wrote does not.
(4) I wrote changes.
I did not write that a ray along the normal keeps the same speed.

Also accept

(1) towards (2) away from (3) does not (4) changes

Watch for

“Light bends towards the normal only when it comes from air.” — What matters is the change of speed. Light is slower in glass than in water, so entering glass from water it slows down and bends towards the normal.
“Along the normal the speed stays the same, because the direction does.” — The speed changes at every boundary. Along the normal, only the direction stays the same.

5. NQ8 · Practice

Model answer. The angle in Y, 53°, is bigger than the angle in X, 38°, so the ray is further from the normal after the boundary: it bends away from the normal. A ray bends away from the normal when it passes into a medium in which it travels faster, so light is faster in Y. Row D.

A full answer includes

I chose row D.
I saw that 53° is bigger than 38°.
I concluded that the ray bends away from the normal.
I used the rule: a ray that bends away from the normal is speeding up.
I concluded that light is faster in Y.
I did not choose row B: bending away means the second material is the faster one.

Also accept

Row D

Watch for

“Row A.” — The ray is further from the normal in Y, so it bends away from the normal, not towards it. And a ray that slows down bends towards the normal.
“Row B.” — The ray does bend away from the normal, but bending away means it speeds up: light is faster in Y, the second material.
“Row C.” — Light is faster in Y, but a ray that speeds up bends away from the normal: 53° is further from the normal than 38°.

6. NQ6 · Practice

Model answer. (a) The ray goes straight across the block, and the two sides are parallel, so the normals at the two sides are parallel. The ray meets the second side at 33° to the normal, the same angle it made with the first normal. (b) Leaving the glass, the ray bends away from the normal, and the change is the same as going in, run backwards: it leaves at 54°. (c) The ray that leaves is at 54° to a normal parallel to the first one, so it is parallel to the ray that went in, but shifted sideways.

Checked automatically

(a) Angle of incidence at the opposite side (°): 33 (within ±0.5)
- Common wrong answer 54 : Inside the glass the ray is at 33° to the normal, and the normals at the two parallel sides are parallel.
(b) Angle in the air at the opposite side (°): 54 (within ±0.5)
- Common wrong answer 33 : That is the angle inside the glass. Leaving the glass, the ray bends away from the normal.

A full answer includes

(a) I wrote 33°.
(a) I gave the reason: the normals at the two parallel sides are parallel.
(b) I wrote 54°.
(b) I gave the reason: leaving the glass, the ray bends away from the normal.
(b) I wrote that it bends back by the same amount as it bent going in.
(c) I wrote that it is parallel to the ray that went in.
(c) I wrote that it is shifted sideways.
I did not write that the ray leaves along the same line as the ray that went in.

Also accept

(a) 33° (b) 54° (c) parallel to the ray in, shifted sideways

Watch for

“The ray leaves along the same line as the ray that went in.” — It is parallel to the ray that went in, but shifted sideways, because it crossed the block at a different angle.
“The ray leaves at 33°.” — Leaving the glass it bends away from the normal, so the angle in the air is bigger: 54°.

7. NQ7 · Practice

Options as numbered on the pupil sheet: 1 — Aim the ray box along a line drawn at a measured angle to that normal, so that the ray enters there.; 2 — Take the prism away, join the crosses and continue the line back to the outline of the prism.; 3 — Draw around the prism, and draw a normal to the first side at the point where the ray will enter.; 4 — Join the entry point to the exit point, draw the normal at the exit point, and measure the angles.; 5 — Mark the ray that leaves the second side of the prism with two crosses, well apart..

Model answer. First draw around the prism on the paper and draw a normal to the first side where the ray will enter. Aim the ray box along a line drawn at a measured angle to that normal, so that the ray enters the prism there. Mark the ray that leaves the second side with two crosses, well apart. Take the prism away, join the crosses and continue the line back to the outline of the second side. Join the point where the ray entered to the point where it left: that is the ray inside the glass. Draw the normal to the second side at the exit point, and measure each angle from its own normal with a protractor.

