Single-Slit DiffractionJEE Advanced

Intensity pattern & fringe widths — interactive Physics simulation for IIT-JEE.

Concept

When light passes a single slit, each part of the wavefront acts as a secondary source; their interference paints a bright central band flanked by dimmer side fringes. Narrowing the slit widens the pattern — diffraction is most dramatic when the slit is comparable to the wavelength.

Key formula

I(\theta) = I_0\left(\frac{\sin\beta}{\beta}\right)^2,\; \beta = \frac{\pi a\sin\theta}{\lambda};\quad a\sin\theta = m\lambda

Derivation

Divide the slit of width $a$ into infinitesimal sources. Their phasors form an arc; summing gives amplitude $\propto \sin\beta/\beta$ with $\beta = \pi a\sin\theta/\lambda$ .

Intensity is the square: $I = I_0(\sin\beta/\beta)^2$ . It vanishes when $\beta = m\pi$ , i.e. $a\sin\theta = m\lambda$ for $m = \pm1, \pm2,\dots$ (but not $m=0$ ).

The central maximum spans the first minima on each side, so its angular width is $2\lambda/a$ and its linear width on a screen at distance $D$ is $2\lambda D/a$ .

Scenarios to explore

Single-Slit Diffraction — Intensity pattern and fringe widths.

Real-world applications

Resolving power of telescopes and microscopes (Rayleigh criterion).
Diffraction gratings and spectroscopy.
Why small apertures blur photographs (diffraction limit).

JEE exam tips

Central maximum is twice as wide as the secondary maxima.
The first secondary maximum has only about 4.5% of the central intensity.

Common mistakes

Using $a\sin\theta = m\lambda$ for maxima — it actually marks the minima.
Including $m = 0$ as a minimum; $m=0$ is the central maximum.
Confusing the single-slit pattern with the double-slit (Young's) fringes.

Exam traps to avoid

Smaller slit → broader central maximum (inverse relationship).
The condition $a\sin\theta = m\lambda$ gives dark fringes, while double-slit $d\sin\theta = m\lambda$ gives bright ones.