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.
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.
Air → glass (less dense → more dense): light slows down and bends towards the normal. So $r < i$.
Glass → air: light speeds up and bends away from the normal.
Along the normal ($i=0$), the ray does not bend; it still changes speed.
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.
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)
Rectangular block — the standard $i$ and $r$ at two parallel faces; ray emerges parallel to the original direction but laterally shifted.
Semi-circular block — ray aimed at the curved face along a radius hits the flat face from inside glass, so you can find the critical angle.
Triangular prism — deviation and, with white light, dispersion into a spectrum. Also TIR in reflecting prisms.
Investigate $n$ of glass (3.19)
Apparatus: ray box, single slit, rectangular glass block, white paper, sharp pencil, protractor, ruler.
Method I mark
Draw around the block. Mark the point of incidence. Draw the normal.
Send a narrow ray in. Mark two dots on the incident ray and two on the emergent ray.
Remove the block. Join the dots. Draw the path inside the glass.
Measure $i$ and $r$ from the normal. Calculate $n=\sin i/\sin r$.
Repeat for at least five angles. Average $n$.
Errors and how to reduce them
Thick pencil lines — use a sharp pencil and a narrow slit.
Protractor not on the normal — draw a long, faint normal first.
Block slips — hold it or use Blu Tack.
Very small $i$ — $\sin i$ and $\sin r$ are both small, so percentage error is large. Use a range such as $20^\circ$–$60^\circ$.
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
The ray travels from a denser medium towards a rarer medium (glass → air).
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?
Denser to rarer, and $i>c$. Fibres carry light pulses (information), not a physical lamp. (A joke check — the serious point is TIR of light along the fibre.)