Answers · Refraction and total internal reflection

Refraction and total internal reflection — final check key

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

1. NK1 · Final check

Model answer. (a) The angle of incidence is the angle between the incident ray and the normal, the line at 90° to the boundary where the ray meets it. (b) She marks the ray that leaves each block with two pencil crosses, well apart, takes the block away and joins the crosses, so the line shows the direction of the ray that left. The two sides of the rectangular block that the ray crosses are parallel, so the bend out undoes the bend in: the ray leaves parallel to the ray that went in, shifted sideways. The two sides of the prism are not parallel, so the bends do not cancel: the ray leaves turned through an angle. (c) The refractive index is the ratio of the speeds: the speed of light in air divided by the speed of light in the material. (d) n = sin i ÷ sin r = sin 43° ÷ sin 26° = 0.682 ÷ 0.438 = 1.56.

A full answer includes

(a) I wrote: the angle between the incident ray and the normal.
(b) I wrote that she marks the ray that leaves with two crosses, well apart.
(b) I wrote that she joins the crosses once the block has been taken away.
(b) I wrote that the ray leaving the rectangular block is parallel to the ray that went in.
(b) I wrote that it is shifted sideways.
(b) I wrote that the ray leaving the prism is not parallel to the ray that went in.
(b) I wrote that the prism turns it through an angle.
(c) I wrote: the speed of light in air (or a vacuum) divided by the speed of light in the material.
(d) Method: I divided the sines: sin 43° ÷ sin 26° = 0.682 ÷ 0.438.
(d) Answer: I wrote 1.56 (or 1.6).
I did not write that the ray leaving the prism is parallel to the ray that went in.

Also accept

(a) between the incident ray and the normal (b) two crosses, joined once the block is removed; the block: parallel, shifted sideways; the prism: turned through an angle (c) speed in air ÷ speed in the material (d) 1.56

Watch for

“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, so the ray is turned.
“n = speed in the material ÷ speed in air.” — Light is slower in the material, so that ratio is less than 1. The speed in air goes on top.

2. NK2 · Final check

Model answer. (a) Below the critical angle, most of the light refracts out into the air, bending away from the normal, and a weak ray is reflected back inside the bead: internal reflection. (b) At the critical angle the angle of refraction is 90°, so the ray in the air runs along the surface of the bead. (c) Above the critical angle no light gets out: all of it is reflected back inside the bead, with the angle of reflection equal to the angle of incidence. This is total internal reflection, used in optical fibres, endoscopes and the prisms of periscopes and binoculars. (d) sin c = 1 ÷ n = 1 ÷ 1.46 = 0.685, so c = sin⁻¹ 0.685 = 43.2°. (e) The call is coded digitally into very fast pulses of light, usually infrared, as a pattern of on and off. The pulses are totally internally reflected along the core of the fibre, and at the far end a detector changes them back into an electrical signal.

A full answer includes

(a) I wrote that most of the light refracts out into the air.
(a) I wrote that it bends away from the normal.
(a) I wrote that a weak ray is reflected back inside the bead.
(a) I called this weak reflection internal reflection.
(b) I wrote that the ray in the air runs along the surface.
(b) I wrote that its angle of refraction is 90°.
(c) I wrote that no light gets out.
(c) I wrote that all of it is reflected back inside the bead.
(c) I called this total internal reflection.
(c) I identified a device, such as an optical fibre, an endoscope or a prism periscope.
(d) Method: I found sin c = 1 ÷ 1.46 = 0.685.
(d) Method: I used sin⁻¹ to turn 0.685 into an angle.
(d) Answer: I wrote 43.2° (or 43°).
(e) I wrote that the call travels as very fast pulses of light, usually infrared.
(e) I wrote that the pulses code the call digitally, as a pattern of on and off.
(e) I wrote that at the far end a detector changes the pulses back into an electrical signal.
I did not write that below the critical angle no light is reflected.

Also accept

(a) most refracts out, away from the normal; a weak ray is reflected inside (b) along the surface, at 90° to the normal (c) total internal reflection, e.g. an optical fibre (d) 43.2° (e) digital pulses of light, changed back into an electrical signal by a detector

Watch for

“Below the critical angle, no light is reflected.” — Some light is always reflected inside; it is weak below the critical angle.
“At the critical angle, all of the light is reflected.” — At the critical angle the ray in the air runs along the surface. All of the light is reflected only at bigger angles.
“The call travels along the fibre as an electric current.” — An optical fibre carries pulses of light. The call is changed into light at one end and back into an electrical signal at the other.