Difficulty: Intermediate | Prerequisites: Basic wave concepts (wavelength, frequency, amplitude, superposition) from Chapters 15 and 16.
This is where your course shifts from treating light as a ray (reflection, refraction, lenses) to treating it as a wave. Wave optics explains phenomena that ray optics cannot: interference patterns, diffraction through slits, and the iridescent colours of thin films like soap bubbles. You should already be comfortable with the basic wave properties covered in Chapters 15 and 16, particularly superposition and the relationship between wavelength, frequency, and speed. Everything here builds on the idea that two waves arriving at the same point can reinforce or cancel each other depending on how their crests and troughs line up.
Light is an electromagnetic wave with a very short wavelength (400 to 700 nm) and travels at c = 3.00 x 10^8 m/s in a vacuum but slows down in materials. When light passes through two narrow slits, the overlapping waves create alternating bright and dark bands (interference fringes) on a screen, with positions determined by the slit spacing, wavelength, and distance to the screen.
Wave optics
The study of light behaviour that can only be explained by treating light as a wave, including interference and diffraction. Think of it as the toolkit you reach for when ray diagrams stop working.
Diffraction
The spreading of a wave as it passes through an opening or around an obstacle. In simple terms, waves bend around corners, and this bending becomes noticeable when the opening is comparable in size to the wavelength.
Electromagnetic (EM) wave
A self-sustaining oscillation of electric and magnetic fields that requires no material medium to travel through. Light, radio waves, and X-rays are all EM waves differing only in wavelength.
Speed of light (c)
The speed at which light travels in a vacuum: c = 3.00 x 10^8 m/s. This is a fundamental constant.
Index of refraction (n)
The ratio of the speed of light in a vacuum to the speed of light in a given material: n = c / v. A higher index means light travels more slowly in that material. Think of it as a measure of how much the material slows light down.
Wavelength in a material (lambda_mat)
The wavelength of light inside a transparent material, given by lambda_mat = lambda_vac / n. The wavelength shrinks, but the frequency stays the same.
Interference
The phenomenon that occurs when two or more waves overlap and combine. Constructive interference produces a larger amplitude; destructive interference produces a smaller one or cancellation.
Interference fringes
The alternating bright and dark bands observed on a screen when light from two coherent sources (such as double slits) overlaps. In simple terms, these are the stripes of light and darkness that prove light is a wave.
Constructive interference
Occurs when two waves arrive in phase (crest meets crest), producing maximum brightness. The path-length difference between the two waves is a whole number of wavelengths: delta r = m lambda.
Destructive interference
Occurs when two waves arrive out of phase (crest meets trough), producing darkness. The path-length difference is a whole number of wavelengths plus a half: delta r = (m + 1/2) lambda.
Path-length difference (delta r)
The difference in distance travelled by two waves from their respective slits to a given point on the screen. This quantity determines whether the point is bright or dark.
Fringe spacing (delta y)
The distance between adjacent bright fringes (or adjacent dark fringes) in a double-slit experiment: delta y = lambda L / d. It is the same everywhere on the screen because it does not depend on m.
Light can be modelled three ways: the wave model (interference, diffraction), the ray model (reflection, lenses, mirrors), and the photon model (quantum behaviour). This chapter uses the wave model.
Light waves are self-sustaining oscillations of electric and magnetic fields. They do not need a medium, which is why sunlight reaches Earth through the vacuum of space.
The speed of light in a vacuum is c = 3.00 x 10^8 m/s.
Visible light has wavelengths between roughly 400 nm (violet) and 700 nm (red). Because these wavelengths are extremely short, light oscillates at very high frequencies.
Example: a 600 nm wavelength corresponds to f = c / lambda = (3.00 x 10^8) / (600 x 10^-9) = 5.00 x 10^14 Hz.
Diffraction (wave spreading) becomes noticeable only when the opening or obstacle is comparable in size to the wavelength. Because light's wavelength is so short, you need very narrow slits to see it.
Light slows down when passing through transparent materials like water or glass. The index of refraction n describes how much it slows: n = c / v.
A vacuum has n = 1 exactly. Air is approximately 1.0003 (treat as 1 unless told otherwise). Water is 1.33, glass is 1.50, diamond is 2.42.
When light enters a material, its speed and wavelength change, but its frequency does not.
The wavelength inside a material is: lambda_mat = lambda_vac / n.
The larger the index of refraction, the shorter the wavelength inside that material.
This is a crucial point for thin-film interference later: two waves travelling through different materials accumulate different numbers of wavelengths over the same physical distance, and that difference drives the interference.
A double slit consists of two very narrow slits (typically about 0.1 mm wide) spaced a small distance d apart (typically about 0.5 mm). A viewing screen sits at distance L, where L is very much larger than d.
Because the slits are narrow, light diffracts (spreads out) behind each one. The two spreading waves overlap in the region between the slits and the screen, producing an interference pattern.
Bright fringes (constructive interference)
Bright bands appear where the path-length difference between waves from the two slits is a whole number of wavelengths.
Condition: delta r = d sin(theta) = m lambda, where m = 0, 1, 2, 3, ...
Angles of bright fringes: sin(theta_m) = m (lambda / d)
Positions on the screen: y_m = m lambda L / d
m = 0 is the central maximum, directly opposite the midpoint between the slits.
