Coaching Centre
Snell’s law, total internal reflection and light intensity
Refractive index and the speed of light in a medium, Snell's law, the practical for measuring refractive index, critical angle and total internal reflection, optical fibres, dispersion, and the inverse square law for light intensity.
Almost every byte of internet traffic between Australia and the rest of the world travels as pulses of infrared light inside glass fibres thinner than a hair, laid across the floor of the Pacific and Indian Oceans. The light stays trapped inside the glass for thousands of kilometres, bouncing off the inside wall billions of times without escaping. The trick is a single angle, and you can calculate it in one line.
Refractive index
Light slows down when it enters a transparent material. The absolute refractive index $n$ of a medium compares the speed of light in a vacuum, $c$, with its speed $v$ in the medium:
$n = \dfrac{c}{v}$
So $n$ has no unit and is never less than 1. Air is 1.0003, close enough to 1.00 for HSC purposes.
| Medium | $n$ (yellow light) | Speed of light ($\times 10^8$ m/s) |
|---|---|---|
| Vacuum (air ≈) | 1.00 | 3.00 |
| Water | 1.33 | 2.26 |
| Perspex (acrylic) | 1.49 | 2.01 |
| Crown glass | 1.52 | 1.97 |
| Diamond | 2.42 | 1.24 |
Snell's law
As for water waves (page 3C), a ray entering a medium where light is slower bends towards the normal, and one entering a faster medium bends away. The exact relationship is Snell's law:
$n_1 \sin\theta_1 = n_2 \sin\theta_2$
where both angles are measured from the normal. Using $n = c/v$, this is the same as $\dfrac{\sin\theta_1}{\sin\theta_2} = \dfrac{v_1}{v_2}$. The frequency is unchanged, so the wavelength in the medium is $\lambda_{\text{medium}} = \lambda_{\text{vacuum}}/n$.
A laser beam in air strikes the surface of a pool at 50° to the normal. Find the angle of refraction and the speed of the light in the water ($n = 1.33$).
Light travels from glass ($n = 1.52$) into water ($n = 1.33$) at 30° to the normal. Find the angle of refraction.
Critical angle and total internal reflection
When light travels from a higher to a lower refractive index (glass to air, say), it bends away from the normal. As the angle of incidence increases, the refracted ray gets closer to the surface, and more of the light is reflected. At the critical angle $\theta_c$ the refracted ray travels along the boundary ($\theta_2 = 90°$). Putting $\theta_2 = 90°$ into Snell's law:
$\sin\theta_c = \dfrac{n_2}{n_1} \qquad (n_1 \gt n_2)$
For any angle larger than $\theta_c$ there is no refracted ray at all: all the light reflects back into the first medium. This is total internal reflection (TIR). Two conditions are needed:
- the light goes from a higher to a lower refractive index, and
- the angle of incidence is greater than the critical angle.
Find the critical angle for (a) glass ($n = 1.52$) to air, (b) diamond ($n = 2.42$) to air, and (c) glass to water ($n = 1.33$).
Optical fibres
An optical fibre has a glass core surrounded by cladding of slightly lower refractive index, inside a protective coating. Light sent in near the axis meets the core–cladding boundary at more than the critical angle every time, so it is totally internally reflected all the way along, even around gentle bends.
Why use cladding instead of plain glass in air? It protects the core surface (scratches, dirt or a finger touching bare glass would let light leak out) and stops light crossing between fibres in a bundle. Fibres carry phone and internet data (the Southern Cross and Indigo cables link Australia to the US and Asia), and bundles of them make endoscopes for seeing inside the body without surgery.
A fibre has core index 1.50 and cladding index 1.46. Find the critical angle at the boundary, and the largest angle a ray inside the core can make with the fibre's axis and still be guided.
Dispersion
The refractive index of glass is slightly different for each colour: larger for violet than for red. So violet light slows more and bends more, and a prism spreads white light into a spectrum. This is dispersion. Rainbows form the same way, by refraction and internal reflection in raindrops. In optical fibres, dispersion spreads out pulses, which limits how fast data can be sent.
The inverse square law for light
A small light source radiates its power in all directions. As for sound (page 3E), the power spreads over a sphere of area $4\pi r^2$:
$I = \dfrac{P}{4\pi r^2} \quad \Rightarrow \quad I_1 r_1^2 = I_2 r_2^2$
The intensity of sunlight at Earth (1.00 AU from the Sun) is about 1360 W m$^{-2}$. Mars is 1.52 AU from the Sun. Find the intensity at Mars.
