Simulation · Physics · Class 12
Young's double-slit fringes
From the lesson Young's double-slit experiment in Wave Optics. Change the values and watch what happens.
The idea behind it
NCERT §10.5
- Two sodium lamps lighting two pinholes give no fringes. Light from an ordinary source suffers sudden phase jumps in times of about 10⁻¹⁰ s, so two independent sources never keep a fixed phase relation. They are incoherent, and their intensities add on the screen.
- Thomas Young's way round this: light from one bright pinhole S falls on two pinholes S₁ and S₂ placed very close together in an opaque screen. Both are fed by the same S, so every phase jump in S reaches S₁ and S₂ alike and the two stay locked in phase. They behave as coherent sources, like the two needles in the water trough.
- The waves from S₁ and S₂ overlap on a screen GG′ and produce alternate bright and dark bands, called fringes. The analysis of the previous lesson fixes where they fall.
- Let d be the separation of S₁ and S₂, D the distance to the screen and x the distance of a point from the centre of the pattern. Bright fringes lie at x = nDλ/d, with n = 0, ±1, ±2, ...; the n = 0 fringe at the centre is the central bright fringe.
- Dark fringes lie at x = (n + ½)Dλ/d, with n = 0, ±1, ±2, ...
- Consecutive bright fringes, and consecutive dark ones, are the same distance apart, β = Dλ/d: the fringes are equally spaced. A longer wavelength or a more distant screen spreads them out; moving the slits further apart packs them closer.
- The computer-drawn pattern of Fig. 10.13 uses d = 0.025 mm, D = 5 cm and λ = 5 × 10⁻⁵ cm. Its spacing is Dλ/d = (0.05 × 5 × 10⁻⁷)/(2.5 × 10⁻⁵) = 1 × 10⁻³ m, that is 1 mm.
- Exercise 10.4: d = 0.28 mm, D = 1.4 m, and the fourth bright fringe is 1.2 cm from the centre. From x₄ = 4Dλ/d, λ = x₄d/(4D) = (1.2 × 10⁻² × 0.28 × 10⁻³)/(4 × 1.4) = 6.0 × 10⁻⁷ m = 600 nm.
- Exercise 10.6: with 650 nm and 520 nm together, the third bright fringe of 650 nm is at x = 3 × 650 nm × D/d = 1950 nm × D/d. Bright fringes of the two colours first coincide where n × 650 = m × 520; the smallest match is 4 × 650 = 5 × 520 = 2600 nm, so at x = 2600 nm × D/d (fourth of 650 nm on the fifth of 520 nm).
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