Refraction at Plane Surfaces and Total Internal Reflection – Reflection and Refraction of Light (Senior 4 Physics) | Ganzaa

Reflection and Refraction of Light

Lessons

Senior 4 · Physics

Reflection and Refraction of Light

5 lessons

Lesson 2 of 5

Refraction at Plane Surfaces and Total Internal Reflection

25 min

Why does the bottom of a pool look so close?

Stand at the edge of a clear swimming pool and the bottom looks shallower than it is; a pupil who jumps in expecting water at waist height gets a surprise. A spear-fisher who aims straight at the fish he sees misses. On a hot afternoon the tarmac ahead seems to be covered with water that is never there when you arrive. All three effects come from one fact you met in S3: light changes direction when it crosses from one medium into another at an angle. This lesson turns that fact into formulas you can use: for a glass block, for apparent depth, and for the total internal reflection that carries internet signals through optical fibres.

Snell's law in its general form

When light passes from a medium of refractive index $n_{1}$ into one of index $n_{2}$ , with angle of incidence $\theta_1$ and angle of refraction $\theta_2$ (both measured from the normal),

n_1 \sin\theta_1 = n_2 \sin\theta_2

The absolute refractive index of a medium is $n = \frac{c}{v}$ , the speed of light in a vacuum divided by its speed in the medium, so $n \ge 1$ . For air $n = 1.0003$ , which we take as 1.00. The relative refractive index from medium 1 to medium 2 is ${}_1n_2 = \frac{n_2}{n_1} = \frac{v_1}{v_2}$ . Light bends towards the normal when it slows down ( $n_2 \gt n_1$ ) and away from it when it speeds up. The frequency never changes; the wavelength changes in proportion to the speed.

Medium	Refractive index $n$ (yellow light)	Speed of light ( $\times 10^{8}\,\mathrm{m\,s^{-1}}$ )
air	1.00	3.00
water	1.33	2.26
crown glass	1.52	1.97
diamond	2.42	1.24

In this lesson "glass" means a typical laboratory glass with $n = 1.50$ unless a value is given.

Definition

The absolute refractive index $n$ of a medium is the ratio $n = \frac{c}{v}$ , where $c$ is the speed of light in a vacuum and $v$ its speed in the medium. It has no unit. The relative refractive index ${}_1n_2 = \frac{n_2}{n_1}$ compares two media. Example: from water to glass, ${}_wn_g = \frac{1.50}{1.33} = 1.13$ .

Experiment 2.1: a ray through a rectangular glass block

Apparatus: a rectangular glass block, a sheet of white paper on a soft board, four optical pins, a protractor and a ruler.

Draw round the block. Draw a normal at a point P on one long side and a line at $i = 20^\circ$ to it. Stick two pins on this line.
Looking through the opposite face, stick two more pins in line with the images of the first two. Remove the block and join the points to trace the ray through the block; measure $r$ .
Repeat for $i = 30^\circ, 40^\circ, \ldots, 70^\circ$ , and plot $\sin i$ against $\sin r$ . The gradient is $n$ .

You will also see that the ray leaves the block parallel to the ray that entered, but shifted sideways (Figure 2.1).

Figure 2.1: Lateral displacement d by a parallel-sided block. The ray at Q makes the same angle i with the normal as the ray at P.

Why parallel? At P, $\sin i = n \sin r$ . The two faces are parallel, so the ray meets the bottom face at the same angle $r$ , and there $n \sin r = \sin e$ , where $e$ is the angle of emergence. So $\sin e = \sin i$ and $e = i$ .

How far is it shifted? The path inside the block is $PQ = \frac{t}{\cos r}$ , and the ray turns through $(i - r)$ at P. The lateral displacement is the perpendicular from Q to the incident ray produced:

d = PQ \sin(i - r) = \frac{t \sin(i - r)}{\cos r}

It grows with the thickness $t$ , the refractive index and the angle of incidence, and is zero at normal incidence.

