Diffraction and polarisation

What you'll learn

What diffraction means: waves spreading after passing through a gap or around an obstacle.
How the size of the gap compared with the wavelength controls how much spreading you see.
How to use a diffraction grating: a many-slit device used to measure light wavelengths.
Why polarisation shows that electromagnetic waves are transverse.

Prerequisite wave language

A wavefront is a line joining points on a wave that are in the same phase, meaning they are at the same point in their vibration cycle. For example, a line through several crests is a wavefront.

The wavelength, $λ\lambda$ , is the distance between two adjacent points in phase, such as crest to crest. The frequency, $f$ , is the number of complete waves passing a point per second, measured in hertz, Hz. For all waves:

v=fλv=f\lambda

A transverse wave has oscillations perpendicular to the direction of energy transfer. A longitudinal wave has oscillations parallel to the direction of energy transfer. Electromagnetic waves, including light, are transverse.

Diffraction: waves spreading out

Definition

Diffraction

Diffraction is the spreading of waves when they pass through a gap or around the edge of an obstacle.

Diffraction happens for all waves: water waves, sound waves, microwaves and light. The key question is not “does diffraction happen?”, but “is the diffraction noticeable?”

Diffraction is greatest when the gap width is similar to the wavelength. If the gap is much larger than the wavelength, the wave mostly travels straight on, with only slight spreading at the edges.

$Diffraction comparison for narrow and wide gaps$

Key Idea

Gap size compared with wavelength

Diffraction is strongest when the gap width $a$ is about the same size as the wavelength $λ\lambda$ . If $\gg \lambda$ , diffraction is much less noticeable.

Why sound diffracts more obviously than light

You can often hear someone around a doorway, but you cannot usually see around the doorway. That is because sound wavelengths are often comparable with everyday openings, while visible light wavelengths are tiny.

Visible light has wavelengths of roughly 400 nm to 700 nm, so a doorway is enormous compared with the wavelength of light.

Example

Comparing diffraction through a doorway

A sound wave has wavelength 0.20 m. Visible light has wavelength $m5.0\times10^{-7}\ \text{m}$ . Both pass through a doorway of width 0.25 m. Compare the diffraction.

Compare the gap width with the wavelength using the ratio $aλ\frac{a}{\lambda}$ .
For the sound wave:
$m=1.25\frac{a}{\lambda}=\frac{0.25\ \text{m}}{0.20\ \text{m}}=1.25$
The gap is only slightly bigger than the wavelength, so strong diffraction occurs.
For the light:
$m=5.0×105\frac{a}{\lambda}=\frac{0.25\ \text{m}}{5.0\times10^{-7}\ \text{m}}=5.0\times10^{5}$
The gap is hundreds of thousands of wavelengths wide, so the light diffracts only very slightly.

Common Mistake

Wider gap does not mean more diffraction

A wider gap lets more wave energy through, but it usually gives less spreading. Strong diffraction happens when the gap is narrow compared with everyday objects, but comparable with the wavelength.

Diffraction patterns and evidence for waves

When light passes through a very narrow slit, it produces a diffraction pattern: a bright central region with weaker bright regions either side, separated by dark regions.

A maximum is a bright region where waves arrive in phase and reinforce. A minimum is a dark region where waves arrive out of phase and cancel. This is an example of interference, which means the superposition of waves to produce reinforcement or cancellation.

This is important evidence that light behaves as a wave. A simple particle or ray model predicts a sharp shadow, but diffraction produces spreading and alternating bright and dark regions.

Diffraction gratings

Definition

Diffraction grating

A diffraction grating is an optical component with many equally spaced slits. It diffracts light and produces sharp bright maxima at particular angles.

The separation between adjacent slits is called the slit spacing, $d$ . The bright spots are labelled by their order, $n$ . The central maximum is $n = 0$ , the first-order maxima are $n = 1$ , the second-order maxima are $n = 2$ , and so on.

For a diffraction grating:

nλ=dsin⁡θn\lambda=d\sin\theta

where:

$n$ is the order number, with no unit
$λ\lambda$ is the wavelength in metres, m
$d$ is the slit spacing in metres, m
$θ\theta$ is the angle from the central maximum to that order

$Diffraction grating experiment showing orders and angle$

Where the grating equation comes from

Light from neighbouring slits travels slightly different distances to reach the same point on the screen. The path difference is the extra distance travelled by one wave compared with another.

For a bright maximum, the path difference between adjacent slits must be a whole number of wavelengths:

difference=nλ\text{path difference}=n\lambda

Geometry gives the path difference as $dsin⁡θd\sin\theta$ , so:

nλ=dsin⁡θn\lambda=d\sin\theta

Tip

Converting lines per millimetre

If a grating has 500 lines per millimetre, first convert to lines per metre: $m−1500\ \text{mm}^{-1}=5.00\times10^{5}\ \text{m}^{-1}$ . Then use $d=1Nd=\frac{1}{N}$ , where $N$ is the number of lines per metre.

Example

Finding a laser wavelength with a grating

A laser shines through a diffraction grating with 500 lines per millimetre. The first-order maximum is 0.310 m from the central maximum on a screen 1.20 m away. Find the wavelength of the laser light.

