Lesson C · 3.14–3.22

Light: reflection, refraction and TIR

Ray diagrams, Snell’s law, the glass-block practicals, critical angle and optical fibres.

The rule I write on the board first

Every angle in this lesson is measured from the normal — the dashed line at $90^\circ$ to the surface. Students who measure from the glass face get a “correct” protractor reading and a completely wrong $n$. I will keep saying “from the normal” until you are sick of it.

3.14–3.16

Light is transverse — and it reflects

Light waves are transverse and can be reflected and refracted (3.14). The law of reflection (3.15) is

$$i = r$$

Angle of incidence equals angle of reflection, both from the normal.

normal i r incident reflected plane mirror / surface

A full ray diagram (3.16) needs: a ruler-straight surface, a dashed normal at the point of incidence, arrows on both rays, and $i$ and $r$ marked from the normal — not from the mirror.

3.16–3.18

Refraction and Snell’s law

Refraction is a change of speed when a wave crosses a boundary. If it hits at an angle, the direction also changes.

Refractive index of the material (for air–glass) is

$$n = \frac{\sin i}{\sin r}$$

$i$ is in air, $r$ is in glass, both from the normal. Glass is typically about $1.5$. If you get $n=0.67$, you have inverted the fraction.

Board example

$i = 40^\circ$, $r = 25^\circ$.

$$n = \sin 40^\circ / \sin 25^\circ = 0.6428 / 0.4226 = 1.52$$

Degree mode on the calculator. I watch people lose both marks on radian mode every year.

3.17, 3.19 · Required practicals

Blocks, prisms, and measuring $n$

What each shape is for (3.17)

Investigate $n$ of glass (3.19)

Apparatus: ray box, single slit, rectangular glass block, white paper, sharp pencil, protractor, ruler.

Method I mark

  1. Draw around the block. Mark the point of incidence. Draw the normal.
  2. Send a narrow ray in. Mark two dots on the incident ray and two on the emergent ray.
  3. Remove the block. Join the dots. Draw the path inside the glass.
  4. Measure $i$ and $r$ from the normal. Calculate $n=\sin i/\sin r$.
  5. Repeat for at least five angles. Average $n$.

Errors and how to reduce them

3.20–3.22

Critical angle and total internal reflection

When light is in glass, heading for air, $r$ is larger than $i$. Increase $i$ and $r$ grows until $r=90^\circ$ — the ray skims along the boundary. That angle of incidence is the critical angle $c$ (3.21).

$$\sin c = \frac{1}{n}$$

For $n=1.50$, $\sin c = 0.667$, so $c = 41.8^\circ$ (use $\sin^{-1}$).

TIR happens only if both are true

  1. The ray travels from a denser medium towards a rarer medium (glass → air).
  2. The angle of incidence is greater than the critical angle ($i > c$).

Then there is no refracted ray — the light is totally reflected inside, and $i=r$ still holds.

Optical fibres (3.20)

Light enters one end of a glass fibre. At the glass–air (or core–cladding) boundary, $i > c$, so TIR keeps the pulse inside [1]. The pulse reflects along the fibre and carries information (telephone, broadband, endoscope) [1]. A 45° prism uses TIR to turn a ray through $90^\circ$ without a silvered mirror [1].

Traps

TIR cannot happen for a ray going air → glass. Writing only “$i > c$” without the denser-to-rarer condition loses a mark. Do not calculate potential-divider-style cladding ratios — not on this course.

Check you can

$i=48^\circ$, $r=30^\circ$. Calculate $n$. Then find $c$ for that $n$.

State the two conditions for TIR. Why are house lamps not sent down optical fibres as white-hot metal?

Worked examples →