$i$ (°)	20	30	40	50	60	70
$r$ (°)	13.0	19.5	25.5	30.5	35.5	39.0
$\sin i$	0.342	0.500	0.643	0.766	0.866	0.940
$\sin r$	0.225	0.334	0.431	0.508	0.581	0.629

Table 2.1: Illustrative results for one block.

Worked example 1: from the table to $n$

The points of Table 2.1 lie close to a straight line through the origin. Taking two well-separated points on the best-fit line, (0.200, 0.304) and (0.600, 0.897):

n = \text{gradient} = \frac{0.897 - 0.304}{0.600 - 0.200} = \frac{0.593}{0.400} = 1.48

A straight line through the origin confirms Snell's law ( $\sin i \propto \sin r$ ). A pin method locates each ray to about ±0.5°, so the answer $n = 1.48$ agrees with the expected 1.50 within the experimental uncertainty.

Worked example 2: lateral displacement

A ray strikes a glass block 6.0 cm thick ( $n = 1.50$ ) at $50^\circ$ to the normal. Find the lateral displacement.

\sin r = \frac{\sin 50^\circ}{1.50} = 0.511 \;\Rightarrow\; r = 30.7^\circ

d = \frac{t\sin(i - r)}{\cos r} = \frac{6.0 \times \sin 19.3^\circ}{\cos 30.7^\circ} = \frac{6.0 \times 0.330}{0.860} = 2.3\,\mathrm{cm}

Real and apparent depth

In Figure 2.2 an object O lies a real depth $N O$ below the surface. The vertical ray ON passes straight out. A second ray OM, close to the vertical, bends away from the normal as it leaves the water, and the eye traces it straight back to I. The eye sees the object at I, at the apparent depth NI.

Figure 2.2: Rays from O seem to come from I, nearer the surface. The angles are drawn larger than in real viewing to make them visible.

Light travels from water to air, so $n \sin i = \sin r$ (here $i$ is in the water and $r$ in the air). In the two right-angled triangles, $\tan i = \frac{NM}{NO}$ and $\tan r = \frac{NM}{NI}$ . Dividing,

\frac{NO}{NI} = \frac{\tan r}{\tan i}

When you look almost straight down, both angles are small, and $\tan\theta \approx \sin\theta$ . Then $\frac{\tan r}{\tan i} \approx \frac{\sin r}{\sin i} = n$ , so

n = \frac{\text{real depth}}{\text{apparent depth}} \qquad\text{and the apparent shift is}\qquad s = t\left(1 - \frac{1}{n}\right)

for a layer of thickness $t$ . For several layers stacked on top of each other, the shifts add: the apparent depth is $\frac{t_1}{n_1} + \frac{t_2}{n_2} + \cdots$ . Viewed at an angle, the formula no longer holds and the object appears even shallower; that is why the spear-fisher misses.

Measuring it. Focus a travelling microscope on a mark on the bench (reading $x_{1}$ ); place a glass block over the mark and refocus on the mark seen through the block ( $x_{2}$ ); sprinkle a little chalk dust on the top of the block and focus on it ( $x_{3}$ ). Then real depth $= x_{3} - x_{1}$ , apparent depth $= x_{3} - x_{2}$ , and $n = \frac{x_3 - x_1}{x_3 - x_2}$ .

Worked example 3: a fish and a two-layer tank

(a) A fish swims 1.20 m below the surface of a lake ( $n = 1.33$ ). How deep does it appear to a person looking straight down?

\text{apparent depth} = \frac{1.20\,\mathrm{m}}{1.33} = 0.902\,\mathrm{m} \approx 0.90\,\mathrm{m}

(b) A mark on the bottom of a tank is covered by a 3.0 cm glass slab ( $n = 1.50$ ) and, on top of it, 4.0 cm of water ( $n = 1.33$ ). Find its apparent depth below the water surface.

\text{apparent depth} = \frac{3.0}{1.50} + \frac{4.0}{1.33} = 2.0 + 3.0 = 5.0\,\mathrm{cm}

The mark is really 7.0 cm down, so it appears to be raised by 2.0 cm.

