Young's Double SlitJEE Main

Concept

Two coherent slits send overlapping waves to a screen. Where their path difference is a whole number of wavelengths they reinforce (bright fringe); where it is a half-integer they cancel (dark). The bright bands are evenly spaced by the fringe width β.

Key formula

Derivation

The path difference between the two slits to a point at angle $\theta$ is $d\sin\theta$. Constructive interference needs $d\sin\theta = n\lambda$.

For small angles $\sin\theta \approx y/D$, so consecutive maxima sit $\beta = \lambda D/d$ apart — wider fringes for longer wavelength, larger screen distance, or closer slits.

Scenarios to explore

• Young's Double Slit — Interference fringes and fringe width β = λD/d.

Real-world applications

• Measuring the wavelength of light.
• Demonstrating the wave nature of light (and matter).
• Thin-film and interferometer metrology.

JEE exam tips

• β ∝ λ, so red fringes are wider than violet.
• Immersing the apparatus in a medium divides β by the refractive index.
• Central fringe (n = 0) is bright and white in white light.

Common mistakes

• Mixing units — keep λ, d and D consistent.
• Confusing fringe width with slit separation.
• Using $d\sin\theta = (n+\tfrac12)\lambda$ for bright fringes (that's dark).

Exam traps to avoid

• Doubling the slit separation halves the fringe width.
• Coherence is essential — two independent bulbs show no fringes.