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E3.2 · Explain diffraction, refraction, polarization, and interference of light

Learn to explain diffraction, refraction, polarization, and interference of light through clear examples and targeted practice.

Ontario Grade 12 Physics

The Wave Nature of Light

Ontario Grade 12 Physics — E3.2

In SPH3U, you learned that waves transfer energy and can be described by their wavelength, frequency, and speed. A wave can also spread, change direction, or combine with another wave. Light shows all of these behaviours. In this lesson, the physical system is light travelling through air or a transparent material, passing an opening, or meeting another light wave. We describe positions relative to the optical equipment or boundary, with the positive direction chosen along the direction of travel being examined. The direction of a light ray is a vector; wavelength, frequency, and speed are scalars. The wave model helps explain the four behaviours in E3.2.

What you will learn

1. Diffraction: light spreads at an opening or edge

Diffraction is the spreading of a wave as it passes through an opening or around an edge. Light diffracts too, even though its spreading is often hard to notice. The effect becomes more noticeable when the opening is about the same size as the wavelength. Visible-light wavelengths are very small, so a narrow opening is needed to produce a clear pattern.
For a single narrow slit, the first dark bands occur at angles described by the slit relationship below. Here, aa is the slit width, λ\lambda is the light’s wavelength, and θ\theta is the angle from the straight-ahead direction to a dark band. The integer mm identifies the dark band; the first one has m=1m=1. The relationship applies when the light passes through a narrow slit and the dark-band angle is measured from the central direction.
Diffraction is not light bending because of a change in material. It is a spreading effect associated with an opening or edge. A diagram of the pattern would show a bright central region with alternating dark and bright regions on either side.
asin⁡θ=mλa\sin\theta=m\lambda

2. Refraction: light changes direction in a new material

Refraction is a change in the direction of light as it crosses a boundary between materials. It occurs because light travels at different speeds in different materials. A ray that enters a material along the boundary does not change direction at that boundary; a ray that enters at an angle usually does.
To describe the geometry, draw a normal: an imaginary line perpendicular to the boundary where the ray meets it. Measure both angles from this normal, not from the boundary. The incident ray approaches the boundary; the refracted ray travels into the second material. The angles and the normal lie in the same plane.
The refractive index, nn, compares the speed of light in vacuum, cc, with its speed in the material, vv. It has no units. Snell’s law relates the refractive indices and the ray angles. When light enters a material with a larger refractive index, it bends toward the normal. When it enters one with a smaller refractive index, it bends away from the normal. At a boundary, the frequency stays the same; the wavelength changes because the speed changes.
n1sin⁡θ1=n2sin⁡θ2,n=cvn_1\sin\theta_1=n_2\sin\theta_2,\qquad n=\frac{c}{v}

3. Polarization: selecting a direction of vibration

Light is a transverse wave: its vibrations are perpendicular to its direction of travel. For ordinary unpolarized light, those vibrations are not restricted to one direction. Polarization is the restriction of a transverse wave’s vibrations to a particular direction.
A polarizing filter has a transmission axis, the direction along which it allows the light’s vibration to pass. The part of the light vibrating along that axis is transmitted; vibration across the axis is blocked. A second filter can reduce the transmitted light further if its axis is turned relative to the first. If the axes are at right angles, an ideal second filter blocks the light transmitted by the first.
Polarization is evidence that light behaves as a transverse wave. It is different from diffraction and refraction: polarization changes the allowed vibration direction, while diffraction concerns spreading and refraction concerns a change in direction at a material boundary.

4. Interference: waves combine

Interference occurs when waves overlap. At a point where two light waves meet, their displacements combine. If their displacements reinforce one another, the result is constructive interference and the light is bright. If they oppose one another, the result is destructive interference and the light is dim or dark.
For two light waves of the same wavelength arriving from two slits, the path difference is the difference between the distances each wave travels to a point. A whole-number multiple of the wavelength gives constructive interference. A half-wavelength offset, or an odd number of half-wavelengths, gives destructive interference. The bright and dark bands form an interference pattern.
The path-difference rules connect the wave model to the observed pattern. They apply when the two waves overlap and have a stable relationship that allows a clear pattern to form. Do not confuse interference with diffraction: diffraction describes spreading at an opening or edge; interference describes the result when waves overlap. A pattern from two narrow slits involves both ideas.
ΔL=mλ (bright),ΔL=(m+12)λ (dark)\Delta L=m\lambda\ \text{(bright)},\qquad \Delta L=\left(m+\frac{1}{2}\right)\lambda\ \text{(dark)}

Worked example

Refraction into glass

Light travels from air into glass. Let n1=1.00n_1=1.00, n2=1.50n_2=1.50, and the incident angle be 30.0∘30.0^\circ. Find the refracted angle.
  1. Set the system and directions
    The system is the light ray at the air–glass boundary. Use the normal as the reference line and measure both angles from it. The ray travels from medium 1, air, into medium 2, glass. The unknown is the refracted angle, θ2\theta_2.
  2. Choose the relationship
    Apply Snell’s law because the ray crosses a boundary between two materials. The refractive indices are unitless.
    n1sin⁡θ1=n2sin⁡θ2n_1\sin\theta_1=n_2\sin\theta_2
  3. Substitute and solve
    Rearrange for the sine of the refracted angle, then use the inverse sine. Keep the angles measured from the normal.
    θ2=sin⁡−1 ⁣(1.00sin⁡30.0∘1.50)=19.5∘\theta_2=\sin^{-1}\!\left(\frac{1.00\sin 30.0^\circ}{1.50}\right)=19.5^\circ
Answer: The refracted angle is 19.5∘19.5^\circ from the normal, on the glass side of the boundary.
Check: The angle is smaller than the incident angle, so the ray bends toward the normal as expected when it enters the higher-index material. Angles and refractive indices are dimensionless, so no unit is required.