Convert the grating line density into slit spacing:
$m−1N=5.00\times10^{5}\ \text{m}^{-1}$ $md=\frac{1}{N}=\frac{1}{5.00\times10^{5}\ \text{m}^{-1}}=2.00\times10^{-6}\ \text{m}$
Use the screen geometry to find the angle:
$θ=tan⁡−1(xL)\theta=\tan^{-1}\left(\frac{x}{L}\right)$ $m)=14.5∘\theta=\tan^{-1}\left(\frac{0.310\ \text{m}}{1.20\ \text{m}}\right)=14.5^\circ$
Rearrange the grating equation for $λ\lambda$ using $n = 1$ :
$λ=dsin⁡θn\lambda=\frac{d\sin\theta}{n}$
Substitute the values:
$m)sin⁡(14.5∘)1\lambda=\frac{\left(2.00\times10^{-6}\ \text{m}\right)\sin\left(14.5^\circ\right)}{1}$ $nm\lambda=5.00\times10^{-7}\ \text{m}=500\ \text{nm}$
This is in the visible range, so the answer is sensible.

Common Mistake

Using the wrong angle

In $nλ=dsin⁡θn\lambda=d\sin\theta$ , $θ\theta$ is measured from the straight-through central maximum to the chosen order. It is not the angle between the first-order spots on opposite sides.

Maximum possible order

Because $sin⁡θ\sin\theta$ cannot be greater than 1, not every order is possible. The largest order must satisfy:

nλ≤dn\lambda\le d

So a quick check is:

n≤dλn\le\frac{d}{\lambda}

The maximum order is the largest whole number that fits this condition.

Practical notes: using a diffraction grating

A common practical method is to use a low-power laser, a diffraction grating and a screen. The laser light is monochromatic, meaning it has one wavelength, and coherent, meaning the waves have a constant phase relationship.

Measure the distance $L$ from grating to screen, then measure the distance $x$ from the central maximum to each bright order. Calculate $θ\theta$ using:

tan⁡θ=xL\tan\theta=\frac{x}{L}

Then use $nλ=dsin⁡θn\lambda=d\sin\theta$ .

To improve the measurement, measure distances to matching orders on both sides of the central maximum and average them. You can also plot $sin⁡θ\sin\theta$ against $n$ ; since:

sin⁡θ=λdn\sin\theta=\frac{\lambda}{d}n

the gradient is $λd\frac{\lambda}{d}$ .

Common Mistake

Laser safety

Never look directly into a laser beam or at strong reflected beams. Keep the beam low, directed away from eyes, and use a screen to view the pattern.

Polarisation

Definition

Polarisation

Polarisation is the restriction of the oscillations of a transverse wave to one plane.

Unpolarised light has electric field oscillations in many different planes perpendicular to the direction of travel. A polarising filter only transmits oscillations parallel to its transmission axis.

The light leaving the filter is plane-polarised, meaning its oscillations are in one plane only. A second polarising filter used to test the polarisation is called an analyser.

Polarisation using a polarising filter and crossed analyser

Key Idea

Polarisation proves transverse waves

Only transverse waves can be polarised, because their oscillations can occur in different planes. Longitudinal waves cannot be polarised in this way.

Example

Predicting transmission through polarisers

Unpolarised light passes through one polarising filter, then through a second filter acting as an analyser. Describe what happens as the analyser is rotated.

After the first filter, the light is plane-polarised along the transmission axis of that filter.
If the analyser axis is parallel to the first filter, the transmitted electric field oscillations are parallel to the analyser axis, so light is transmitted strongly.
If the analyser is rotated by 90 degrees, the axes are crossed. The polarised light has no component along the analyser axis, so ideally no light is transmitted.
At angles between parallel and crossed, only part of the oscillation is transmitted, so the intensity decreases as the analyser is rotated towards 90 degrees.

Polarising sunglasses use this idea. Reflected glare from roads or water is often partially polarised, so a polarising lens can reduce the transmitted glare.

Common Mistake

Polarisation is not caused by longitudinal waves

If a wave can be polarised, it must be transverse. Sound waves in air are longitudinal, so they cannot be polarised by a polarising filter.

Exam technique

In the exam

For diffraction questions, always compare the wavelength with the gap size: strongest spreading occurs when they are similar.
For grating calculations, convert line spacing carefully, use $θ\theta$ from the central maximum, and check that $sin⁡θ≤1\sin\theta\le1$ .
For polarisation explanations, explicitly say that only transverse waves can be polarised, so polarisation is evidence that light is transverse.

Self review

Check yourself

Why is sound heard around a doorway more easily than light is seen around a doorway?
A grating has more lines per millimetre. What happens to the slit spacing $d$ ?
What observation with two polarising filters shows that light is transverse?

Diffraction and polarisation

What you'll learn

Prerequisite wave language

Diffraction: waves spreading out

Why sound diffracts more obviously than light

Diffraction patterns and evidence for waves

Diffraction gratings

Where the grating equation comes from

Maximum possible order

Practical notes: using a diffraction grating

Polarisation

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Diffraction and polarisation Revision Guide