E3.3 Explain colour separation using wave-optics concepts

Ontario Grade 12 Physics

The Wave Nature of Light

Diffraction, interference, and refraction

A beam of white light contains many visible wavelengths. A wave-optics device can send those wavelengths in different directions, making separate colours visible. In this lesson, the physical system is light travelling from a source through a grating or prism. The reference frame is the stationary optical setup: the grating or prism and screen are at rest. For grating angles, take the central direction, perpendicular to the grating, as zero; angles to the right are positive and angles to the left are negative. An angle is a scalar measurement, while a direction such as “to the right” specifies orientation. We will use wave behaviour to explain the separation.

What you will learn

Describe how wavelength affects the direction of light in a diffraction grating.
Use the grating relationship to predict the angles of bright fringes.
Explain how a prism separates colours through wavelength-dependent refraction.
Distinguish a colour-separation pattern from the light source's original spectrum.

1. Prerequisite bridge: light as a wave

A wave transfers energy without carrying matter from the source to the screen. Light is an electromagnetic wave. Its wavelength, written as

\lambda

, is the distance between matching points on successive waves. Frequency,

f

, is the number of wave cycles passing a point each second. Wavelength is measured in metres and frequency in hertz.

In a uniform material, wave speed is related to wavelength and frequency by

v=f\lambda

. When light enters a different material, its speed and wavelength can change. Its frequency is set by the source and stays the same at the boundary. This distinction helps explain refraction in a prism.

White light is a mixture of visible wavelengths. A single-wavelength source is called monochromatic light. Real sources may have a range of wavelengths, so the boundaries between colours are not always sharp.

v=f\lambda

Wavelength is a length; frequency is measured in hertz.
In a material, wavelength can change when wave speed changes, while frequency remains constant at a boundary.
Different wavelengths can interact with an optical device in different ways.

2. Diffraction gratings: different wavelengths, different angles

A diffraction grating has many narrow, evenly spaced openings. The spacing between neighbouring openings is the grating spacing,

d

, measured in metres. Light spreads after passing through each opening. This spreading is diffraction.

The spreading waves overlap. Where wave crests arrive together, they reinforce one another; this is constructive interference. Where a crest meets a trough, they can partly or fully cancel. The bright directions occur when the waves from neighbouring openings reinforce. For light arriving perpendicular to the grating, the condition is

d\sin\theta=m\lambda

. Here,

\theta

is measured from the central direction, and

m

is an integer called the order. The central bright image has order zero; the first bright band on either side has orders

+1

and

-1

.

For a fixed grating and order, a larger wavelength requires a larger value of

\sin\theta

, so it appears at a larger angle. In visible light, red generally has a longer wavelength than violet. Thus, in the same non-zero order, red appears farther from the centre than violet. The two sides are mirror images: the sign of the angle changes, but its magnitude is the same.

The equation applies only when the sine value is physically possible:

|\sin\theta|\leq 1

. Therefore, not every order is present for every wavelength. A grating separates colours because each wavelength has its own bright-angle condition, not because the grating changes the light's colour.

d\sin\theta=m\lambda

Bright fringes result from constructive interference of light from the openings.
The zero order is central; non-zero orders spread wavelengths by angle.
For one grating and order, longer wavelength means a larger angle from the centre.
The grating spacing and wavelength must use matching length units.

3. Prisms: wavelength-dependent refraction

A prism is a transparent object with angled surfaces. Refraction is the change in direction that can occur when light crosses between materials because its speed changes. At an angled boundary, different parts of a wavefront enter the second material at different times. This change in speed turns the wavefront and changes the ray's direction.

The refractive index,

n

, describes how much slower light travels in a material than in vacuum. In a material,

n=c/v

, where

c

is the speed of light in vacuum and

v

is the speed in that material. For many transparent materials in the visible range, the index depends on wavelength. This wavelength dependence is called dispersion.

At each surface, the direction can be related using Snell's law:

n_1\sin\theta_1=n_2\sin\theta_2

. The angles are measured from the normal, an imaginary line perpendicular to the surface. In common glass prisms, violet light usually has a larger refractive index than red light. It is bent more at the surfaces, so the emerging colours spread apart. The precise amount depends on the material and prism shape.

