3.2.2 Refraction of light

Learning objectives

3.2.2.1Terms normal, angle of incidence and• Define and use the terms normal, angle of incidence and angle of refraction 3.2.2.2An experiment to show refraction of• Describe an experiment to show refraction of light by transparent blocks of different shapes 3.2.2.3Passage of light through a transparent• Describe the passage of light through a transparent material (limited to the boundaries between two mediums only)3.2.2.4Meaning of critical angle• State the meaning of critical angle 3.2.2.5Internal reflection and total internal• Describe internal reflection and total internal reflection using both experimental and everyday examples 3.2.2.6Refractive index, n, as the ratio of• Define refractive index, n, as the ratio of the speeds of a wave in two different regions 3.2.2.7Refractive index: n = sin i / sin r• Recall/use refractive index: n = sin i / sin r 3.2.2.8Critical angle relation: n = 1 / sin c• Recall/use critical angle relation: n = 1 / sin c 3.2.2.9Use of optical fibres, particularly in• Describe the use of optical fibres, particularly in telecommunications

At the point where a ray meets a boundary, the normal is an imaginary line drawn at right angles to the boundary.

Term	Meaning
incident ray	ray travelling towards the boundary
refracted ray	transmitted ray travelling in the second region
angle of incidence, $i$	angle between the incident ray and the normal
angle of refraction, $r$	angle between the refracted ray and the normal

Draw the normal through the exact point of incidence and place the centre of a protractor there. Measure $i$ in the first region and $r$ in the second region, both from the normal.

If the incident ray makes $50^\circ$ with the boundary, then $i=90^\circ-50^\circ=40^\circ$ .

An angle drawn against the boundary is not $i$ or $r$ ; it is the complement of the corresponding angle to the normal.

A narrow ray and a traced transparent block make the change of direction at each boundary measurable.

Step	Method
1	place a rectangular, triangular or semicircular transparent block on plain paper and trace its outline
2	direct one narrow ray from a ray box at a chosen face and mark two points on the incident path and two on the emergent path
3	remove the block, join the marked points with a ruler and reconstruct the path through the outline
4	draw a normal at every boundary crossing and measure the angles from each normal
5	repeat with a different incidence angle and with blocks of different shapes

The ray bends towards the normal when it enters a region in which it travels more slowly and away from the normal when it enters one in which it travels faster. A ray entering along the normal does not change direction. A parallel-sided block gives an emergent ray parallel to the incident ray but shifted sideways; a non-parallel face can change the final direction.

Changing the block shape changes the orientation of its boundaries; it does not by itself change the refractive behaviour of the material. Keep the ray narrow and mark points far enough apart for accurate ruler lines.

Refraction is a change in wave direction caused by a change in wave speed at a boundary between two regions.

Passage across the boundary	Ray direction	Other wave properties
into a region where light travels more slowly	bends towards the normal, so $r<i$	frequency stays constant; wavelength decreases
into a region where light travels faster	bends away from the normal, so $r>i$	frequency stays constant; wavelength increases
along the normal, $i=0^\circ$	does not bend	speed and wavelength may still change

At the first face of a parallel-sided block the ray bends towards the normal; at the second, parallel face it bends away by the matching amount. The emerging ray is parallel to the incident ray but laterally displaced.

The ray bends at the boundary, not gradually throughout a uniform material. Frequency is fixed by the source and does not change when the ray crosses the boundary.

The critical angle $c$ is the angle of incidence in the region where light travels more slowly for which the refracted ray in the faster region is at $90^\circ$ to the normal.

At $i=c$ , the refracted ray travels along the boundary. The critical angle is measured from the normal inside the slower region, not from the boundary and not in the faster region.

Incident angle in the slower region	What happens at the boundary
$i<c$	a transmitted ray refracts into the faster region, with some reflection possible
$i=c$	the transmitted ray travels along the boundary
$i>c$	no transmitted ray emerges; total internal reflection occurs

The ray at the critical angle is not an example of total internal reflection because a transmitted ray still exists along the boundary.

