Virtual laboratory/Physics/Optics/Diffraction and interference of light

Diffraction and interference of light

What we'll do: We'll send light through one slit, two slits and a grating, and watch how it diffracts and forms interference fringes.

What you see on screen: Wavefronts passing through the slits and diffracting, plus the pattern on the screen — for two slits, regular bright and dark fringes; the sliders change the slit width and spacing and the wavelength, spreading the fringes apart. Readouts: slit configuration, wavelength and fringe spacing.

Diffraction and interference of light
Physics · Optics · Diffraction & interference (FIZ.3.4.4)
Slit configuration
Slit width (smaller → wider fringes)30 µm
Wavelengthλ = 550 nm
Light passes through a VERY narrow slit. What do we see on the screen?
configuration
1 slit
slit width
30 µm
wavelength
550 nm
fringe spacing
—
🎯 Challenge: hit the fringe spacing
streak: 0

I'll give you a target fringe spacing — adjust the wavelength and slit spacing to reach it (Δy = λ·L/d).

What just happened?

After passing through a narrow slit, light did not give a sharp point but a blurred band of fringes — it diffracted. This is proof that light is a WAVE.

Diffraction:

a wave meeting a narrow slit or obstacle DIFFRACTS and spreads even "around the corner". The narrower the slit (or the longer the wave), the stronger the diffraction.

Interference (Young's experiment):

when light passes through TWO close slits, the waves from them superpose. Where "crest meets crest" they reinforce → a BRIGHT fringe; where "crest meets trough" they cancel → a DARK fringe. A regular fringe pattern forms.

Diffraction grating:

many slits give sharp, bright maxima in directions satisfying the condition:

d sin⁡θ=m λd\,\sin\theta = m\,\lambda

(d — slit spacing, m — order, λ — wavelength). From this the wavelength is determined.

Why you need this for the exam:

diffraction, interference and the grating-maxima condition are the core of wave optics (advanced level). The trick: a bright fringe = waves reinforcing, dark = cancelling. Also worth remembering: d sin⁡θ=m λd\,\sin\theta = m\,\lambda.

Remember: light as a wave diffracts and superposes on itself (interference), giving bright and dark fringes; for a grating the maxima satisfy d·sinθ = m·λ.

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Theory & formulas

The wave nature of light

At a narrow slit light diffracts, and behind two slits the waves from them superpose (interference), producing bright and dark fringes on the screen.

How fringes form

  • waves reinforcing (in phase) → a bright fringe,
  • cancelling (out of phase) → a dark fringe.

The condition for maxima

For two slits / a diffraction grating, bright fringes appear when: d sin⁡θ=m λd\,\sin\theta = m\,\lambda A longer wavelength λ\lambda and a smaller slit spacing dd spread the fringes apart.

Notation: dd — slit spacing (grating constant), θ\theta — diffraction angle, mm — order of the maximum (0,1,2,…0, 1, 2, \ldots), λ\lambda — wavelength.

Frequently asked questions

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What is diffraction of light?

The bending of light (a wave) at narrow slits or obstacles.

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Where do bright and dark interference fringes come from?

From the superposition of waves: reinforcement gives a bright fringe, cancellation a dark one.

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What is the condition for the maxima of a diffraction grating?

d·sinθ = m·λ, where d is the slit spacing, m the order of the maximum, and λ the wavelength.

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