A grating separates light through diffraction and interference. A prism separates light through refraction and dispersion. Both use wave behaviour, but the governing models are different.

n_1\sin\theta_1=n_2\sin\theta_2

Refraction changes a wave's direction when its speed changes at a boundary.
The normal is perpendicular to the surface; refraction angles are measured from it.
In typical visible-light glass, the refractive index is greater for violet than for red.

4. Reading colour-separation patterns

A colour pattern shows where different wavelength components travel after interacting with an optical device. The pattern is not a motion graph: the horizontal screen position records where light arrives, not the path of an object moving through time. For a grating, the central maximum is the undeviated reference direction, and matching colours appear on both sides.

A useful prediction follows from the model. If grating spacing is reduced while wavelength and order stay the same, the angle must increase. If wavelength is increased while spacing and order stay the same, the angle also increases. In a prism, the direction and amount of separation depend on how the material's refractive index varies with wavelength and on the prism's geometry.

In calculations, define the known values, the unknown, and the sign convention before substituting. Keep the units consistent, report a sensible number of significant figures, and check whether the result is physically possible. An angle from an inverse sine must be compatible with a sine value between negative one and positive one.

|\sin\theta|\leq 1

The screen position represents an arrival location, not a moving object's displacement.
Check that a calculated grating angle is possible and that its direction matches the chosen sign convention.
Keep the grating and prism explanations distinct.

Worked example

Finding a first-order angle

A grating has spacing

2.00\times10^{-6}\,\mathrm{m}

. Monochromatic light of wavelength

600\,\mathrm{nm}

shines perpendicular to it. Find the angle of the first bright fringe on the right.

Set the system and unknown
The system is the light and grating, with the screen direction measured from the grating's central normal. Right is positive. The known values are $d=2.00\times10^{-6}\,\mathrm{m}$ , $\lambda=600\,\mathrm{nm}$ , and $m=+1$ . The unknown is $\theta$ .
Convert and apply the model
Convert nanometres to metres, then use the bright-fringe condition. The units of $d$ and $\lambda$ now match.
$\lambda=600\,\mathrm{nm}=6.00\times10^{-7}\,\mathrm{m}$
Calculate the direction
Divide by the grating spacing and use the inverse sine. The positive result places the fringe to the right.
$\theta=\sin^{-1}\left(\frac{m\lambda}{d}\right)=\sin^{-1}\left(\frac{(+1)(6.00\times10^{-7}\,\mathrm{m})}{2.00\times10^{-6}\,\mathrm{m}}\right)=+17.5^\circ$

Answer: The first-order fringe is at

17.5^\circ

to the right of the central direction.

Check: The metre units cancel inside the sine. The ratio is

0.300

, which is between zero and one, so the angle is possible. The positive sign agrees with the stated direction.

Worked example

Comparing red and violet fringes

A grating has spacing

1.50\times10^{-6}\,\mathrm{m}

. It is illuminated with red light of wavelength

650\,\mathrm{nm}

and violet light of wavelength

400\,\mathrm{nm}

. Find both first-order angles on the right and identify which colour is farther from the centre.

Define the setup
The grating and light form the system. The central normal is zero, and the right side is positive. For both colours, $d=1.50\times10^{-6}\,\mathrm{m}$ and $m=+1$ . The unknowns are the two angles.
Find the red angle
Convert the red wavelength to metres and substitute it into the grating condition. The ratio is dimensionless.
$\theta_{\mathrm{red}}=\sin^{-1}\left(\frac{(1)(6.50\times10^{-7}\,\mathrm{m})}{1.50\times10^{-6}\,\mathrm{m}}\right)=25.7^\circ$
Find the violet angle
Use the same grating and order, changing only the wavelength. Comparing the two angles shows the colour separation.
$\theta_{\mathrm{violet}}=\sin^{-1}\left(\frac{(1)(4.00\times10^{-7}\,\mathrm{m})}{1.50\times10^{-6}\,\mathrm{m}}\right)=15.5^\circ$

Answer: The red fringe is at

25.7^\circ

and the violet fringe at

15.5^\circ

on the right. Red is farther from the centre.

Check: Both sine ratios are less than one, and both angles are positive as required. The longer red wavelength produces the larger angle, matching the grating model.

Worked example

Refraction at a prism surface

At an air-to-glass surface, a violet ray has an incident angle of

40.0^\circ

. Use

n_{\mathrm{air}}=1.00

and

n_{\mathrm{glass}}=1.52

for this wavelength. Find its refracted angle inside the glass, measured from the normal.

