Answers · Refraction and total internal reflection

Refraction and total internal reflection — return questions key

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

1. NR1 · Return

Model answer. (a) The angle of refraction is the angle between the refracted ray and the normal, the line at 90° to the boundary where the ray crosses it. (b) Along the normal, the angle of incidence is 0°: the ray goes straight on into the glass without changing direction, although it slows down. (c) The ray that leaves is not parallel to the ray that went in. The two sides of the prism are not parallel, so their normals point in different directions and the bend out does not undo the bend in: the ray is turned through an angle. (d) n = sin i ÷ sin r, so sin i = n × sin r = 1.42 × sin 24° = 1.42 × 0.407 = 0.578, and i = sin⁻¹ 0.578 = 35.3°. It is bigger than 24°, as it should be: the ray in the air is further from the normal than the ray in the plastic.

A full answer includes

(a) I wrote: the angle between the refracted ray and the normal.
(b) I wrote that it goes straight on, without changing direction.
(b) I wrote that it still slows down.
(c) I wrote that it is not parallel.
(c) I gave the reason: the two sides it crosses are not parallel.
(c) I wrote that the two bends do not cancel.
(d) Method: I found sin i = n × sin r = 1.42 × sin 24° = 1.42 × 0.407 = 0.578.
(d) Method: I used sin⁻¹ to turn 0.578 into an angle.
(d) Answer: I wrote 35.3° (or 35°).
I did not divide sin 24° by 1.42 (that gives 16.6°, smaller than the angle of refraction).

Also accept

(a) between the refracted ray and the normal (b) straight on, slower (c) not parallel: the sides are not parallel (d) 35.3°

Watch for

“The angle of refraction is the angle between the refracted ray and the surface.” — It is measured from the normal.
“sin i = sin 24° ÷ 1.42.” — That gives 16.6°, smaller than the angle in the plastic. The angle in the air is the bigger one: sin i = n × sin r.
“i = 1.42 × 24° = 34.1°.” — Multiply the sine of the angle by n, then take sin⁻¹.
“The ray leaving a prism is parallel to the ray that went in, as for a block.” — Only parallel sides make the two bends cancel. The sides of a prism are at an angle to each other.

2. NR2 · Return

Model answer. (a) The critical angle is the angle of incidence, inside a medium such as glass, at which the angle of refraction is 90°: the refracted ray runs along the boundary. (b) The light must be inside the medium in which it travels more slowly, meeting a boundary with one in which it travels faster, and the angle of incidence must be bigger than the critical angle. (c) The light inside the plastic meets the back faces of the prisms, with air behind them, at angles of incidence bigger than the critical angle of the plastic. So it is totally internally reflected, from one back face to another, and turned back out through the front towards the car. Total internal reflection reflects all of the light, so no silver layer is needed. (d) n = 1 ÷ sin c = 1 ÷ sin 39° = 1 ÷ 0.629 = 1.59. The speed of light in the plastic = speed in air ÷ n = (3.0 × 10⁸ m/s) ÷ 1.59 = 1.9 × 10⁸ m/s. (e) Inside each fibre the light meets the side of the core at 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 follows the gentle bends of the endoscope and comes out of the far end, with very little lost on the way.

A full answer includes

(a) I wrote: the angle of incidence, inside the medium, at which the angle of refraction is 90°.
(b) I wrote the first condition: the light is inside the slower medium, meeting a faster one.
(b) I wrote the second condition: the angle of incidence is bigger than the critical angle.
(c) I wrote that the light inside the plastic meets the back faces, with air behind them.
(c) I wrote that it meets them at more than the critical angle.
(c) I wrote that it is totally internally reflected.
(c) I wrote that total internal reflection reflects all of the light, so no silver layer is needed.
(d) Method: I found n = 1 ÷ sin 39° = 1 ÷ 0.629.
(d) Answer: I wrote n = 1.59 (or 1.6).
(d) Method: I divided the speed in air by n: 3.0 × 10⁸ ÷ 1.59.
(d) Answer: I wrote 1.9 × 10⁸ m/s (or 1.89 × 10⁸ m/s).
(e) I wrote that the light meets the side of each fibre’s core at more than the critical angle.
(e) I wrote that it is totally internally reflected.
(e) I wrote that this happens again and again, so the light follows the gentle bends.
I did not write that the critical angle is the angle at which all the light is reflected.

Also accept

(a) angle of incidence giving an angle of refraction of 90° (b) slower to faster medium, bigger than c (c) total internal reflection at the back faces, at more than c (d) 1.59; 1.9 × 10⁸ m/s (e) total internal reflection again and again along each fibre

Watch for

“The critical angle is when all the light is reflected.” — At the critical angle the ray in the air runs along the boundary; all the light is reflected only beyond it.
“The reflector works because its back is shiny, like a mirror.” — There is no silver layer. The light is totally internally reflected at the boundary between the plastic and the air behind it.
“n = sin 39° = 0.63.” — That is sin c. The refractive index is 1 ÷ sin c, bigger than 1.
“Light goes round the bends because the fibres are mirrors inside.” — The light is totally internally reflected at the boundary of each core, again and again; there is no mirror.