A full answer includes

I put first: draw around the prism and draw the normal where the ray will enter.
I put second: aim the ray box along a line at a measured angle to that normal.
I put third: mark the ray that leaves the second side with two crosses.
I put fourth: take the prism away and join the crosses back to the outline.
I put last: join the entry point to the exit point, draw the second normal and measure.
I did not measure the angles through the glass with the prism still in place.

Also accept

outline and first normal; aim the ray; mark the ray out; remove the prism, join; join entry to exit, second normal, measure

Watch for

“Measure the angle of refraction through the side of the prism, with the prism in place.” — The ray inside the glass is found by joining where it entered to where it left, once the prism is taken away; then each angle is measured from its normal.
“The ray leaving a prism is parallel to the ray that went in, so it need not be marked.” — The two sides of a prism are not parallel, so the ray leaves turned through an angle. Mark it with two crosses to find its direction.

8. NQ9 · Practice

Model answer. (a) Going from air into glass, the ray slows down and bends towards the normal at the first side. Leaving the glass at the second side, it speeds up and bends away from the normal there. (b) In a rectangular block the two sides are parallel, so the normals are parallel and the bend out exactly undoes the bend in: the ray leaves parallel to the ray in. The sides of a prism are at an angle to each other, so the two normals point in different directions, the two bends do not undo each other, and the ray leaves turned through an angle.

A full answer includes

(a) I wrote that at the first side the ray bends towards the normal.
(a) I wrote that at the second side it bends away from the normal.
(b) I wrote that the two sides of the prism are not parallel.
(b) I wrote that their normals point in different directions.
(b) I wrote that the two bends do not cancel.
(b) I wrote that the ray leaves turned through an angle.
(b) I wrote that the two sides of a rectangular block are parallel.
(b) I wrote that, in the block, the bend out undoes the bend in.
I did not write that a prism bends light by a different rule from a block.

Also accept

(a) towards the normal, then away from the normal (b) the sides are not parallel, so the bends do not cancel

Watch for

“A prism bends light in a different way from a block.” — The same rules apply at every boundary. What is different is the angle between the sides.
“The ray bends towards the normal at both sides.” — At the second side the light leaves the glass and speeds up, so it bends away from the normal.

9. NQ11 · Practice

Model answer. n = speed in air ÷ speed in ice = (3.00 × 10⁸ m/s) ÷ (2.29 × 10⁸ m/s) = 1.310…, which is 1.31 to 3 significant figures. The units cancel, so n has no unit.

Checked automatically

Refractive index of ice: 1.31 (3 significant figures)
- Common wrong answer 0.763 : That is the speed in ice ÷ the speed in air. Put the speed in air on top.

A full answer includes

Method: I divided the speed in air by the speed in ice: 3.00 × 10⁸ ÷ 2.29 × 10⁸.
Answer: I wrote 1.31, to 3 significant figures.
I gave no unit, because n is one speed divided by another.
I did not divide the speed in ice by the speed in air (0.763).

Also accept

1.31

Watch for

“n = 2.29 ÷ 3.00 = 0.763.” — Light is slower in ice, so the speed in air goes on top and n is bigger than 1.

10. NQ28 · Practice

Model answer. The refractive index is the ratio of the speeds of a wave in the two regions: the speed in the region the wave comes from ÷ the speed in the region it passes into. n = 0.42 m/s ÷ 0.30 m/s = 1.4. The units cancel, so n has no unit. The waves slow down over the sandbank, so n is bigger than 1, as for light going from air into glass.

Checked automatically

Refractive index, deep water to the sandbank: 1.4 (within ±0.02)
- Common wrong answer 0.71 : That is the slower speed ÷ the faster. Put the speed in the deep water, where the waves come from, on top.
- Common wrong answer 0.12 : You subtracted the speeds. Divide them.

A full answer includes

Method: I divided the speed in the deep water by the speed over the sandbank: 0.42 ÷ 0.30.
Answer: I wrote 1.4.
I gave no unit, because n is one speed divided by another.
I did not divide the slower speed by the faster one (0.71).