Critical angle and total internal reflection

When light goes from a denser medium (higher $n$ ) into a less dense one, it bends away from the normal, so the angle of refraction is always the larger. As the angle of incidence grows, the refracted ray swings towards the surface and grows weaker, while the reflected ray grows stronger (Figure 2.3).

Figure 2.3: Light inside a semicircular glass block, aimed at the centre of the flat face, so it enters the curved face without bending.

At the critical angle $C$ the refracted ray grazes the surface ( $\theta_2 = 90^\circ$ ). Snell's law gives $n_1 \sin C = n_2 \sin 90^\circ$ , so

\sin C = \frac{n_2}{n_1} \qquad\text{and, into air,}\qquad \sin C = \frac{1}{n}

For any larger angle there is no refracted ray at all: all the light is reflected. Two conditions must hold together: light must travel from the higher to the lower refractive index, and the angle of incidence must exceed $C$ . Total internal reflection reflects essentially 100% of the light, more than any silvered mirror, and the reflecting surface cannot tarnish.

Boundary	$\sin C$	$C$
water to air	$\frac{1}{1.33} = 0.752$	48.8°
glass to air	$\frac{1}{1.50} = 0.667$	41.8°
diamond to air	$\frac{1}{2.42} = 0.413$	24.4°
glass to water	$\frac{1.33}{1.50} = 0.887$	62.5°

A diamond's small critical angle traps light inside it through many reflections before it escapes from the top facets, which is why a well-cut diamond sparkles.

Definition

The critical angle $C$ is the angle of incidence, in the medium of higher refractive index, for which the angle of refraction is $90^\circ$ : $\sin C = \frac{n_2}{n_1}$ . Total internal reflection is the complete reflection of light at a boundary with a medium of lower refractive index when the angle of incidence is greater than $C$ . Example: for glass in air, $C = 41.8^\circ$ , so light meeting a glass–air face at $45^\circ$ from inside is totally reflected.

Applications of total internal reflection

Optical fibres. A fibre has a thin glass core of index $n_{1}$ surrounded by cladding of slightly lower index $n_{2}$ (Figure 2.4). Light that enters the end within a cone of half-angle $\theta_a$ , the acceptance angle, meets the core–cladding boundary at more than $C$ and is trapped, reflecting thousands of times per metre with almost no loss. The cladding protects the reflecting surface: two touching bare fibres would leak light into each other.

Figure 2.4: A step-index optical fibre. The angles are exaggerated; real acceptance angles are about 10° to 15°.

To find $\theta_a$ , apply Snell's law at the end face: $1 \times \sin\theta_a = n_1 \sin(90^\circ - C) = n_1 \cos C$ . With $\sin C = \frac{n_2}{n_1}$ , $\cos C = \sqrt{1 - \frac{n_2^2}{n_1^2}}$ , so

\sin\theta_a = \sqrt{n_1^2 - n_2^2}

This quantity is the fibre's numerical aperture. Fibres carry telephone calls, internet data and television as pulses of infrared light; bundles of fibres also carry images in medical endoscopes.

Totally reflecting prisms. A right-angled isosceles glass prism has angles of 45°, 45° and 90°. Light that enters a face at right angles meets the next face at 45°, which is more than glass's 41.8°, so it is totally reflected (Figure 2.5). One prism turns light through 90° (periscopes, cameras); with the light entering the long face it turns it through 180° (binoculars, cycle reflectors, road studs). A prism made of a material with $n \lt \sqrt{2} = 1.414$ would not work, because its critical angle would exceed 45°.

Figure 2.5: Totally reflecting prisms. The light enters and leaves at right angles to the faces, so it is not deviated there.