Dark fringes (destructive interference)
Dark bands appear where the path-length difference is a whole number plus half a wavelength.
Condition: delta r = (m + 1/2) lambda
Positions: y_m = (m + 1/2) lambda L / d
Dark fringes sit exactly halfway between adjacent bright fringes.
Fringe spacing
The spacing between any two adjacent bright fringes is: delta y = lambda L / d.
This spacing is independent of m, so the bright fringes are equally spaced across the screen.
A longer wavelength, a greater slit-to-screen distance, or a smaller slit spacing all produce wider fringe spacing.
c = 3.00 \times 10^8 \text{ m/s}f = \frac{c}{\lambda}n = \frac{c}{v}\lambda_{\text{mat}} = \frac{\lambda_{\text{vac}}}{n}\Delta r = d \sin\theta = m\lambda \quad (\text{bright fringes, } m = 0, 1, 2, \ldots)y_m = \frac{m\lambda L}{d} \quad (\text{positions of bright fringes})\Delta r = \left(m + \tfrac{1}{2}\right)\lambda \quad (\text{dark fringes})y_m = \left(m + \tfrac{1}{2}\right)\frac{\lambda L}{d} \quad (\text{positions of dark fringes})\Delta y = \frac{\lambda L}{d} \quad (\text{fringe spacing})Double-slit interference is the classic demonstration that light behaves as a wave. The same principle underlies anti-reflective coatings on camera lenses, the rainbow sheen on a compact disc (a diffraction grating in disguise), and the precise measurement of wavelengths in spectroscopy. Understanding the index of refraction is essential for designing optical fibres, which guide light via total internal reflection through glass with a carefully chosen refractive index.
Students often think that both frequency and wavelength change when light enters a new material. Only the wavelength changes; the frequency stays the same.
Students sometimes confuse path-length difference with physical distance. The path-length difference delta r is the extra distance one wave travels compared to the other, not the total distance from slit to screen.
It is common to forget that the fringe-spacing formula delta y = lambda L / d is independent of m. All adjacent bright fringes are equally spaced, not just the ones near the centre.
Students occasionally assume that a higher index of refraction means light speeds up. The opposite is true: a higher n means slower light and a shorter wavelength inside that material.
Expect to be asked to calculate the position or angle of a specific bright or dark fringe given lambda, d, and L.
Know how to work backwards: given a fringe position and slit spacing, solve for the wavelength (this is how the double-slit experiment is used to measure wavelength).
The relationship n = c / v and lambda_mat = lambda_vac / n are frequently tested as conceptual questions.
Be ready for problems that give you the fringe number m indirectly (e.g., "the third bright fringe from centre") and require you to identify m = 3.
True or false: The frequency of light changes when it enters glass from air. (False)
Fill in the blank: The index of refraction of a vacuum is exactly ___. (1)
True or false: Dark fringes in a double-slit experiment are located exactly halfway between bright fringes. (True)
Fill in the blank: If slit spacing d is halved while everything else stays the same, the fringe spacing delta y is ___. (Doubled)
True or false: A larger index of refraction means a longer wavelength inside the material. (False, it means a shorter wavelength)
Q: Light from a helium-neon laser (lambda = 633 nm) illuminates two slits spaced 0.40 mm apart. A viewing screen is 2.0 m behind the slits. A bright fringe is observed at 9.5 mm from the centre. What is the fringe number m?
A: m = y_m d / (lambda L) = (9.5 x 10^-3)(0.40 x 10^-3) / ((633 x 10^-9)(2.0)) = 3. This is the third-order bright fringe.
Q: In the same setup, how much farther does the wave from one slit travel compared to the wave from the other to reach that point?
A: delta r = m lambda = 3 x 633 nm = 1.9 x 10^-6 m (about 1.9 micrometres).
Q: A double-slit pattern is observed on a screen 1.0 m away. The slits are spaced 0.30 mm apart. The distance from one fringe to the ninth bright fringe beyond it is 1.6 cm. What is the wavelength?
A: Fringe spacing delta y = 1.6 cm / 9 = 1.78 x 10^-3 m. Then lambda = (d / L) x delta y = (3.0 x 10^-4 / 1.0) x (1.78 x 10^-3) = 5.3 x 10^-7 m, or about 530 nm.
Q: What happens to the fringe spacing if you replace the light source with one of a longer wavelength?
A: The fringe spacing increases. Since delta y = lambda L / d, a larger lambda produces wider spacing between fringes.
The principle of superposition covered in Chapter 16 is the foundation for everything here: constructive and destructive interference are direct applications of adding wave displacements. The index of refraction connects to Snell's law and total internal reflection from ray optics (Chapter 16), and will return in the thin-film interference section of this chapter. Double-slit interference extends naturally to diffraction gratings (covered in Part 2 of these notes), which use the same equation but with many more slits for sharper fringes.
Wave optics, physical optics, electromagnetic wave, EM wave, speed of light, c = 3 x 10^8, index of refraction, refractive index, n = c/v, wavelength in a material, lambda material, double slit, Young's double slit experiment, two-slit interference, interference fringes, bright fringes, dark fringes, constructive interference, destructive interference, path length difference, path difference, fringe spacing, delta y, visible light spectrum, 400 nm to 700 nm, wave model of light, diffraction, Chapter 17, General Physics