Define the system and angles
The system is the violet light crossing the air-glass boundary. Angles are measured from the normal. The known values are the two indices and the incident angle; the unknown is the refracted angle.
Substitute in Snell's law
Use the index for the stated violet wavelength. Solve for the sine of the refracted angle.
$\sin\theta_2=\frac{n_1\sin\theta_1}{n_2}=\frac{(1.00)\sin(40.0^\circ)}{1.52}$
Find the refracted angle
Taking the inverse sine gives the angle inside the glass. It is smaller than the incident angle because the ray enters a material with a larger index.
$\theta_2=\sin^{-1}\left(\frac{(1.00)\sin(40.0^\circ)}{1.52}\right)=25.0^\circ$

Answer: The refracted violet ray is at

25.0^\circ

from the normal inside the glass.

Check: Refractive indices and the sine ratio are dimensionless. The result is a possible angle and is smaller than the incident angle, consistent with bending toward the normal on entering the higher-index material.

Common mistakes and how to avoid them

Using the grating equation with wavelength in nanometres and spacing in metres without converting.

Correction: Convert both lengths to the same unit before dividing. Their ratio must be dimensionless.

Measuring a prism angle from the surface instead of from the normal.

Correction: Draw the normal perpendicular to the boundary. Measure both incident and refracted angles from that line.

Claiming that a grating's first-order red fringe is closer to the centre than violet.

Correction: For a fixed grating and order, a longer wavelength gives a larger angle. Red is generally farther from the centre than violet.

Treating a grating and a prism as if they separate colours by the same process.

Correction: A grating uses diffraction and interference. A prism uses refraction whose amount depends on wavelength.

Assuming every possible order must appear for every wavelength.

Correction: Check the sine condition. An order is possible only when the required sine value is no greater than one in magnitude.

Lesson summary

Light's wavelength and frequency are related to speed by $v=f\lambda$ .
A grating's bright directions satisfy $d\sin\theta=m\lambda$ . For a fixed grating and order, longer wavelengths appear at larger angles.
A prism separates wavelengths because its refractive index can vary with wavelength, changing the refraction angle.
Use matched units, measure angles from the correct reference line, and check the direction and physical possibility of each result.

Check your understanding

Question 1

A grating is illuminated with two wavelengths in the same non-zero order. Which one appears farther from the centre?

The longer wavelength
The shorter wavelength
Both always appear at the same angle
The answer depends only on the screen distance

Show answer and explanation

The longer wavelength

For fixed grating spacing and order,

\sin\theta

is proportional to wavelength. A longer wavelength therefore has a larger angle.

Question 2

In a grating pattern, what produces a bright fringe?

Constructive interference of light from the openings
The light stopping at each opening
A change in the source frequency at the screen
Refraction at a prism surface

Show answer and explanation

Constructive interference of light from the openings

At a bright fringe, waves from neighbouring openings reinforce one another.

Question 3

What is the main cause of colour separation in a typical glass prism?

The refractive index depends on wavelength
The grating spacing changes with colour
The light frequency becomes zero inside the prism
All wavelengths travel at the same speed in every material

Show answer and explanation

The refractive index depends on wavelength

Wavelength-dependent refractive index means different colours refract by different amounts.

Key terms

Diffraction: The spreading of a wave as it passes through an opening or around an obstacle.
Interference: The overlapping of waves, which can reinforce or reduce the resulting disturbance.
Diffraction grating: An optical device with many evenly spaced openings that produces interference patterns.
Order: An integer identifying a bright fringe in a grating pattern.
Refraction: A change in a wave's direction as it crosses into a material where its speed changes.
Normal: An imaginary line perpendicular to a surface at the point where a ray meets it.
Refractive index: A number that compares the speed of light in vacuum with its speed in a material.
Dispersion: The dependence of a material's refractive index on the light's wavelength.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation E3.3. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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E3.3 · Explain colour separation using wave-optics concepts

Ontario Grade 12 Physics

Diffraction, interference, and refraction

What you will learn

1. Prerequisite bridge: light as a wave

2. Diffraction gratings: different wavelengths, different angles

3. Prisms: wavelength-dependent refraction

4. Reading colour-separation patterns

Finding a first-order angle

Comparing red and violet fringes

Refraction at a prism surface

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH4U

About this lesson and its review