Mirages. Over a hot road, the air near the surface is hotter and less dense than the air above, so its refractive index is slightly lower. A ray from the sky or a distant object heading down at a shallow angle passes through layers of ever lower index, bends further from the normal at each, and finally turns upwards: a gradual form of total internal reflection (Figure 2.6). The brain assumes light travels in straight lines, so it sees an inverted image apparently below the road, like a reflection in a pool.

Figure 2.6: A road mirage: rays curve upwards in the hot air near the road.

A mirage over a hot road. Below each vehicle you can see an upside-down image of it, as if the road were wet; the "water" is really light from the vehicles and the bright sky bent upwards by the hot air just above the tarmac. Photo: Yuri Khristich, CC0, via Wikimedia Commons.

Prism binoculars from the First World War, seen from the eyepiece end. The box-shaped housings behind the eyepieces hold right-angled glass prisms that turn the light back on itself by total internal reflection, so long light paths fit into a short instrument. Photo: Torfaen Corvine, CC BY-SA 4.0, via Wikimedia Commons.

Worked example 4: an optical fibre

A fibre has a core of index 1.48 and cladding of index 1.46. Find (a) the critical angle at the core–cladding boundary, (b) the acceptance angle in air, and (c) the extra time taken over 1.00 km by a ray that zigzags at the critical angle, compared with a ray along the axis.

\text{(a)}\;\; \sin C = \frac{1.46}{1.48} = 0.9865 \;\Rightarrow\; C = 80.6^\circ

\text{(b)}\;\; \sin\theta_a = \sqrt{1.48^2 - 1.46^2} = \sqrt{0.0588} = 0.242 \;\Rightarrow\; \theta_a = 14.0^\circ

(c) The axial ray takes $t_0 = \frac{L n_1}{c} = \frac{1000 \times 1.48}{3.00 \times 10^{8}} = 4.93 \times 10^{-6}\,\mathrm{s}$ . The zigzag path is longer by the factor $\frac{1}{\sin C} = \frac{n_1}{n_2}$ , so

\Delta t = t_0\left(\frac{n_1}{n_2} - 1\right) = 4.93 \times 10^{-6} \times 0.0137 = 6.8 \times 10^{-8}\,\mathrm{s} = 68\,\mathrm{ns}

A short pulse therefore spreads out as it travels. This limits how many pulses per second the fibre can carry, which is why long-distance links use very thin "single-mode" fibres in which only the axial path survives.

Warning

Common mistakes. (1) Expecting total internal reflection when light goes from air into glass or water: it can only happen going towards the lower index. (2) Measuring angles from the surface instead of from the normal. (3) Using $\sin C = \frac{1}{n}$ at a glass–water boundary: use $\sin C = \frac{n_2}{n_1}$ . (4) Applying real depth ÷ apparent depth = $n$ to oblique viewing: it holds only for near-normal viewing. Safety: in a ray-box or laser experiment never look along the beam; use a class 2 laser pointer or a ray box, and keep the beam below eye level.

Tip

For any refraction question, first write $n_1 \sin\theta_1 = n_2 \sin\theta_2$ with the medium labels. Total internal reflection is just the case where this equation asks for $\sin\theta_2 \gt 1$ : no refracted ray can exist.

Total internal reflection (MITx, with Professor Yen-Jie Lee). Watch the laser beam in the water tank: as the angle of incidence at the water surface grows, the refracted beam disappears and all the light is reflected.

✦

Practice

Light travels at $2.26 \times 10^{8}\,\mathrm{m\,s^{-1}}$ in water. Calculate the refractive index of water.
A ray in air strikes glass ( $n = 1.50$ ) at $35^\circ$ to the normal. Find the angle of refraction.
A coin lies at the bottom of a beaker containing 15.0 cm of water ( $n = 1.33$ ). Find the apparent depth and the apparent shift of the coin when viewed from directly above.
In a travelling-microscope experiment the readings are $x_1 = 2.24\,\mathrm{cm}$ (mark on bench), $x_2 = 3.96\,\mathrm{cm}$ (mark through block) and $x_3 = 7.38\,\mathrm{cm}$ (top of block). Find the refractive index of the block.
Calculate the critical angle for light going from glass ( $n = 1.52$ ) into water ( $n = 1.33$ ). Can a ray inside the glass meeting the glass–water face at $55^\circ$ escape into the water?
A glass block 8.0 cm thick ( $n = 1.50$ ) receives a ray at $60^\circ$ . Calculate the lateral displacement.
An optical fibre has a core of index 1.50 and an acceptance angle in air of $20^\circ$ . Find the refractive index of the cladding.
A student proposes making a periscope with right-angled isosceles prisms of a clear plastic with $n = 1.38$ . Evaluate the proposal.

