Q1. Light of wavelength 500 nm falls on two slits 0.2 mm apart. Find the fringe width on a screen 1 m away.
β = λD/d
β = (500×10⁻⁹ × 1) / (0.2×10⁻³)
β = 5×10⁻⁷ / 2×10⁻⁴ = 2.5×10⁻³ m = 2.5 mm
Light is a wave — and waves interfere. Learn Young's experiment and diffraction, and compute the fringe width for any set-up.
Set the wavelength, screen distance and slit separation to find the fringe width in Young's experiment.
Fringes get wider (easier to see) with longer wavelength, a bigger screen distance, or slits placed closer together.
Fringe-width problems with steps.
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Q1. Light of wavelength 500 nm falls on two slits 0.2 mm apart. Find the fringe width on a screen 1 m away.
β = λD/d
β = (500×10⁻⁹ × 1) / (0.2×10⁻³)
β = 5×10⁻⁷ / 2×10⁻⁴ = 2.5×10⁻³ m = 2.5 mm
Q2. State two conditions for sustained interference of light.
The two sources must be COHERENT (constant phase difference).
They should have the same frequency and nearly equal amplitude, and be narrow and close together.
| Quantity | Formula | SI unit |
|---|---|---|
| Fringe width | β = λ D / d | metre (m) |
| Constructive interference | path diff = n λ | — |
| Destructive interference | path diff = (n + ½) λ | — |
| Single-slit minima | a sinθ = n λ | — |
The distance between two consecutive bright (or dark) fringes, β = λD/d, where D is the screen distance and d the slit separation.
Interference is superposition of waves from two (or more) coherent sources; diffraction is the bending/spreading of a wave from a single slit or edge.
Two sources that emit waves of the same frequency with a constant phase difference — essential for a steady interference pattern.
Every point on a wavefront is a source of secondary spherical wavelets, and the new wavefront is the forward envelope (tangent) of these wavelets.
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