Also accept

Watch for

“n = 0.30 ÷ 0.42 = 0.71.” — The waves pass from the deep water, where they are faster, into the shallow water. The speed they come from goes on top, as the speed in air does for light.
“A refractive index is only for light.” — It is the ratio of the speeds of any wave in two regions. Here the wave is a water wave and the two regions are deep and shallow water.
“n = 0.42 − 0.30 = 0.12.” — A refractive index is a ratio: divide one speed by the other.

11. NQ13 · Practice

Model answer. n = speed in air ÷ speed in glass, so the speed in glass = speed in air ÷ n = (3.0 × 10⁸ m/s) ÷ 1.65 = 1.82 × 10⁸ m/s, which is 1.8 × 10⁸ m/s to 2 significant figures. It is less than the speed in air, and less than the 2.0 × 10⁸ m/s in ordinary glass of refractive index 1.5: the bigger the refractive index, the slower the light.

Checked automatically

Speed of light in the glass: 1.82e8 m/s (within ±2%)
- Common wrong answer 4.95e8 : You multiplied by n. Light is slower in glass, so divide the speed in air by n.

A full answer includes

Method: I divided the speed in air by n: 3.0 × 10⁸ ÷ 1.65.
Answer: I wrote 1.8 × 10⁸ m/s (or 1.82 × 10⁸ m/s).
I did not multiply by n: light is slower in glass, not faster.

Also accept

1.8 × 10⁸ m/s
1.82 × 10⁸ m/s

Watch for

“Speed in glass = 3.0 × 10⁸ × 1.65 = 4.95 × 10⁸ m/s.” — Light is slower in glass than in air, so divide by n.

12. NQ10 · Practice

Model answer. n = sin i ÷ sin r = sin 56° ÷ sin 34° = 0.829 ÷ 0.559 = 1.48. A refractive index is a ratio, so it has no unit.

Checked automatically

Refractive index of the oil: 1.48 (within ±0.02)
- Common wrong answer 1.65 : That is 56 ÷ 34. Divide the sines: sin 56° ÷ sin 34°.
- Common wrong answer 0.67 : That is sin r ÷ sin i. Put the angle in the air on top.

A full answer includes

Method: I divided the sines: sin 56° ÷ sin 34° = 0.829 ÷ 0.559.
Answer: I wrote 1.48 (or 1.5).
I did not divide the angles themselves (56 ÷ 34 = 1.65).

Also accept

1.48

Watch for

“n = 56 ÷ 34 = 1.65.” — Divide the sines of the angles, not the angles themselves.
“n = sin 34° ÷ sin 56° = 0.67.” — Put the angle in the air on top: n = sin i ÷ sin r is bigger than 1.

13. NQ12 · Practice

Model answer. sin r = sin i ÷ n = sin 70° ÷ 1.55 = 0.940 ÷ 1.55 = 0.606, so r = sin⁻¹ 0.606 = 37.3°. It is smaller than 70°, as it should be for light going into glass.

Checked automatically

Angle of refraction (°): 37.3 (within ±0.5)
- Common wrong answer 45.2 : That is 70 ÷ 1.55. Divide the sine: sin r = sin 70° ÷ 1.55, then take sin⁻¹.
- Common wrong answer 0.606 : That is sin r. Use sin⁻¹ to turn it into an angle.

A full answer includes

Method: I found sin r = sin 70° ÷ 1.55 = 0.940 ÷ 1.55 = 0.606.
Method: I used sin⁻¹ to turn 0.606 into an angle.
Answer: I wrote 37.3° (or 37°).
I did not divide the angle itself by 1.55 (that gives 45.2°).

Also accept

37.3°

Watch for

“r = 70° ÷ 1.55 = 45.2°.” — Divide the sine of the angle, then take sin⁻¹; dividing the angle itself gives the wrong answer.
“sin r = 1.55 × sin 70°.” — That gives 1.46, and no angle has a sine bigger than 1. Going into glass, sin r = sin i ÷ n.