Internal reflection is light reflected back into its original transparent region; total internal reflection (TIR) is the limiting case in which all the light is reflected and no refracted ray crosses the boundary.

TIR requires both conditions: light approaches a boundary from a region where it travels more slowly into one where it travels faster, and its angle of incidence is greater than the critical angle.

Experimental move	Observation
send a narrow ray into the curved face of a semicircular block along a radius	it reaches the flat face without bending at the curved face
increase $i$ at the flat block–air boundary	the refracted ray moves farther from the normal and the internally reflected ray remains
set $i=c$	the refracted ray runs along the flat surface
increase to $i>c$	the refracted ray disappears and only the internally reflected ray remains

A swimmer can see the water surface act like a mirror for rays striking it at sufficiently large angles. Right-angle glass prisms can also turn a ray through $90^\circ$ by TIR, as used in optical viewing devices.

A large incidence angle alone is not enough: light approaching from the faster region cannot undergo TIR at that boundary.

For a wave passing from region 1 into region 2, the relative refractive index is the ratio of its speed before the boundary to its speed after the boundary.

$n=\frac{v_1}{v_2}$

$v_1$ is the wave speed in the incident region and $v_2$ is its speed in the refracting region. For light travelling from air into a material, $v_1\approx3.0\times10^8\,\text{m s}^{-1}$ and $n\approx c/v$ .

If light travels at $3.0\times10^8\,\text{m s}^{-1}$ in air and $2.0\times10^8\,\text{m s}^{-1}$ in glass, then $n=(3.0\times10^8)/(2.0\times10^8)=1.5$ .

Refractive index has no unit because it is a ratio of two speeds with the same unit. Reversing the direction swaps the speed ratio, so identify regions 1 and 2 before substituting.

For light travelling from air into a transparent material, refractive index connects the two angles measured from the normal.

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

$i$ is the angle in air and $r$ is the angle in the material for this form of the equation. Check that both are measured from the normal, and put the calculator in degree mode.

For $i=46^\circ$ and $r=26^\circ$ , $n=\sin46^\circ/\sin26^\circ=1.64$ (3 s.f.). To find an unknown angle, first isolate its sine and then apply $\sin^{-1}$ .

Do not replace $\sin i/\sin r$ with $i/r$ . If the ray travels from the material into air, the angle in air still belongs in the numerator when calculating the material's refractive index.

For light at a material–air boundary, the material's refractive index $n$ and its critical angle $c$ are related because the refracted angle is $90^\circ$ at the critical condition.

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

To find the critical angle, rearrange to $\sin c=1/n$ , so $c=\sin^{-1}(1/n)$ . Use degree mode and give the angle with appropriate precision.

For glass with $n=1.56$ , $c=\sin^{-1}(1/1.56)=39.9^\circ$ . At exactly this incidence angle, the refracted ray travels along the glass–air boundary.

This equation is for the critical boundary between the material and air. TIR occurs only for incidence from the material side with $i>c$ , not at $i=c$ .

An optical fibre carries information as pulses of visible or infrared light along a transparent core.

The core has a higher refractive index than the surrounding cladding. A suitably launched ray meets the core–cladding boundary from the slower core at an incidence angle greater than the critical angle, so repeated total internal reflection keeps it inside the core.

Stage	Role in telecommunications
transmitter	converts information into a timed pattern of light pulses
fibre	guides the pulses over distance by repeated TIR
detector	converts the received pulse pattern into an electrical signal that can be decoded

Optical fibres can carry large amounts of data at high rates with little signal loss, so they are used for internet, telephone and cable-television links.

The fibre does not trap every possible ray. If a ray reaches the core boundary below the required incidence angle, some light escapes into the cladding and the transmitted signal weakens.

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—3.2.3 Thin lensesRead

3.2.2 Refraction of light

Learning objectives

Name the angles at a refracting boundary

Show refraction with transparent blocks

Predict how light bends at a boundary

Recognise the critical angle

Distinguish internal reflection from total internal reflection

Define refractive index from wave speed

Use the sine rule for refractive index

Link refractive index to critical angle

Carry information through optical fibres

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