Show answersHide answers

$n = \frac{3.00 \times 10^{8}}{2.26 \times 10^{8}} = 1.33$
$\sin r = \frac{\sin 35^\circ}{1.50} = 0.382$ , so $r = 22.5^\circ$ .
Apparent depth $= \frac{15.0}{1.33} = 11.3\,\mathrm{cm}$ ; shift $= 15.0 - 11.3 = 3.7\,\mathrm{cm}$ .
Real depth $= 7.38 - 2.24 = 5.14\,\mathrm{cm}$ ; apparent depth $= 7.38 - 3.96 = 3.42\,\mathrm{cm}$ ; $n = \frac{5.14}{3.42} = 1.50$ .
$\sin C = \frac{1.33}{1.52} = 0.875$ , so $C = 61.0^\circ$ . $55^\circ \lt 61.0^\circ$ , so yes, the ray refracts out into the water (with a weak reflection).
$\sin r = \frac{\sin 60^\circ}{1.50} = 0.577$ , $r = 35.3^\circ$ ; $d = \frac{8.0 \sin 24.7^\circ}{\cos 35.3^\circ} = \frac{8.0 \times 0.418}{0.816} = 4.1\,\mathrm{cm}$ .
$\sin 20^\circ = \sqrt{1.50^2 - n_2^2}$ : ${0.342}^{2} = 0.117 = 2.25 - n_{2}^{2}$ , so $n_{2}^{2} = 2.133$ and $n_{2} = 1.46$ .
For $n = 1.38$ , $\sin C = \frac{1}{1.38} = 0.725$ and $C = 46.4^\circ$ . The light meets the hypotenuse at $45^\circ$ , which is less than $C$ , so most of it refracts out instead of being totally reflected; the periscope would give a dim image. The prisms need $n \gt 1.414$ (ordinary glass works); otherwise the hypotenuse must be silvered.

Quick check

Light travelling from air into water can be totally internally reflected if the angle of incidence is large enough.

Key points

$n_1\sin\theta_1 = n_2\sin\theta_2$ , with $n = \frac{c}{v}$ .
A parallel-sided block displaces a ray sideways by $d = \frac{t\sin(i - r)}{\cos r}$ without changing its direction.
For near-normal viewing, $n = \frac{\text{real depth}}{\text{apparent depth}}$ ; layers add as $\sum \frac{t_k}{n_k}$ .
Total internal reflection needs light going towards a lower index at more than $C$ , where $\sin C = \frac{n_2}{n_1}$ . It works optical fibres ( $\sin\theta_a = \sqrt{n_1^2 - n_2^2}$ ), prisms in binoculars and periscopes, and mirages.

Sources and credits

25.3: The Law of Refraction and 25.4: Total Internal Reflection, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Mirage over a hot road.jpg by Yuri Khristich, Wikimedia Commons, CC0 1.0.
Photo: French Binoculars WW1, No D.13, 8x24.jpg by Torfaen Corvine, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: Total Internal Reflection, MITx Videos, embedded from YouTube.

The experiments and their illustrative results, the six diagrams, the derivations, the worked examples and the practice exercises are written by Ganzaa.

1Why does the bottom of a pool look so close?

2Snell's law in its general form

3Experiment 2.1: a ray through a rectangular glass block

4Real and apparent depth

5Critical angle and total internal reflection

6Applications of total internal reflection