14. NQ3 · Practice

Model answer. (a) Work out sin i ÷ sin r for each row. 30°: 0.500 ÷ 0.342 = 1.46. 45°: 0.707 ÷ 0.469 = 1.51. 55°: 0.819 ÷ 0.643 = 1.27. 70°: 0.940 ÷ 0.629 = 1.49. Three ratios are close together, and 1.27 is far from them, so the reading at 55° is anomalous: its angle of refraction was probably misread, and should be about 33°. (b) Leave that row out and take the mean of the other three: (1.462 + 1.506 + 1.493) ÷ 3 = 1.487, which is 1.49 to 2 decimal places.

Checked automatically

(a) Angle of incidence of the anomalous reading (°): 55 (within ±0.5)
- Common wrong answer 70 : Check sin i ÷ sin r for each row. The 70° row gives 1.49, like the others.
(b) Refractive index of the plastic, to 2 decimal places: 1.49 (within ±0.01)
- Common wrong answer 1.43 : You included the 55° reading. Its ratio, 1.27, is far from the others, so leave it out of the mean.
- Common wrong answer 1.63 : You divided the angles. Divide their sines.
- Common wrong answer 1.54 : You took the mean of the angles first. Work out sin i ÷ sin r for each row, then take the mean.
- Common wrong answer 0.67 : That is sin r ÷ sin i. Put the angle in the air on top.

A full answer includes

(a) Method: I worked out sin i ÷ sin r for every row.
(a) Answer: I wrote 55°.
(b) Method: I left out the 55° row.
(b) Method: I took the mean of the other three ratios, 1.46, 1.51 and 1.49.
(b) Answer: I wrote 1.49.
I did not include the anomalous ratio, 1.27, in the mean.

Also accept

(a) 55° (b) 1.49

Watch for

“The 70° row is anomalous, because its angle of refraction, 39°, is smaller than the one before.” — Check sin i ÷ sin r for each row. The 70° row gives 1.49, like the 30° and 45° rows; the 55° row gives 1.27.
“Every reading goes into the mean.” — An anomalous reading is left out of the mean. With it, the mean drops to 1.43.
“n = i ÷ r, the ratio of the angles.” — The ratio of the angles changes from row to row. It is the ratio of their sines that stays the same.

15. NT1 · Apply

Model answer. (a) Light passes from water into air, so it speeds up and bends away from the normal. (b) In the empty mug, light travels in straight lines, and the straight line from the coin to your eye is blocked by the rim. In water, the ray leaves the coin at a steeper angle, reaches the surface and bends away from the normal, so it passes over the rim and into your eye. (c) The angle in the air is 52°, and n = sin(angle in air) ÷ sin(angle in water), so sin(angle in water) = sin 52° ÷ 1.33 = 0.788 ÷ 1.33 = 0.592, and the angle in the water is sin⁻¹ 0.592 = 36.3°.

Checked automatically

(c) Supplement: Angle to the normal in the water (°): 36.3 (within ±0.5)
- Common wrong answer 39.1 : That is 52 ÷ 1.33. Divide the sine, then take sin⁻¹.
- Common wrong answer 0.592 : That is the sine of the angle. Use sin⁻¹ to turn it into an angle.

A full answer includes

(a) I wrote away from the normal.
(a) I gave the reason: light speeds up as it passes from water into air.
(b) I wrote that the ray from the coin bends at the water surface.
(b) I wrote that the bent ray now passes over the rim into my eye.
(b) I wrote that before, the straight line from the coin to my eye was blocked by the rim.
(c) Method: I found sin(angle in water) = sin 52° ÷ 1.33 = 0.788 ÷ 1.33 = 0.592.
(c) Method: I used sin⁻¹ to turn 0.592 into an angle.
(c) Answer: I wrote 36.3° (or 36°).
I did not write that the water lifts the coin up.

Also accept

(a) away from the normal (b) the ray bends at the surface and passes over the rim (c) 36.3°

Watch for

“The water lifts the coin up.” — The coin does not move. The light from it changes direction at the surface.
“The angle in the water is bigger than 52°.” — Light is slower in water, so the ray in the water is closer to the normal: smaller than 52°.