To test the law, measure the light intensity from a small bulb with a light meter (or phone light sensor) in a dark room at several distances, subtract the background reading, and plot $I$ against $1/r^2$. The law only works for a source that is small compared with the distance; very close to a long fluorescent tube, it fails.
- At the start (air into glass at 40°) the readout shows θ₂ = 25.0° and a speed in glass of 1.97 × 10⁸ m/s.
- Choose from Glass into Air. At 40° the light leaves at 77.7°; the critical angle readout shows 41.1°. At 45° there is no refracted ray.
- Find the critical angles for water to air (48.8°) and diamond to air (24.4°).
- In Measure n, record at least five angles and read the gradient of the sin i against sin r graph (close to 1.50 for the first block). Press New block and measure again.
- In Optical fibre (core 1.50, cladding 1.46) a 10° launch angle is guided; at 30° light leaks out. Find the largest launch angle that still works (about 20°).
- With a 10 W lamp the meter reads 0.796 W/m² at 1.0 m. Drag the meter to 2.0 m: 0.199 W/m², a quarter.
- Turn on Plot I against 1/r². What shape is the graph, and what does its gradient represent?
- Set 100 W at 2.0 m: 1.99 W/m². Switch to Speaker. Does anything about the law change?
Test yourself
Light passes from air into a liquid at 60° to the normal and refracts at 40°. The refractive index of the liquid is
$n = \sin 60°/\sin 40° = 0.866/0.643 = 1.35$.
The critical angle of a plastic in air is 42°. Its refractive index is
$n = 1/\sin 42° = 1/0.669 = 1.49$.
Which situation can produce total internal reflection?
TIR needs light heading into a lower index at more than $\theta_c$. Water to air has $\theta_c = 48.8°$, and 60° exceeds it. Water to glass is into a higher index.
A light meter reads 36 lux at 0.50 m from a small lamp. At 1.5 m it should read
Three times the distance gives one ninth of the intensity: $36/9 = 4.0$.
Red light of wavelength 650 nm in air enters glass of $n = 1.50$. In the glass its
Frequency is fixed by the source; $\lambda = 650/1.50 = 433$ nm. (It still looks red: colour depends on frequency.)
Label the optical fibre.
Order these media from slowest to fastest speed of light.
The refractive index of a medium is the speed of light in a vacuum divided by the of light in the medium. Light entering a medium with a higher refractive index bends the normal. Total internal reflection needs light travelling into a medium of refractive index at an angle greater than the angle. A prism splits white light because light has the largest refractive index.
A student shines a ray into a semicircular perspex block and records these angles: incidence in air 20°, 40°, 60°; refraction in the block 13.3°, 25.6°, 35.5°. (a) Use the data to determine the refractive index of perspex. (b) Calculate the critical angle for perspex in air. (c) Explain why the ray is not refracted where it leaves the curved surface. (d) Evaluate the use of a semicircular block for this experiment and explain how a graph would improve the result.
(a) $n = \dfrac{\sin i}{\sin r}$: $\dfrac{\sin 20°}{\sin 13.3°} = 1.49$, $\dfrac{\sin 40°}{\sin 25.6°} = 1.49$, $\dfrac{\sin 60°}{\sin 35.5°} = 1.49$. So $n = 1.49$.
(b) $\sin\theta_c = \dfrac{1.00}{1.49} = 0.671$, so $\theta_c = 42.2°$.
(c) The ray enters at the centre of the flat face, so inside the block it travels along a radius of the semicircle. A radius is perpendicular to the curved surface, so the ray meets that surface along the normal. With an angle of incidence of 0°, Snell's law gives an angle of refraction of 0°, so it leaves without bending, and the angle measured outside is the true angle of refraction inside.
(d) The semicircular block is a good choice: only one refraction needs to be measured, and the same block can be used to find the critical angle directly by aiming at the flat face from inside. Its weakness is that the ray must hit the exact centre of the flat face, or it bends again at the curved face and the measured angle is wrong; reading angles of a wide ray with a protractor also adds random error. Plotting $\sin i$ against $\sin r$ for many angles and drawing a line of best fit averages out random errors, gives $n$ as the gradient, and shows whether the line passes through the origin as Snell's law predicts, so a systematic error (such as an off-centre block) would show up.