Worked example

First dark band from a slit

Monochromatic light with wavelength 5.00×10−7 m5.00\times10^{-7}\ \mathrm{m} passes through a slit 2.00×10−5 m2.00\times10^{-5}\ \mathrm{m} wide. Find the angle to the first dark band.
  1. Define the setup
    The system is light passing through a single slit. Measure the angle from the central, straight-ahead direction. The slit width and wavelength are known; the first dark-band angle is unknown.
  2. Use the dark-band relationship
    For the first dark band, use m=1m=1 in the single-slit relationship. The ratio of wavelength to slit width is unitless, as required for a sine.
    sin⁡θ=mλa\sin\theta=\frac{m\lambda}{a}
  3. Substitute and calculate
    Substitute the values in metres. The metre units cancel before taking the inverse sine.
    θ=sin⁡−1 ⁣((1)(5.00×10−7 m)2.00×10−5 m)=1.43∘\theta=\sin^{-1}\!\left(\frac{(1)(5.00\times10^{-7}\ \mathrm{m})}{2.00\times10^{-5}\ \mathrm{m}}\right)=1.43^\circ
Answer: The first dark band is 1.43∘1.43^\circ from the central direction. There is a matching first dark band on the opposite side.
Check: The angle is small because the wavelength is much smaller than the slit width. The ratio inside the inverse sine is unitless, and the result is a direction angle.

Worked example

Bright interference from path difference

Two slits send light of wavelength 6.00×10−7 m6.00\times10^{-7}\ \mathrm{m} to a point where the path difference is 1.20×10−6 m1.20\times10^{-6}\ \mathrm{m}. Is the point bright or dark?
  1. Identify the comparison
    The system is the pair of light waves arriving at one point. Compare their path difference with the wavelength. The path difference is a scalar length; the question is whether the waves reinforce or cancel.
  2. Test the bright condition
    Constructive interference occurs when the path difference is a whole-number multiple of the wavelength.
    ΔLλ=1.20×10−6 m6.00×10−7 m=2\frac{\Delta L}{\lambda}=\frac{1.20\times10^{-6}\ \mathrm{m}}{6.00\times10^{-7}\ \mathrm{m}}=2
  3. Interpret the result
    The path difference is two wavelengths, a whole-number multiple. The waves arrive in step and reinforce one another.
    ΔL=2λ\Delta L=2\lambda
Answer: The point is bright because the path difference is two wavelengths.
Check: The ratio is unitless, and its whole-number value matches the constructive-interference condition.

Common mistakes and how to avoid them

Measuring a refraction angle from the boundary.
Correction: Measure the incident and refracted angles from the normal, which is perpendicular to the boundary.
Saying that light speeds up or slows down when its frequency changes at a boundary.
Correction: For refraction at a stationary boundary, frequency stays the same. Speed and wavelength change in the new material.
Treating diffraction and interference as the same process.
Correction: Diffraction is spreading at an opening or edge. Interference is the result of overlapping waves.
Claiming that polarization changes the direction of travel of light.
Correction: Polarization restricts the direction of vibration. It does not, by itself, describe a change in the ray’s travel direction.

Lesson summary

Check your understanding

Question 1

A light ray enters a material with a larger refractive index. If it reaches the boundary at an angle, which way does it bend?
  1. Toward the normal
  2. Away from the normal
  3. Along the boundary in every case
  4. It stops at the boundary
Show answer and explanation
Toward the normal
The ray bends toward the normal when it enters a material with a larger refractive index.

Question 2

Two light waves meet with a path difference of 32λ\frac{3}{2}\lambda. What kind of interference occurs?
  1. Constructive, because the difference is a whole number of wavelengths
  2. Destructive, because the difference is an odd number of half-wavelengths
  3. No interference, because the waves have the same wavelength
  4. Diffraction, because the path difference is not zero
Show answer and explanation
Destructive, because the difference is an odd number of half-wavelengths
An odd number of half-wavelengths gives destructive interference, so the waves oppose one another.

Question 3

What does a polarizing filter select?
  1. The light’s frequency
  2. The light’s direction of travel
  3. A direction of vibration
  4. The material’s refractive index
Show answer and explanation
A direction of vibration
A polarizer transmits the component vibrating along its transmission axis.

Key terms

Diffraction
The spreading of a wave as it passes an opening or edge.
Refraction
A change in a wave’s direction as it crosses into a material where its speed differs.
Normal
An imaginary line perpendicular to a surface at the point where a ray meets it.
Polarization
Restriction of a transverse wave’s vibrations to a particular direction.
Interference
The result when waves overlap and their displacements combine.
Path difference
The difference between the distances two waves travel to reach the same point.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation E3.2. 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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